Skip to main content
Logo image

Exploring Differential Equations An Interactive, Student-Centered Approach

Geoffrey W. Cox, Ph.D.

Appendix G Selected Solutions

I Getting Started
1 Differential Equation Basics
Chapter 1 Exercises

✍🏻 Classification

13. Determine the Linearity of Each Term.
Solution.
To classify terms, focus on the dependent variable \(y\) and its derivatives. The table below summarizes the linearity of each term:
πŸ”—
Term Coefficient Dependent Variable Linearity
\(e^t y^{(7)}\) \(e^t\) \(y^{(7)}\) Linear
\((t+1)y'y'''\) \((t+1)\) \(y'y'''\) Nonlinear
\(t \ln y''\) \(t\) \(\ln y''\) Nonlinear
\(y' \sin t\) \(\sin t\) \(y'\) Linear
\(\tan y\) \(1\) \(\tan y\) Nonlinear
\(\frac{4}{y}\) \(4\) \(\frac{1}{y}\) Nonlinear
\(\frac{3}{t}\) \(-\) \(-\) Linear
Note the last term, \(\frac{3}{t}\text{,}\) is a free term since it does not involve the dependent variable \(y\) or its derivatives. These terms are always linear in the context of differential equations.
πŸ”—
πŸ”—

2 Solutions to Differential Equations
Chapter 2 Exercises

πŸ•ΉοΈ Solutions to Differential Equations

9.
Solution.
We proceed as in the previous question. First we take derivatives so that we can substitute into the differential equation.
\begin{align*} y \amp = t^k \\ \Rightarrow \quad y' \amp = kt^{k-1} \\ \Rightarrow \quad y'' \amp = k(k-1)t^{k-2} \end{align*}
Now we can substitute into the differential equation:
\begin{align*} 3t^2 \frac{d^2y}{dt^2} \amp = -11t \frac{dy}{dt} + 3y \\ 3t^2\Big[ k(k-1)t^{k-2} \Big] \amp = -11t\Big[ kt^{k-1} \Big] + 3\Big[ t^k \Big] \\ 3k(k-1)t^2 t^{k-2} \amp = -11kt^1\cdot t^{k-1} + 3t^k \\ (3k^2 - 3k)t^{2+(k-2)} \amp = -11k t^{1 + (k-1)} + 3t^k \\ (3k^2 - 3k)t^{k} \amp = -11k t^{k} + 3t^k \\ (3k^2 - 3k)t^{k} + 11k t^{k} - 3t^k \amp = 0 \\ (3k^2 - 3k + 11k - 3)t^k \amp = 0 \\ (3k^2 +8k - 3)t^k \amp = 0 \\ (3k -1)(k+3)t^k \amp = 0 \end{align*}
Note that there is no value of \(k\) that can guarantee that \(t^k\) is 0. Hence we have
\begin{align*} 3k-1 = 0 \amp \quad\text{or}\quad k+3 = 0 \\ k = \frac{1}{3}, \amp \quad\text{or}\quad k = -3 \end{align*}
So the two solutions are \(k = -3, \frac{1}{3}.\)
πŸ”—
πŸ”—
14. Answer the following.
Solution.
  1. To show this, let’s first move all terms to one side of the equation, so
    πŸ”—
    \begin{equation*} \os{\large{\text{LHS}}}{\overline{x^2y^\prime + xy - 1}} = 0 \end{equation*}
    πŸ”—
    Now, we will compute \(y^\prime\) and plug it and \(y\) into the left hand side (LHS) of the equation to see if it simplifies to zero.
    πŸ”—
    \begin{align*} \os{\large{\text{LHS}}}{\overline{x^2y^\prime + xy - 1}} \amp = x^2\left(\frac{1 - \ln x - c}{x^2}\right) + x\left(\frac{\ln x + c}{x}\right) - 1\\ \amp = 1 - \ln x - c + \ln x + c - 1 \\ \amp = 0 \quad βœ… \end{align*}
    πŸ”—
    Applying \(8=y(9)\) gives
    \begin{equation*} 8 = \frac{\ln 9 + c}{9} \quad \implies \quad c = 72-\ln(9) \end{equation*}
    and so the particular solution is
    \begin{equation*} \ds y = \frac{\ln x + 72-\ln(9)}{x} \end{equation*}
    πŸ”—
    πŸ”—
πŸ”—
πŸ”—
15. Verifying Particular Solutions of Initial-Valued Problems Takes 2-Steps.
Solution.
To verify that the function \(y = -3e^{x^2} + 3\) is a particular solution to the initial-value problem
\begin{equation*} \frac{dy}{dx} = 2xy - 6x, \qquad y(0) = 0\text{,} \end{equation*}
we must show the following two things:
πŸ”—
Let’s start by showing that the function \(y = -3e^{x^2} + 3\) satisfies the differential equation.
πŸ”—
\begin{align*} \frac{dy}{dx} \amp = 2xy - 6x \\ \frac{d}{dx}\left[-3e^{x^2} + 3\right] \amp = 2x(-3e^{x^2} + 3) - 6x \\ -3\frac{d}{dx}\left[e^{x^2}\right]+ 0 \amp = -6xe^{x^2} + 6x - 6x \\ -3\left(e^{x^2} \cdot 2x\right) \amp = -6xe^{x^2} \\ -6xe^{x^2} \amp = -6xe^{x^2}\quadβœ… \end{align*}
πŸ”—
Now, let’s show that the function satisfies the initial condition \(y(0) = 0\text{.}\)
πŸ”—
\begin{align*} y(0) \amp = -3e^{0^2} + 3 \\ 0 \amp = -3e^0 + 3 \\ 0 \amp = -3 + 3 \\ 0 \amp = 0\quadβœ… \end{align*}
πŸ”—
πŸ”—
16.
Solution.
We must show that \(y = 3x^2 - 2x + 1\) satisfies both the differential equation and the initial condition.
πŸ”—
First, we check if \(y = 3x^2 - 2x + 1\) satisfies the differential equation.
πŸ”—
\begin{align*} \frac{dy}{dx} \amp = 6x - 2 \\ \frac{d}{dx}\left[3x^2 - 2x + 1\right] \amp = 6x - 2 \\ 6x - 2 \amp = 6x - 2 \quadβœ… \end{align*}
πŸ”—
Now, we check if \(y = 3x^2 - 2x + 1\) satisfies the initial condition \(y(0) = 1\text{.}\)
πŸ”—
\begin{align*} y(0) \amp = 3(0)^2 - 2(0) + 1 \\ 1 \amp = 1 \quadβœ… \end{align*}
πŸ”—
Since \(y = 3x^2 - 2x + 1\) satisfies both the differential equation and the initial condition, it is a particular solution to the initial-valued problem.
πŸ”—
πŸ”—
17.
Solution.
To verify that the function \(y = -3e^{x^2} + 3\) is a particular solution to the initial-value problem
\begin{equation*} \frac{dy}{dx} = 2xy - 6x, \qquad y(0) = 0\text{,} \end{equation*}
we must show the following two things:
πŸ”—
Let’s start by showing that the function \(y = -3e^{x^2} + 3\) satisfies the differential equation.
πŸ”—
\begin{align*} \frac{dy}{dx} \amp = 2xy - 6x \\ \frac{d}{dx}\left[-3e^{x^2} + 3\right] \amp = 2x(-3e^{x^2} + 3) - 6x \\ -3\frac{d}{dx}\left[e^{x^2}\right]+ 0 \amp = -6xe^{x^2} + 6x - 6x \\ -3\left(e^{x^2} \cdot 2x\right) \amp = -6xe^{x^2} \\ -6xe^{x^2} \amp = -6xe^{x^2}\quadβœ… \end{align*}
πŸ”—
Now, let’s show that the function satisfies the initial condition \(y(0) = 0\text{.}\)
πŸ”—
\begin{align*} y(0) \amp = -3e^{0^2} + 3 \\ 0 \amp = -3e^0 + 3 \\ 0 \amp = -3 + 3 \\ 0 \amp = 0\quadβœ… \end{align*}
πŸ”—
πŸ”—

✍🏻 Solving Differential Equations

16.
Solution.
Start by isolating the derivative, \(y'\text{,}\) on one side of the equation
\begin{align*} y' \amp = 1 + 2 \sin x \end{align*}
Integrate both sides with respect to \(x\) to leave us with \(y\) on the left side
\begin{align*} \int y'\ dx \amp = \int \left(1 + 2 \sin x\right) \ dx \\ y + c_1 \amp = x - 2 \cos x + c_2 \\ y \amp = x - 2 \cos x + \boxed{c_2 - c_1}\leftarrow c \\ y(x) \amp = x - 2 \cos x + c \end{align*}
πŸ”—
Finally, apply the initial condition \(y(0) = 5\) to find \(c\)
\begin{align*} y(0) \amp = 5 \\ (0) - 2 \cos(0) + c \amp = 5 \\ 0 - 2 + c \amp = 5 \\ c \amp = 7 \end{align*}
Thus, the solution to the differential equation is \(\quad y = x - 2 \cos x + 7 \text{.}\)
πŸ”—
πŸ”—
18. Solve the Equation.
Solution.
Start by isolating the derivative, \(y'\text{,}\) on one side of the equation
\begin{align*} y' \amp = 1 + 2 \sin x \end{align*}
Integrate both sides with respect to \(x\) to leave us with \(y\) on the left side
\begin{align*} \int y'\ dx \amp = \int \left(1 + 2 \sin x\right) \ dx \\ y + c_1 \amp = x - 2 \cos x + c_2 \\ y \amp = x - 2 \cos x + \boxed{c_2 - c_1}\leftarrow c \\ y(x) \amp = x - 2 \cos x + c \end{align*}
πŸ”—
Finally, apply the initial condition \(y(0) = 5\) to find \(c\)
\begin{align*} y(0) \amp = 5 \\ (0) - 2 \cos(0) + c \amp = 5 \\ 0 - 2 + c \amp = 5 \\ c \amp = 7 \end{align*}
Thus, the solution to the differential equation is \(\quad y = x - 2 \cos x + 7 \text{.}\)
πŸ”—
πŸ”—

II First-Order Differential Equations
3 Separation of Variables
Chapter 3 Exercises

πŸ—οΈ Warm-ups & Drills

1. Verify the solution.
Solution.
\begin{align*} z' - 1 \amp = z^2\\ \frac{d}{dt}\Big( \tan(t+C) \Big) - 1 \amp = \Big( \tan(t+C) \Big)^2\\ \sec^2(t+C)\cdot \frac{d}{dt}(t+C) - 1 \amp = \tan^2(t+C)\\ \sec^2(t+C)\cdot (1+0) - 1 \amp = \sec^2 t - 1\\ \sec^2(t+C) - 1 \amp = \sec^2 t - 1 \quad βœ…οΈ \end{align*}
πŸ”—
Now, let’s find \(C\) with the initial condition \(z(4) = 9\text{:}\)
\begin{equation*} 9 = z(4) = \tan(4+C)\quad\Rightarrow\quad C = tan^{-1}(9) - 4 \approx 79.66 \end{equation*}
πŸ”—
So the (approximate) particular solution is:
\begin{equation*} z = \tan(t + 79.66) \end{equation*}
πŸ”—
πŸ”—
2. Show the differential equations is separable.
Solution.
To see that this equation is separable, we need to check that it can be rearranged in separable form. Isolating \(s' \) we get
\begin{equation*} s' = \frac{s + t}{s} \end{equation*}
and it should be clear that there is no way to separate \(t\) and \(s\) by multiplication in the numerator, so this equation is not separable.
πŸ”—
πŸ”—
3.
Solution.
not separable; there’s no way to separate th \(\ds z \) an \(\ds t \) inside the sine term.
πŸ”—
πŸ”—
4.
Solution.
This doesn’t look separable; but if we use a property of logarithms we can re-write as follows.
\begin{align*} s' \amp = t\ln(s^{2t}) + 8t^2 \\ \frac{ds}{dt} \amp = t \cdot 2t\ln(s) + 8t^2 \\ \amp = 2t^2 \ln(s) + 8t^2 \\ \amp = \big(t^2\big)\big(2\ln(s) + 8\big) \end{align*}
Hence, this DE is separable.
πŸ”—
πŸ”—
5.
Solution.
We can rearrange as follows.
\begin{align*} \frac{dy}{dx} \amp = \big(1\big)\big(2y^3 + y + 4\big) \end{align*}
Hence, this DE is separable.
πŸ”—
πŸ”—
6.
Solution.
We can rearrange as follows.
\begin{align*} y' - xy \amp = 0 \\ y' \amp = \big(x\big)\big(y\big) \end{align*}
Hence, this DE is separable.
πŸ”—
πŸ”—
7.
Solution.
not separable; there’s no way to separate \(\ds x \) and \(\ds y \) via multiplication/division
πŸ”—
πŸ”—
8.
Solution.
not separable; this DE is second order, and DEs must be first order to be separable
πŸ”—
πŸ”—

✍🏻 Solve the Differential Equations

2.
Solution.
This equation is not obviously separable as written, so we begin by solving for \(y'\text{:}\)
\begin{align*} y^2 y' \amp = 6x^2 + 1 \\ \frac{dy}{dx} \amp = \frac{6x^2 + 1}{y^2} = \big( 6x^2 + 1 \big)\cdot \left( \frac{1}{y^2} \right) \end{align*}
This is now in separable form.
πŸ”—
We now separate and integrate:
\begin{align*} y^2 dy \amp = (6x^2 + 1) dx \\ \int y^2 dy \amp = \int (6x^2 + 1) dx \\ \frac{y^3}{3} \amp = 2x^3 + x + c \end{align*}
Multiplying through and isolating \(y\) gives the general solution:
\begin{equation*} y = \sqrt[3]{6x^3 + 3x + c} \end{equation*}
πŸ”—
πŸ”—
12.
Solution.
\begin{align*} \frac{dy}{y} \amp = x dx \\ \int \frac{dy}{y} \amp = \int x dx \\ \ln|y| \amp = \frac{1}{2}x^2 + C_1 \\ y \amp = Ce^{\frac{1}{2}x^2} \end{align*}
πŸ”—
πŸ”—
13.
Solution.
\begin{align*} y dy \amp = (x^2 + 1) dx \\ \int y dy \amp = \int (x^2 + 1) dx \\ \frac{1}{2}y^2 \amp = \frac{1}{3}x^3 + x + C_1 \\ y^2 \amp = \frac{2}{3}x^3 + 2x + 2C_1 \end{align*}
πŸ”—
πŸ”—
14.
Solution.
\begin{align*} \frac{dy}{dx} \amp = \frac{1 - x^2}{y^2} \\ y^2 dy \amp = (1-x^2)dx \\ \int y^2 dy \amp = \int (1-x^2)dx \\ \frac{1}{3}y^3 \amp = x - \frac{1}{3}x^3 + C_1 \\ y^3 \amp = 3x - x^3 + 3C_1 \\ y \amp = \sqrt[3]{3x - x^3 + 3C_1}, \mbox{ or} \\ y \amp = \sqrt[3]{3x - x^3 + C_2} \end{align*}
Note: since, in general \(\ds \sqrt[3]{a+b} \ne \sqrt[3]{a} + \sqrt[3]{b} \) , we CANNOT simplify the solution to \(\ds y = \sqrt[3]{3x - x^3} + C.\)
πŸ”—
πŸ”—
15.
Solution.
\begin{align*} y' - 2y \amp = y\sin x \\ y' \amp = 2y + y\sin x \\ \frac{dy}{dx} \amp = y(2 + \sin x) \\ \frac{1}{y} dy \amp = (2 + \sin x)dx \\ \int \frac{1}{y} dy \amp = \int (2 + \sin x)dx \\ \ln|y| \amp = 2x - \cos x + C_1 \\ e^{\ln|y|} \amp = e^{2x - \cos x + C_1} \\ |y| \amp = e^{2x - \cos x}\cdot e^{C_1} \\ y \amp = \pm e^{2x - \cos x}\cdot e^{C_1} \\ y \amp = \underbrace{\pm e^{C_1}}_{\scriptsize = C_2}\cdot e^{2x - \cos x} \\ y \amp = C_2e^{2x - \cos x} \end{align*}
πŸ”—
πŸ”—
16.
Solution.
\begin{align*} x\frac{dv}{dx} \amp = \frac{1-4v^2}{3v} \\ \frac{dv}{dx} \amp = \frac{1-4v^2}{3v}\cdot \frac{1}{x} \\ \frac{3v}{1-4v^2} dv \amp = \frac{1}{x} dx \\ \int \frac{3v}{1-4v^2} dv \amp = \int \frac{1}{x} dx \end{align*}
For the integral on the left hand side, we will need to do a substitution. If we le \(\ds u = 1-4v^2, \) the \(\ds du = -8vdv, \) and therefor \(\ds -\frac{1}{8}du = vdv. \) Then we have the following:
\begin{align*} 3\int \frac{v}{1-4v^2} dv \amp = \int \frac{1}{x} dx \\ 3\int \underbrace{\frac{1}{1-4v^2}}_{\scriptsize 1/u} \underbrace{vdv}_{\scriptsize -\frac{1}{8}du} \amp = \int \frac{1}{x} dx \\ 3\int \frac{1}{u}\cdot -\frac{1}{8}du \amp = \int \frac{1}{x} dx \\ -\frac{3}{8} \int \frac{1}{u}du \amp = \int \frac{1}{x} dx \\ -\frac{3}{8} \ln|u| \amp = \ln|x| + C_1 \\ -\frac{3}{8} \ln|1 - 4v^2| \amp = \ln|x| + C_1 \\ \ln|1 - 4v^2| \amp = -\frac{8}{3} \left(\ln|x| + C_1\right) \\ \ln|1 - 4v^2| \amp = -\frac{8}{3}\ln|x| + -\frac{8}{3}C_1 \\ \ln|1 - 4v^2| \amp = \ln\left(|x|^{-8/3}\right) + -\frac{8}{3}C_1 \\ \ln|1 - 4v^2| \amp = \ln\left(x^{-8/3}\right) + -\frac{8}{3}C_1 \mbox )} \\ e^{\ln|1 - 4v^2|} \amp = e^{\ln\left(x^{-8/3}\right) + -\frac{8}{3}C_1} \\ |1 - 4v^2| \amp = e^{\ln\left(x^{-8/3}\right)}\cdot e^{-\frac{8}{3}C_1} \\ |1 - 4v^2| \amp = x^{-8/3}\cdot e^{-\frac{8}{3}C_1} \\ 1 - 4v^2 \amp = \pm e^{-\frac{8}{3}C_1}\cdot x^{-8/3} \\ 1 - 4v^2 \amp = \underbrace{\pm e^{-\frac{8}{3}C_1}}_{\scriptsize = C_2}\cdot x^{-8/3} \\ 1 - 4v^2 \amp = C_2x^{-8/3} \end{align*}
\begin{align*} - 4v^2 \amp = C_2x^{-8/3} - 1 \\ v^2 \amp = \frac{C_2}{-4}x^{-8/3} + \frac{1}{4} \\ v^2 \amp = \underbrace{\frac{C_2}{-4}}_{\scriptsize = C_3}x^{-8/3} + \frac{1}{4} \\ v^2 \amp = C_3x^{-8/3} + \frac{1}{4} \\ v \amp = \pm \sqrt{C_3x^{-8/3} + \frac{1}{4}} \end{align*}
πŸ”—
πŸ”—
17.
Solution.
When we separate the variables and integrate both sides, we get
\begin{equation*} \int (3y^2-5)dy = \int (4-2x)dx \end{equation*}
\begin{equation*} y^3-5y=4x-x^2+C \end{equation*}
Since the equation cannot readily be solved for \(y\) as an explicit function of \(x\text{,}\) we are done.
πŸ”—
πŸ”—
18.
Solution.
Separation of variables gives:
\begin{align*} \ds\int \frac{1}{3(y-1)^2/3} dy = \int 2xdx\\ (y-1)^1/3\amp = \ds x^2+C\\ y(x)\amp = \ds 1+(x^2+C)^3 \end{align*}
πŸ”—
πŸ”—
19.
Solution.
Informally, we divide both side of the differential equation by \(y\) and multiply each side by \(dx\) to get
\begin{equation*} \frac{dy}y = -6xdx. \end{equation*}
\begin{equation*} \int \frac{dy}y = \int (-6x)dx \end{equation*}
\begin{equation*} \ln|y|=-3x^2+C \end{equation*}
\begin{equation*} \ln|y|=-3x^2+C \end{equation*}
\begin{equation*} |y|=e^{-3x^2+C} \end{equation*}
\begin{equation*} y=\pm e^C e^{-3x^2} \end{equation*}
and hence
\begin{equation*} y(x) = Ae^{-3x^2} \qquad (A = \pm e^C) \end{equation*}
The condition \(y(0) = 7\) yields \(A =7\text{,}\) so the desired solution is
\begin{equation*} y(x) = 7e^{-3x^2} \end{equation*}
πŸ”—
πŸ”—
20.
Solution.
First we find the general solution:
\begin{align*} \frac{dz}{dt} \amp = 2tz^2 \\ \frac{1}{z^2}dz \amp = 2tdt \\ \int \frac{1}{z^2}dz \amp = \int 2tdt \\ -\frac{1}{z} \amp = t^2 + C_1 \\ z \amp = -\frac{1}{t^2 + C_1} \end{align*}
Using the initial condition \(z(1) = 2\text{:}\)
\begin{align*} 2 \amp = -\frac{1}{1^2 + C_1} \\ 2 \amp = -\frac{1}{1 + C_1} \\ C_1 \amp = -\frac{1}{2} - 1 \\ C_1 \amp = -\frac{3}{2} \end{align*}
So the solution to the initial value problem is:
\begin{align*} z(t) \amp = -\frac{1}{t^2 - \frac{3}{2}} \end{align*}
πŸ”—
πŸ”—
21.
Solution.
First we find the general solution:
\begin{align*} \frac{dy}{d\theta} \amp = y\sin\theta \\ \frac{1}{y}dy \amp = \sin\theta d\theta \\ \int \frac{1}{y}dy \amp = \int \sin\theta d\theta \\ \ln|y| \amp = -\cos\theta + C_1 \\ e^{\ln|y|} \amp = e^{-\cos\theta + C_1} \\ |y| \amp = e^{-\cos\theta}\cdot e^{C_1} \\ y \amp = \pm e^{C_1}e^{-\cos\theta} \\ \amp = \underbrace{\pm e^{C_1}}_{\scriptsize = C_2}e^{-\cos\theta} \\ \amp = C_2e^{-\cos\theta} \end{align*}
Then we use the initial conditio \(\ds y(\pi) = -3 \) .
\begin{align*} -3 \amp = y(\pi) \\ \amp = C_2e^{-\cos\pi} \\ \amp = C_2e^{-(-1)} \\ \amp = C_2e \\ -\frac{3}{e} \amp = C_2, \mbox{ or} \\ C_2 \amp = -3e^{-1} \end{align*}
Hence, the solution i \(\ds y = -3e^{-1}e^{-\cos\theta} \) , or \(\ds y = -3e^{-\cos\theta-1} \) .
πŸ”—
πŸ”—
22.
Solution.
First we find the general solution:
\begin{align*} \frac{dy}{dx} - 8x^3e^{-2y} \amp = 0 \\ \frac{dy}{dx} \amp = 8x^3e^{-2y} \\ \frac{1}{e^{-2y}}dy \amp = 8x^3dx \\ e^{2y}dy \amp = 8x^3dx \\ \int e^{2y}dy \amp = \int 8x^3dx \\ \frac{1}{2}e^{2y} \amp = 2x^4 + C_1 \\ e^{2y} \amp = 4x^4 + 2C_1 \\ \ln(e^{2y}) \amp = \ln(4x^4 + 2C_1) \\ 2y \amp = \ln(4x^4 + 2C_1) \\ y \amp = \frac{1}{2}\ln(4x^4 + 2C_1), \mbox{ or} \\ \amp = \frac{1}{2}\ln(4x^4 + C_2), \mbox{ or} \\ \amp = \ln\Big[(4x^4 + C_2)^{1/2}\Big], \mbox{ or} \\ \amp = \ln\Big[\sqrt{4x^4 + C_2}\Big] \end{align*}
Then we use the initial conditio \(\ds y(1) = 0 \) .
\begin{align*} 0 \amp = y(1) \\ 0 \amp = \ln\Big[\sqrt{4\cdot 1^4 + C_2}\Big] \\ 0 \amp = \ln\Big[\sqrt{4 + C_2}\Big] \\ e^0 \amp = e^{\ln\Big[\sqrt{4 + C_2}\Big]} \\ 1 \amp = \sqrt{4 + C_2} \\ 1^2 \amp = \Big[\sqrt{4 + C_2}\Big]^2 \\ 1 \amp = 4+C_2 \\ -3 \amp = C_2 \end{align*}
Hence, the solution i \(\ds y = \frac{1}{2}\ln(4x^4 - 3) \) , or \(\ds y = \ln\Big[\sqrt{4x^4 - 3}\Big] \text{.}\)
πŸ”—
πŸ”—
23.
Solution.
We want to find a non-zero function \(\mu(x)\) such that
\begin{equation*} \frac{d}{dx}\left[\mu\sin x\right] = \mu \cos x + 6 \mu \sin x. \end{equation*}
Applying the product rule to the left side of this equation gives
\begin{equation*} \mu \cos x + \frac{d\mu}{dx} \sin x = \mu \cos x + 6 \mu \sin x. \end{equation*}
Subtracting \(\mu \cos x\) from both sides gives
\begin{equation*} \frac{d\mu}{dx} \sin x = 6 \mu \sin x\text{,} \end{equation*}
which implies that
\begin{equation*} \frac{d\mu}{dx} = 6 \mu. \end{equation*}
This is a separable differential equation that we can solve with separation of variables:
\begin{align*} \frac{d\mu}{\mu} \amp = 6 dx \\ \int \frac{1}{\mu} d\mu \amp = \int 6 dx \\ \ln|\mu| \amp = 6x + c_1 \\ |\mu| \amp = e^{6x + c_1} = e^{c_1} e^{6x} \\ \mu \amp = \pm e^{c_1} e^{6x} \end{align*}
But, \(\pm e^c_1\) is just a constant, so we can write this as
\begin{equation*} \mu = ce^{6x}. \end{equation*}
Since the question asks for a non-zero function, we can take \(c = 1\text{.}\) Thus, \(\mu = e^{6x}\) satisfies the given differential equation.
πŸ”—
πŸ”—
24.
Solution.
We want to find a non-zero function \(\mu(x)\) such that
\begin{equation*} \frac{d}{dx}\left[\mu\ln x\right] = \mu \frac{\mu}{x} + 2x \mu \ln x. \end{equation*}
Applying the product rule to the left side of this equation gives
\begin{equation*} \mu \frac{1}{x} + \frac{d\mu}{dx} \ln x = \mu \frac{\mu}{x} + 2x \mu \ln x. \end{equation*}
Subtracting \(\mu \frac{\mu}{x}\) from both sides gives
\begin{equation*} \frac{d\mu}{dx} \ln x = 2x \mu \ln x\text{,} \end{equation*}
which implies that
\begin{equation*} \frac{d\mu}{dx} = 2x \mu. \end{equation*}
This is a separable differential equation that we can solve with separation of variables:
\begin{align*} \frac{d\mu}{\mu} \amp = 2x\ dx \\ \int \frac{1}{\mu} d\mu \amp = \int 2x\ dx \\ \ln|\mu| \amp = x^2 + c_1 \\ |\mu| \amp = e^{x^2 + c_1} = e^{c_1} e^{x^2} \\ \mu \amp = \pm e^{c_1} e^{x^2} \end{align*}
But, \(\pm e^c_1\) is just a constant, so we can write this as
\begin{equation*} \mu = ce^{x^2}. \end{equation*}
Since the question asks for a non-zero function, we can take \(c = 1\text{.}\) Thus, \(\mu = e^{x^2}\) satisfies the given differential equation.
πŸ”—
πŸ”—
25.
Solution.
We want to find a non-zero function \(\mu(x)\) such that
\begin{equation*} \frac{d}{dx}\left[\mu y\right] = \mu \frac{dy}{dx} + 5 \mu y. \end{equation*}
Applying the product rule to the left side of this equation gives
\begin{equation*} \mu \frac{dy}{dx} + \frac{d\mu}{dx} y = \mu \frac{dy}{dx} + 5 \mu y. \end{equation*}
Subtracting \(\mu \frac{dy}{dx}\) from both sides gives
\begin{equation*} \frac{d\mu}{dx} y = 5 \mu y\text{,} \end{equation*}
which implies that
\begin{equation*} \frac{d\mu}{dx} = 5 \mu. \end{equation*}
This is a separable differential equation that we can solve with separation of variables:
\begin{align*} \frac{d\mu}{\mu} \amp = 5 dx \\ \int \frac{1}{\mu} d\mu \amp = \int 5 dx \\ \ln|\mu| \amp = 5x + c_1 \\ |\mu| \amp = e^{5x + c_1} = e^{c_1} e^{5x} \\ \mu \amp = \pm e^{c_1} e^{5x} \end{align*}
But, \(\pm e^c_1\) is just a constant, so we can write this as
\begin{equation*} \mu = ce^{5x}. \end{equation*}
Since the question asks for a non-zero function, we can take \(c = 1\text{.}\) Thus, \(\mu = e^{5x}\) satisfies the given differential equation.
πŸ”—
πŸ”—
26.
Solution.
We take time \(t\) in minutes, with \(t = 0\) corresponding to 5:00 PM We also assume (somewhat unrealistically) that at any instant the temperature \(T(t)\) of the roast is uniform throughout. We have \(T(t) \lt A = 375\text{,}\) \(T(0)=50\text{,}\) and \(T(75) =125\text{.}\) Hence
\begin{align*} \ds \frac{dT}{dt} = k(375-T)\\ \int \frac{1}{375-T}dT\amp = \ds \int kdt\\ y(x)\amp = \ds 1+(x^2+C)^3\\ -\ln(375-T)=\amp kt+C \end{align*}
Now \(T(0)=50\text{,}\) implies that \(B = 325\text{,}\) so \(T(t) = 375-325e^{kt}\text{.}\) We also know that \(T = 125\) when \(t = 75\text{.}\) Substitution of these values in the preceding equation yields
\begin{equation*} k=-\frac{1}{75}\ln\left(\frac{250}{325}\right)\approx 0.0035. \end{equation*}
Hence we finally solve the equation
\begin{equation*} 150=375-325e^{(-0.0035)t} \end{equation*}
for \(t = -[\ln(225/325)]/[0.0035] \approx 105\) (min), the total cooking time required.Because the roast was placed in the oven at 5:00 PM, it it should be removed at about 6:45 PM
πŸ”—
πŸ”—

4 Integrating Factor
Chapter 4 Exercises

πŸ—οΈ Warm-ups & Drills

1.
Solution.
This DE is first order and linear, so it can be solved using an integrating factor.
πŸ”—
It is not separable. After isolating the derivative:
\begin{equation*} \dfrac{dy}{dx} = \dfrac{y - \cos x}{x^2}, \end{equation*}
we see that \(y\) and \(x\) cannot be separated by multiplication or division.
πŸ”—
Therefore, this DE can be solved using an integrating factor, but not by separating variables.
πŸ”—
πŸ”—
2.
Solution.
This DE is first order but not linear because the dependent variable \(x\) appears inside the nonlinear term \(e^x\text{.}\)
πŸ”—
It is also not separable. After isolating the derivative:
\begin{equation*} \frac{dx}{dt} = e^x - xt, \end{equation*}
there’s no way to write the right-hand side as a product of a function of \(x\) and a function of \(t\text{.}\)
πŸ”—
Therefore, this DE cannot be solved using either technique.
πŸ”—
πŸ”—
3.
Solution.
This DE is first order and linear, so it can be solved using an integrating factor.
πŸ”—
It is also separable. Solving for the derivative:
\begin{align*} (t^2 + 1) \frac{dy}{dt} \amp = yt - y\\ \frac{dy}{dt} \amp = \frac{yt - y}{t^2+1}\\ \frac{dy}{dt} \amp = \frac{y(t - 1)}{t^2+1}\\ \frac{dy}{dt} \amp = y \cdot \frac{t - 1}{t^2+1} \end{align*}
Here, we can separate the variables: \(\dfrac{dy}{y} = \dfrac{t - 1}{t^2 + 1} dt\text{.}\)
πŸ”—
Therefore, this DE can be solved both by using an integrating factor and by separating variables.
πŸ”—
πŸ”—
4.
Solution.
This DE is first order but not linear because the dependent variable \(y\) is raised to a power (\(y^2\)).
πŸ”—
It is separable. Solving for the derivative:
\begin{align*} \frac{dy}{dt} - y^2 t \amp = t\\ \frac{dy}{dt} \amp = t + y^2 t\\ \frac{dy}{dt} \amp = t \cdot (1 + y^2) \end{align*}
allows us to separate variables: \(\dfrac{dy}{1+y^2} = t \, dt\text{.}\)
πŸ”—
Therefore, this DE can be solved by separating variables, but not using an integrating factor.
πŸ”—
πŸ”—
5.
Solution.
This DE is first order and linear, so it can be solved using an integrating factor.
πŸ”—
It is not separable. Solving for the derivative:
\begin{equation*} \frac{dr}{d\theta} = 3r + \theta^3, \end{equation*}
shows no way to separate \(r\) and \(\theta\) into separate multiplicative terms.
πŸ”—
Therefore, this DE can be solved using an integrating factor, but not by separating variables.
πŸ”—
πŸ”—

✍🏻 Solve the Differential Equations

1.
Solution.
Standard form: \(y' + (-4)y = x\text{,}\) so \(P(x) = -4\text{.}\) Integrating factor: \(\mu(x) = e^{\int -4 dx} = e^{-4x}\text{.}\)
πŸ”—
\begin{align*} e^{-4x}y' - 4e^{-4x}y \amp = xe^{-4x}\\ \frac{d}{dx}[e^{-4x}y] \amp = xe^{-4x} \end{align*}
Integrate:
πŸ”—
\begin{align*} e^{-4x}y \amp = \int xe^{-4x} dx = -\frac{1}{4}xe^{-4x} - \frac{1}{16}e^{-4x} + C\\ y \amp = -\frac{1}{4}x - \frac{1}{16} + Ce^{4x} \end{align*}
πŸ”—
2.
Solution 1.
Standard form: \(y' - \frac{1}{x^2}y = \frac{1}{x^2}\text{.}\) Integrating factor: \(\mu(x) = e^{\int -\frac{1}{x^2}dx} = e^{1/x}\text{.}\)
πŸ”—
\begin{align*} \frac{d}{dx}[e^{1/x}y] \amp = \frac{e^{1/x}}{x^2}\\ e^{1/x}y \amp = -e^{1/x} + C\\ y \amp = -1 + Ce^{-1/x} \end{align*}
πŸ”—
Solution 2. (detailed)
We put the equation in standard form to identify \(P(x)\) and compute the integrating factor.
πŸ”—
\begin{equation*} y' + \us{P}{\boxed{-\frac{1}{x^2}}}\ y = \frac{1}{x^2} \quad \implies \quad \mu(x) = e^{\large \int -\sfrac{1}{x^2} dx} = e^{1/x} \end{equation*}
πŸ”—
Multiply by \(e^{1/x}\) and reverse the product rule on the left side.
\begin{align*} e^{1/x}y' - \frac{e^{1/x}}{x^2} y \amp = \frac{e^{1/x}}{x^2} \\ \frac{d}{dx}\left[e^{1/x} y\right] \amp = \frac{e^{1/x}}{x^2} \end{align*}
πŸ”—
Next, integrate both sides and use \(u\)-substitution on the right side.
πŸ”—
\begin{align*} \int \frac{d}{dx}\left[e^{1/x} y\right] dx \amp = \int \frac{e^{1/x}}{x^2} dx \qquad \rightarrow\\ e^{1/x} y \amp = \int -e^u du \\ e^{1/x} y \amp = -e^u + c \\ e^{1/x} y \amp = -e^{1/x} + c \end{align*}
πŸ”—
\begin{align*} u \amp = \frac{1}{x} \\ du \amp = -\frac{1}{x^2} dx \end{align*}
πŸ”—
πŸ”—
Finally, solve for \(y\text{:}\)
\begin{gather*} y = -1 + ce^{-1/x} \end{gather*}
πŸ”—
πŸ”—
3.
Solution.
Already in standard form: \(y' + (-1)y = e^{3x}\text{.}\) Integrating factor: \(\mu(x) = e^{\int -1 dx} = e^{-x}\text{.}\)
πŸ”—
\begin{align*} \frac{d}{dx}[e^{-x}y] \amp = e^{2x}\\ e^{-x}y \amp = \frac{1}{2}e^{2x} + C\\ y \amp = \frac{1}{2}e^{3x} + Ce^{x} \end{align*}
πŸ”—
4.
Solution.
Standard form: \(y' - \frac{1}{x}y = 2x + 1\text{.}\) Integrating factor: \(\mu(x) = e^{\int -\frac{1}{x}dx} = \frac{1}{x}\text{.}\)
πŸ”—
\begin{align*} \frac{d}{dx}\left[\frac{1}{x}y\right] \amp = 2 + \frac{1}{x}\\ \frac{1}{x}y \amp = 2x + \ln x + C\\ y \amp = 2x^2 + x \ln x + Cx \end{align*}
πŸ”—
5.
Solution.
Treat \(x\) as dependent: \(x' + \frac{2}{y}x = 5y^2\) with \(x' = \frac{dx}{dy}\text{.}\) Integrating factor: \(\mu(y) = e^{\int \frac{2}{y}dy} = y^2\text{.}\)
πŸ”—
\begin{align*} \frac{d}{dy}[y^2 x] \amp = 5y^4\\ y^2 x \amp = y^5 + C\\ x \amp = y^3 + \frac{C}{y^2} \end{align*}
πŸ”—
6.
Solution 1.
Rewrite: \(M' - \frac{4}{t}M = t^6 + t\) with \(M' = \frac{dM}{dt}\text{.}\) Integrating factor: \(\mu(t) = e^{\int -\frac{4}{t} dt} = t^{-4}\text{.}\)
πŸ”—
\begin{align*} \frac{d}{dt}[t^{-4}M] \amp = t^2 + t^{-3}\\ t^{-4}M \amp = \frac{1}{3}t^{3} - \frac{1}{2}t^{-2} + C\\ M \amp = \frac{1}{3}t^{7} - \frac{1}{2}t^{2} + Ct^{4} \end{align*}
πŸ”—
Solution 2. (detailed)
As always, start by putting the equation in standard form:
\begin{align*} \frac{1}{t} \frac{dM}{dt} - \frac{4}{t^2} \cdot M \amp = t^5 + 1 \\ \frac{dM}{dt} + \us{P}{\boxed{-\frac{4}{t}}} \cdot M \amp = t^6 + t \text{.} \end{align*}
πŸ”—
Find Integrating Factor, \(\mu\)
πŸ”—
\begin{equation*} \mu = e\vphantom{\large|}^{\int \sfrac{-4}{t}\ dt} = e^{-4\ln t} = e^{\ln t^{-4}} = t^{-4} \end{equation*}
πŸ”—
Multiply by \(\mu\)
πŸ”—
\begin{align*} t^{-4} \frac{dM}{dt} - t^{-4} \frac{4}{t} M \amp = t^{-4}(t^6 + t) \end{align*}
πŸ”—
Complete the Product Rule
πŸ”—
\begin{align*} \frac{d}{dt}\left[ \frac{1}{t^4} M \right] \amp = t^2 + t^{-3} \end{align*}
πŸ”—
Integrate Both Sides
πŸ”—
\begin{gather*} \int \frac{d}{dt}\left[ \frac{1}{t^4} M \right]\ dt = \int \left(t^2 + t^{-3}\right) dt \end{gather*}
πŸ”—
Isolate \(M\)
πŸ”—
\begin{gather*} \frac{1}{t^4} M = \frac{1}{3}t^{3} - \frac{1}{2}t^{-2} + c \end{gather*}
πŸ”—
General Solution
πŸ”—
\begin{equation*} M = \frac{1}{3}t^{7} - \frac{1}{2}t^2 + ct^4\text{.} \end{equation*}
πŸ”—
πŸ”—
7.
Solution.
Standard form with \(P(x) = 4x^3\text{.}\) Integrating factor: \(\mu(x) = e^{\int 4x^3 dx} = e^{x^4}\text{.}\)
πŸ”—
\begin{align*} \frac{d}{dx}[e^{x^4}y] \amp = 5x\\ e^{x^4}y \amp = \frac{5}{2}x^2 + C\\ y \amp = \frac{5}{2}x^2 e^{-x^4} + Ce^{-x^4} \end{align*}
πŸ”—
8.
Solution.
Standard form: \(r' + \tan\theta r = \sec\theta\) with \(r' = \frac{dr}{d\theta}\text{.}\) Integrating factor: \(\mu(\theta) = e^{\int \tan\theta d\theta} = \sec\theta\text{.}\)
πŸ”—
\begin{align*} \frac{d}{d\theta}[\sec\theta r] \amp = \sec^2\theta\\ \sec\theta r \amp = \tan\theta + C\\ r \amp = \sin\theta + C\cos\theta \end{align*}
πŸ”—
9.
Solution.
Standard form: \(y' - \frac{1}{x} y = x e^x\text{.}\) Integrating factor: \(\mu(x) = e^{\int -\frac{1}{x} dx} = x^{-1}\text{.}\)
πŸ”—
\begin{align*} \frac{d}{dx}\left[\frac{1}{x}y\right] \amp = e^x\\ \frac{1}{x}y \amp = \int e^x dx = e^x + C\\ y \amp = x(e^x + C) \end{align*}
Apply \(y(1) = e - 1\text{:}\)
πŸ”—
\begin{equation*} e - 1 = 1(e + C) \Rightarrow C = -1. \end{equation*}
Final solution:
πŸ”—
\begin{equation*} y(x) = x(e^x - 1). \end{equation*}
πŸ”—
10.
Solution.
Divide by \(e^t\text{:}\) \(z' + 4z = e^{-t}\text{.}\) Integrating factor: \(\mu(t) = e^{\int 4 dt} = e^{4t}\text{.}\)
πŸ”—
\begin{align*} \frac{d}{dt}[e^{4t}z] \amp = e^{4t} e^{-t} = e^{3t}\\ e^{4t}z \amp = \frac{1}{3} e^{3t} + C\\ z \amp = \frac{1}{3}e^{-t} + Ce^{-4t} \end{align*}
Apply \(z(0) = \frac{4}{3}\text{:}\)
πŸ”—
\begin{equation*} \frac{4}{3} = \frac{1}{3} + C \Rightarrow C = 1. \end{equation*}
Final solution:
πŸ”—
\begin{equation*} z(t) = \frac{1}{3}e^{-t} + e^{-4t}. \end{equation*}
πŸ”—
11.
Solution.
Standard form with \(P(x) = -3\text{.}\) Integrating factor: \(\mu(x) = e^{\int -3 dx} = e^{-3x}\text{.}\)
πŸ”—
\begin{align*} \frac{d}{dx}[e^{-3x}y] \amp = 6e^{-3x}\\ e^{-3x}y \amp = -2e^{-3x} + C\\ y \amp = -2 + Ce^{3x} \end{align*}
Apply \(y(0) = 5\text{:}\)
πŸ”—
\begin{equation*} 5 = -2 + C \Rightarrow C = 7. \end{equation*}
Final solution:
πŸ”—
\begin{equation*} y(x) = -2 + 7e^{3x}. \end{equation*}
πŸ”—

5 Qualitative Methods
Chapter 5 Exercises

Exercises

5. Find the Equilibrium Solution.
Solution.
To find equilibrium solutions, we set \(f(y) = y^2 = 0\text{.}\) This gives us the solution \(y = 0\text{.}\) So, \(y = 0\) is the only equilibrium solution for this equation.
πŸ”—
This means that if the solution starts at \(y(0) = 0\text{,}\) it will remain at zero for all time. If it starts anywhere else, it will either increase or decrease away from zero.
πŸ”—
πŸ”—
6. Finding Equilibrium Points.
Solution.
To find the equilibrium solutions, we set the right-hand side equal to zero:
πŸ”—
\begin{equation*} y^2 - y^4 = 0. \end{equation*}
Factoring gives us:
πŸ”—
\begin{equation*} y^2(1 - y^2) = 0. \end{equation*}
This has solutions:
πŸ”—
  1. \(\displaystyle y = 0\)
    πŸ”—
  2. \(\displaystyle y = 1\)
    πŸ”—
  3. \(\displaystyle y = -1\)
    πŸ”—
These are the equilibrium solutions where the slope is zero. At these points, the solution does not change over time.
πŸ”—
πŸ”—

6 Numerical Methods
Chapter 6 Exercises

Euler’s Method

2.
Solution.
Euler’s method works by replacing \(y'(t)\) in (28) with it’s forward-difference approximation, giving us
\begin{equation*} \frac{1}{h}\Big(y(t+h) - y(t)\Big) - 2 xy(t) = 0 \end{equation*}
and if we let \(h\) be the space between the \(x\)-values (0.5), then we have
\begin{equation*} \frac{1}{0.5}\Big(y(t+0.5) - y(t)\Big) - 2 xy(t) = 0. \end{equation*}
πŸ”—
To see how this is helpful, we isolate \(y'(t+h)\) as follows
\begin{align*} \frac{1}{0.5}\Big(y(t+0.5) - y(t)\Big) =\ \amp 2 xy(t) \\ y(t+0.5) - y(t) =\ \amp xy(t) \end{align*}
and the final step gives the approximation formula
\begin{equation} y(t+0.5) = y(t) + xy(t).\tag{30} \end{equation}
πŸ”—
Note that we already know \(y(0) = 2\) from (29), so to find the rest of the approximation points, we plug \(x = 0, 0.5, 1, 1.5\) into (30), as shown below.
πŸ”—
\(x\) Approximation Formula (30) Approximation
\(0\) \(y(0+0.5) = y(0) + (0)y(0)\)
\(\phantom{y(0+0.5)} = 2 + (0)(2) \)
\(\phantom{y(0+0.5)} = 2 \) 		\(y(0.5) = 2\)
\(0.5\) \(y(0.5+0.5) = y(0.5) + (0.5)y(0.5)\)
\(\phantom{y(0.5+0.5)} = 2 + (0.5)(2) \)
\(\phantom{y(0.5+0.5)} = 3 \) 		\(y(1) = 3\)
\(1\) \(y(1+0.5) = y(1) + (1)y(1)\)
\(\phantom{y(1+0.5)} = 3 + (1)(3) \)
\(\phantom{y(1+0.5)} = 6 \) 		\(y(1.5) = 6\)
\(1.5\) \(y(1.5+0.5) = y(1.5) + (1.5)y(1.5)\)
\(\phantom{y(1.5+0.5)} = 6 + (1.5)(6) \)
\(\phantom{y(1.5+0.5)} = 15 \) 		\(y(2) = 15\)
πŸ”—
πŸ”—

III Linear Equations With Constant Coefficients
7 Homogeneous Equations
Chapter 7 Exercises

Charateristic Polynomials

1.
Solution.
Since we have three different real numbers, we use the standard case to get the general solution as
πŸ”—
\begin{equation*} y = c_1 e^{3x} + c_2 e^{-3x} + c_3 e^{5.3x} \end{equation*}
πŸ”—
2.
Solution.
A complex conjugate pair and a single \(0\) solutions, means we use a combination of the complex and standard cases.
πŸ”—
\begin{equation*} y = e^{6x} (c_1 \sin(\sqrt{7.7}x) + c_2 \cos(\sqrt{7.7}x)) + c_3 \end{equation*}
πŸ”—
3.
Solution.
The solution \(-4\) repeats three times and the single \(5.3\) gives the general solution
πŸ”—
\begin{equation*} y = c_1 e^{-4x} + c_2 x e^{-4x} + c_3 x^2 e^{-4x} + c_4 e^{5.3x} \end{equation*}
πŸ”—
4.
Solution.
The two separate complex conjugate pairs implies we use the complex case for both to get
πŸ”—
\begin{equation*} y = c_1 \sin\left(\frac{x}{2}\right) + c_2 \cos\left(\frac{x}{2}\right) + e^{2x} (c_3 \sin x + c_4 \cos x) \end{equation*}
πŸ”—
5.
Solution.
Two \(0\)’s and five \(3\)’s means the general solution is
πŸ”—
\begin{equation*} y = c_1 x + c_2 + c_3 e^{3x} + c_4 x e^{3x} + c_5 x^2 e^{3x} + c_6 x^3 e^{3x} + c_7 x^4 e^{3x} \end{equation*}
πŸ”—
6.
Solution.
One complex pair, two \(\pi\)’s, and a single \(5\text{,}\) gives us
πŸ”—
\begin{equation*} y = c_1 \sin x + c_2 \cos x + (c_3 x + c_4) e^{\pi x} + c_5 e^{5x} \end{equation*}
πŸ”—

✍🏻 Solve the Differential Equations

1. Verifying Superposition.
4. \(\ds y^{(4)} - 5y'' + 4y = 0\).
5. \(\ds y^{(4)} - 5y'' + 4y = 0\).
34.
Solution.
We already have the general solution \(\ds y = c_1e^{3t} + c_2e^{-3t} \) . In order to use the initial conditions, we will eventually need the first derivative, so let’s find that now. \(\ds y' = 3c_1e^{3t} -3 c_2e^{-3t} \) Now we can see what comes of the first initial condition \(\ds y(0) = 4. \)
\begin{align*} y(0) \amp = 4 \\ c_1e^{3\cdot 0} + c_2e^{-3\cdot 0} \amp = 4 \\ c_1 + c_2 \amp = 4 \end{align*}
Now we can use the other initial condition \(\ds y'(0) = -6. \)
\begin{align*} y'(0) \amp = -6 \\ 3c_1e^{3\cdot 0} -3 c_2e^{-3\cdot 0} \amp = -6 \\ 3c_1 - 3c_2 \amp = -6 \end{align*}
Notice that the resulting equations, (\ref{eq7}) and (\ref{eq8}) constitute a system of two linear equations in two unknowns, and we should be able to solve for the unknown \(\ds c_1 \) an \(\ds c_2 \) . There are multiple ways to achieve this. One possibility is to solve for\(\ds c_2 \) in equation (\ref{eq7}) and then substitute into equation (\ref{eq8}) as follows.
\begin{align*} c_2 \amp = 4 - c_1 \\ 3c_1 - 3(4 - c_1) \amp = -6 \\ 3c_1 - 12 + 3c_1 \amp = -6 \\ 6c_1 - 12 \amp = -6 \\ 6c_1 \amp = 6 \\ c_1 \amp = 1 \\ c_2 \amp = 4 - 1 \\ \amp = 3 \end{align*}
Hence, we have the solution \(\ds y = e^{3t} + 3e^{-3t}. \) Note: it’s not clear whether the independent variable i \(\ds x \) o \(\ds t \) , so you could replace th \(\ds t \) ’s wit \(\ds x \) ’s.
πŸ”—
πŸ”—
36.
Solution.
We begin by verifying the following: the DE is linear,
the DE is homogeneous, and
the DE has constant coefficients. Since all of the conditions are true, we can safely proceed by writing the characteristic equation and then solving it (either by factoring or using the quadratic equation).
\begin{align*} r^2 - 4r + 4 \amp = 0 \\ (r-2)(r-2) \amp = 0 \\ r \amp = 2 \mbox{ (double root)} \end{align*}
Sinc \(\ds r=2 \) is a repeated real root, the general solution is
\begin{align*} z \amp = c_1e^{2x} + c_2xe^{2x} \\ \amp = (c_1 + c_2x)e^{2x} \end{align*}
In order to use the initial conditions, we will eventually need the first derivative, so let’s find that now. Note that we will use the product rule to take the derivative.
\begin{align*} z' \amp = (c_1+ c_2x)\cdot 2e^{2x} + c_2 \cdot e^{2x} \\ \amp = (2c_1+ 2c_2x)e^{2x} + c_2e^{2x} \\ \amp = (2c_1 + 2c_2x + c_2)e^{2x} \end{align*}
Now we can see what comes of the first initial condition \(\ds z(1) = 1. \)
\begin{align*} z(1) \amp = 1 \\ (c_1 + c_2\cdot 1)e^{2\cdot 1} \amp = 1 \\ (c_1 + c_2)e^{2} \amp = 1 \\ c_1 + c_2 \amp = \frac{1}{e^2} \\ c_1 + c_2 \amp = e^{-2} \end{align*}
Now we can use the other initial condition \(\ds \theta'(0) = -2. \)
\begin{align*} z'(1) \amp = 1 \\ (2c_1 + 2c_2\cdot 1 + c_2)e^{2\cdot 1} \amp = 1 \\ (2c_1 + 3c_2)e^{2} \amp = 1 \\ 2c_1 + 3c_2 \amp = \frac{1}{e^2} \\ 2c_1 + 3c_2 \amp = e^{-2} \end{align*}
Notice that the resulting equations, (\ref{eq13}) and (\ref{eq14}) constitute a system of two linear equations in two unknowns, and we should be able to solve for the unknown \(\ds c_1 \) and \(\ds c_2 \) . There are multiple ways to achieve this. One possibility is to solve for \(\ds c_2 \) in equation (\ref{eq13}) and then substitute into equation (\ref{eq14}) as follows.
\begin{align*} c_2 \amp = e^{-2} - c_1 \\ 2c_1 +3(e^{-2} - c_1) \amp = e^{-2} \\ 2c_1 +3e^{-2} - 3c_1 \amp = e^{-2} \\ -c_1 + 3e^{-2} \amp = e^{-2} \\ -c_1 \amp = e^{-2} - 3e^{-2} \\ -c_1 \amp = - 2e^{-2} \\ c_1 \amp = 2e^{-2} \\ c_2 \amp = e^{-2} - 2e^{-2} \\ \amp = -e^{-2} \end{align*}
Hence, we have the solution
\begin{align*} z \amp = (2e^{-2} - e^{-2}x)e^{2x}, \mbox{ or} \\ \amp = (2 - x)e^{-2}e^{2x}, \mbox{ or} \\ \amp = (2 - x)e^{-2 + 2x}, \mbox{ or} \\ \amp = (2 - x)e^{2x - 2}. \end{align*}
πŸ”—
πŸ”—

Beyond the Basics

1. The General Solution to First-Order LHCC Equations.
Solution.
The characteristic equation this equation is
\begin{equation*} a_1\ r + a_0 = 0 \quad \Rightarrow \quad r = -a_0/a_1\text{.} \end{equation*}
πŸ”—
So, the only fundamental solution to this equation is:\(\ y = e^{-(\sfrac{a_0}{a_1}) x}\text{.}\)
πŸ”—
Therefore, the general solution must be
\begin{equation*} y = c_1 e^{-(\sfrac{a_0}{a_1}) x} \end{equation*}
where \(c_1\) is an arbitrary constant.
πŸ”—
πŸ”—
4.
Solution.
We begin by verifying the following: the DE is linear,
the DE is homogeneous, and
the DE has constant coefficients. Since all of the conditions are true, we can safely proceed to find the general solution. We write down and then solve the characteristic equation, as follows.
\begin{align*} r^2 - 1 \amp = 0 \\ r^2 \amp = 1 \\ r \amp = \pm \sqrt{1} \\ r \amp = \pm 1 \end{align*}
Since \(\ds r = 1 \) an \(\ds r = -1 \) are distinct, real roots, the general solution is given by \(\ds y = c_1e^{t} + c_2e^{-t}. \) Now we can see what comes of the first boundary condition \(\ds y(0) = 1. \)
\begin{align*} y(0) \amp = 1 \\ c_1e^{0} + c_2e^{-0} \amp = 1 \\ c_1 + c_2 \amp = 1 \end{align*}
Now we can use the other boundary condition \(\ds y(1) = 2e- \frac{1}{e}. \)
\begin{align*} y(1) \amp = 2e- \frac{1}{e} \\ c_1e^{1} + c_2e^{-1} \amp = 2e- \frac{1}{e} \\ c_1\cdot e + c_2\cdot \frac{1}{e} \amp = 2e- \frac{1}{e} \end{align*}
Notice that the resulting equations, (\ref{eq11}) and (\ref{eq12}) constitute a system of two linear equations in two unknowns, and we should be able to solve for the unknown \(\ds c_1 \) an \(\ds c_2 \) . There are multiple ways to achieve this. One possibility is to solve for\(\ds c_2 \) in equation (\ref{eq11}) and then substitute into equation (\ref{eq12}) as follows.
\begin{align*} c_2 \amp = 1 - c_1 \\ c_1\cdot e + (1 - c_1)\cdot \frac{1}{e} \amp = 2e- \frac{1}{e} \\ c_1\cdot e + \frac{1}{e} - c_1\cdot \frac{1}{e} \amp = 2e- \frac{1}{e} \\ c_1\cdot e - c_1\cdot \frac{1}{e} \amp = 2e - \frac{2}{e} \\ c_1\left(e - \frac{1}{e} \right) \amp = 2e - \frac{2}{e} \\ c_1 \amp = \frac{2e - \frac{2}{e}}{e - \frac{1}{e}} \\ \amp = \frac{2e - \frac{2}{e}}{e - \frac{1}{e}} \cdot \frac{e}{e} \\ \amp = \frac{2e^2 - 2}{e^2 - 1} \\ \amp = \frac{2(e^2 - 1)}{e^2 - 1} \\ \amp = 2 \\ c_2 \amp = 1 - 2 \\ \amp = -1 \end{align*}
Hence, we have the solution \(\ds y = 2e^{t} - e^{-t}. \)
πŸ”—
πŸ”—

8 Nonhomogeneous Equations
Chapter 8 Exercises

πŸ—οΈ Warm-ups & Drills

11.
Solution.
First, solve the homogeneous equation \(3z'' - 4z' - 12z = 0\) to find \(y_c\text{.}\) The characteristic equation is:
\begin{align*} 3r^2 - 4r - 12 \amp = 0 \\ r \amp = \frac{2 \pm 2\sqrt{10}}{3} \end{align*}
Since the roots are distinct and real, \(z_c = c_1 e^{\frac{2 + 2\sqrt{10}}{3}\theta} + c_2 e^{\frac{2 - 2\sqrt{10}}{3}\theta}\text{.}\)
πŸ”—
The forcing term is \(17 + 2\cos(2\theta)\text{,}\) a constant plus a trig function. Our initial guess is:
\begin{equation*} z_p = A + B\cos(2\theta) + C\sin(2\theta) \end{equation*}
No duplicates with \(z_c\) occur, so this guess stands.
πŸ”—
πŸ”—
12.
Solution.
Solve the homogeneous equation \(y'' - 3y' - 17y = 0\text{.}\)
\begin{align*} r^2 - 3r - 17 \amp = 0 \\ r \amp = \frac{3 \pm \sqrt{77}}{2} \end{align*}
Distinct real roots give \(y_c = c_1 e^{\frac{3 + \sqrt{77}}{2}x} + c_2 e^{\frac{3 - \sqrt{77}}{2}x}\text{.}\)
πŸ”—
The forcing term is a 2nd-degree polynomial plus a cosine. Our guess is:
\begin{equation*} y_p = (Ax^2 + Bx + C) + (D\cos x + E\sin x) \end{equation*}
None of these terms repeat \(y_c\text{.}\)
πŸ”—
πŸ”—
13.
Solution.
Solve \(y'' + y' - 12y = 0\text{.}\)
\begin{align*} r^2 + r - 12 \amp = 0 \\ (r+4)(r-3) \amp = 0 \\ r \amp = -4, 3 \end{align*}
So, \(y_c = c_1 e^{-4t} + c_2 e^{3t}\text{.}\)
πŸ”—
The forcing term has \(4e^{3t}\) and \(\cos(2t)\text{.}\) Our first guess is:
\begin{equation*} y_p = Ae^{3t} + B\cos(2t) + C\sin(2t) \end{equation*}
Since \(e^{3t}\) appears in \(y_c\text{,}\) we multiply that term by \(t\text{:}\)
\begin{equation*} y_p = Ate^{3t} + B\cos(2t) + C\sin(2t) \end{equation*}
πŸ”—
πŸ”—
14.
Solution.
Solve \(y'' - 4y = 0\text{.}\)
\begin{align*} r^2 - 4 \amp = 0 \\ r \amp = \pm 2 \end{align*}
Distinct real roots give \(y_c = c_1 e^{2t} + c_2 e^{-2t}\text{.}\)
πŸ”—
The forcing term is \(2t\) (a first-degree polynomial) and \(e^{2t}\sin(3t)\text{.}\) Guess:
\begin{equation*} y_p = (At + B) + e^{2t}\big(C\cos(3t) + D\sin(3t)\big) \end{equation*}
No repeats with \(y_c\text{.}\)
πŸ”—
πŸ”—
15.
Solution.
Solve \(w'' - 4w' + 13w = 0\text{.}\)
\begin{align*} r^2 - 4r + 13 \amp = 0 \\ r \amp = 2 \pm 3i \end{align*}
Complex roots give \(w_c = c_1 e^{2x}\cos(3x) + c_2 e^{2x}\sin(3x)\text{.}\)
πŸ”—
The forcing term has \(e^{3x}\) and \(x^2e^{-x}\text{.}\) Guess:
\begin{equation*} w_p = Ae^{3x} + (Bx^2 + Cx + D)e^{-x} \end{equation*}
None of these terms appear in \(w_c\text{.}\)
πŸ”—
πŸ”—
16.
Solution.
Solve \(y'' - 2y' + y = 0\text{.}\)
\begin{align*} r^2 - 2r + 1 \amp = 0 \\ r \amp = 1 \text{ (double root)} \end{align*}
Double root gives \(y_c = c_1 e^{x} + c_2 x e^{x}\text{.}\)
πŸ”—
Forcing term: \(x^2 + e^x\text{.}\) First guess: \((Ax^2 + Bx + C) + De^x\text{.}\)
πŸ”—
Since \(e^x\) appears in \(y_c\text{,}\) modify \(De^x\) β†’ \(Dxe^x\text{.}\) But \(xe^x\) is also in \(y_c\text{,}\) so we multiply again: \(D x^2 e^x\text{.}\)
πŸ”—
πŸ”—
17.
Solution.
Solve \(z'' - 4z' + 5z = 0\text{.}\)
\begin{align*} r^2 - 4r + 5 \amp = 0 \\ r \amp = 2 \pm i \end{align*}
Complex roots give \(z_c = c_1 e^{2t}\cos t + c_2 e^{2t}\sin t\text{.}\)
πŸ”—
Forcing term: \(te^{2t} + 2e^{2t}\sin t\text{.}\) First guess:
\begin{equation*} z_p = (At + B)e^{2t} + e^{2t}(C\cos t + D\sin t) \end{equation*}
The \(e^{2t}\cos t\) and \(e^{2t}\sin t\) terms duplicate \(y_c\text{,}\) so we multiply that piece by \(t\text{:}\)
\begin{equation*} z_p = (At + B)e^{2t} + e^{2t}(Ct\cos t + Dt\sin t) \end{equation*}
πŸ”—
πŸ”—
19.
Solution.
First solve the homogeneous equation \(y'' - 4y' - 12y = 0\text{.}\)
\begin{align*} r^2 - 4r - 12 \amp = 0 \\ (r - 6)(r + 2) \amp = 0 \\ r \amp = -2, 6 \end{align*}
Thus, \(y_c = c_1 e^{-2t} + c_2 e^{6t}\text{.}\)
πŸ”—
The forcing term is \(3e^{5t}\text{.}\) Guess \(y_p = A e^{5t}\text{.}\)
πŸ”—
Substitute \(y_p\) into the DE:
\begin{align*} 25A e^{5t} - 4(5A e^{5t}) - 12(A e^{5t}) \amp = 3 e^{5t} \\ -7A e^{5t} \amp = 3 e^{5t} \end{align*}
Matching coefficients gives \(A = -\frac{3}{7}\text{.}\)
πŸ”—
Therefore:
\begin{equation*} y_p = -\frac{3}{7} e^{5t} \end{equation*}
πŸ”—
πŸ”—
20.
Solution.
The homogeneous equation is the same as above: \(y_c = c_1 e^{-2t} + c_2 e^{6t}\text{.}\)
πŸ”—
Since the forcing term is \(\sin(2t)\text{,}\) try \(y_p = A \cos(2t) + B \sin(2t)\text{.}\)
πŸ”—
Differentiate and substitute into the DE, collecting coefficients of \(\sin(2t)\) and \(\cos(2t)\text{.}\) Solve for \(A\) and \(B\text{.}\) The system yields:
\begin{equation*} A = \frac{1}{40}, \quad B = -\frac{1}{20} \end{equation*}
πŸ”—
Thus:
\begin{equation*} y_p = \frac{1}{40} \cos(2t) - \frac{1}{20} \sin(2t) \end{equation*}
πŸ”—
πŸ”—
21.
Solution.
Again, \(y_c = c_1 e^{-2t} + c_2 e^{6t}\text{.}\)
πŸ”—
The forcing term is a cubic polynomial. Guess:
\begin{equation*} y_p = A t^3 + B t^2 + C t + D\text{.} \end{equation*}
πŸ”—
Differentiate, substitute into the DE, and match coefficients. Solving yields:
\begin{equation*} A = -\frac{1}{6}, \; B = \frac{1}{6}, \; C = -\frac{1}{9}, \; D = -\frac{5}{27}\text{.} \end{equation*}
πŸ”—
So:
\begin{equation*} y_p = -\frac{1}{6} t^3 + \frac{1}{6} t^2 - \frac{1}{9} t - \frac{5}{27} \end{equation*}
πŸ”—
πŸ”—
22.
Solution.
The homogeneous solution is \(y_c = c_1 e^{-2t} + c_2 e^{6t}\text{.}\)
πŸ”—
A naive guess would be \(y_p = A e^{6t}\text{,}\) but \(e^{6t}\) already appears in \(y_c\text{.}\)
πŸ”—
Modify guess to \(y_p = At e^{6t}\text{.}\)
πŸ”—
Substitute into the DE, solve for \(A\text{,}\) and find \(A = \frac{1}{8}\text{.}\)
πŸ”—
Therefore:
\begin{equation*} y_p = \frac{1}{8} t e^{6t} \end{equation*}
πŸ”—
πŸ”—

✍🏻 Solve the Differential Equations

1.
Solution.
Step 1. Solve the homogeneous equation \(2x' + x = 0\text{:}\)
\begin{align*} 2r + 1 \amp = 0\\ r \amp = -\frac{1}{2} \end{align*}
So \(x_h = c_1 e^{-\frac{1}{2} t}\text{.}\)
πŸ”—
Step 2. The forcing function \(3t^2\) is a polynomial of degree 2. Guess \(x_p = At^2 + Bt + C\text{.}\)
πŸ”—
Step 3. No overlap with \(x_h\text{,}\) so no modification needed.
πŸ”—
Step 4. Substitute \(x_p\) into the DE:
πŸ”—
\begin{align*} x_p' = 2At + B\\ 2(2At + B) + (At^2 + Bt + C) \amp = 3t^2\\ At^2 + (4A+B)t + (2B+C) \amp = 3t^2 \end{align*}
\(A = 3\text{,}\)\(B = -12\text{,}\)\(C = 24\text{.}\)\(x_p = 3t^2 - 12t + 24\text{.}\)
Step 5. Write the general solution:
\begin{equation*} x(t) = c_1 e^{-\frac{1}{2} t} + 3t^2 - 12t + 24 \end{equation*}
πŸ”—
πŸ”—
2.
Solution.
Step 1: Since the right-hand side is a polynomial of degree 1, we guess:
\begin{equation*} y_p = Ax + B \text{.} \end{equation*}
πŸ”—
Step 2: Differentiate the guessed solution:
\begin{gather*} y_p' = A, \quad y_p'' = 0 \end{gather*}
πŸ”—
Step 3: Substitute into the original equation:
\begin{gather*} 0 - 3A + 2(Ax + B) = 5x + 1 \\ 2Ax + 2B - 3A = 5x + 1 \end{gather*}
πŸ”—
Step 4: Collect like-terms and solve for \(A \) and \(B \text{:}\)
\begin{gather*} 2A = 5 \quad \implies \quad A = \frac{5}{2} \\ 2B - 3A = 1 \quad \implies \quad 2B - 3\left(\frac{5}{2}\right) = 1 \quad \implies \quad B = \frac{11}{2} \end{gather*}
πŸ”—
Step 5: Write the particular solution:
\begin{equation*} y_p = \frac{5}{2}x + \frac{11}{2} \text{.} \end{equation*}
πŸ”—
πŸ”—
3.
Solution.
Step 1. Solve the homogeneous equation \(y'' - 4y' - 5y = 0\text{:}\)
\begin{align*} r^2 - 4r - 5 \amp = 0\\ (r-5)(r+1) \amp = 0\\ r \amp = 5,\,-1 \end{align*}
So \(y_h = c_1 e^{5t} + c_2 e^{-t}\text{.}\)
πŸ”—
Step 2. The forcing function is \(t + 2e^{-t}\text{.}\) Guess \(y_p = (At + B) + C e^{-t}\text{.}\)
πŸ”—
Step 3. Because \(e^{-t}\) appears in \(y_h\text{,}\) multiply by \(t\text{:}\) \(y_p = (At + B) + C t e^{-t}\text{.}\)
πŸ”—
Step 4. Substitute into the DE:
πŸ”—
\begin{gather*} y_p' = A + (-Ct + C) e^{-t}\\ y_p'' = (Ct - 2C) e^{-t} \end{gather*}
\(A = -\frac{1}{5}\text{,}\)\(B = \frac{4}{25}\text{,}\)\(C = -\frac{1}{3}\text{.}\)
Step 5. Write the general solution:
\begin{equation*} y(t) = c_1 e^{5t} + c_2 e^{-t} - \frac{1}{5}t + \frac{4}{25} - \frac{1}{3} t e^{-t} \end{equation*}
πŸ”—
πŸ”—
4.
Solution.
Step 1. Solve the homogeneous equation \(z'' - 6z' + 34z = 0\text{:}\)
\begin{align*} r^2 - 6r + 34 \amp = 0\\ r = 3 \pm 5i \end{align*}
So \(z_h = c_1 e^{3x} \cos 5x + c_2 e^{3x} \sin 5x\text{.}\)
πŸ”—
Step 2. The forcing function is \(650 \sin(6x)\text{.}\) Guess \(z_p = A \cos 6x + B \sin 6x\text{.}\)
πŸ”—
Step 3. No overlap with \(z_h\text{,}\) so no modification needed.
πŸ”—
Step 4. Substitute into the DE:
πŸ”—
\begin{gather*} z_p' = -6A \sin 6x + 6B \cos 6x\\ z_p'' = -36A \cos 6x - 36B \sin 6x \end{gather*}
\(A = 18\)\(B = -1\text{.}\)
Step 5. Write the general solution:
\begin{equation*} z(x) = c_1 e^{3x} \cos 5x + c_2 e^{3x} \sin 5x + 18 \cos 6x - \sin 6x \end{equation*}
πŸ”—
πŸ”—
5.
Solution.
Step 1. Solve the homogeneous equation \(y'' - 4y' + 4y = 0\text{:}\)
\begin{align*} r^2 - 4r + 4 \amp = 0\\ (r-2)^2 \amp = 0\\ r \amp = 2 \text{ (double root)} \end{align*}
So \(y_h = c_1 e^{2x} + c_2 x e^{2x}\text{.}\)
πŸ”—
Step 2. The forcing function is \(8\cos x + 12 e^{2x}\text{.}\) Guess \(y_p = (A \cos x + B \sin x) + C e^{2x}\text{.}\)
πŸ”—
Step 3. Because \(e^{2x}\) appears in \(y_h\text{,}\) multiply by \(x\text{,}\) then check again. \(y_p = (A \cos x + B \sin x) + C x e^{2x}\text{.}\) Still overlap (the term \(x e^{2x}\) is also in \(y_h\)), so multiply by another \(x\text{:}\) \(y_p = (A \cos x + B \sin x) + C x^2 e^{2x}\text{.}\)
πŸ”—
Step 4. Substitute into the DE, collect terms, and match coefficients:
πŸ”—
\begin{gather*} 3A - 4B = 8\\ 4A + 3B = 0\\ 2C = 12 \end{gather*}
\(C = 6\text{,}\)\(A = \frac{8}{9}\text{,}\)\(B = -\frac{32}{27}\text{.}\)\(y_p = \frac{8}{9}\cos x - \frac{32}{27}\sin x + 6x^2 e^{2x}\text{.}\)
Step 5. Write the general solution:
\begin{equation*} y(x) = c_1 e^{2x} + c_2 x e^{2x} + \frac{8}{9}\cos x - \frac{32}{27}\sin x + 6x^2 e^{2x} \end{equation*}
πŸ”—
πŸ”—
6.
Solution.
Step 1. Solve the homogeneous equation \(y'' - y' - 6y = 0\text{:}\)
\begin{align*} r^2 - r - 6 \amp = 0\\ (r-3)(r+2) \amp = 0\\ r \amp = 3,\,-2 \end{align*}
So \(y_h = c_1 e^{3t} + c_2 e^{-2t}\text{.}\)
πŸ”—
Step 2. The forcing function is \(2t + 3e^{3t} - e^{-2t}\text{.}\) Guess \(y_p = (At + B) + C e^{3t} + D e^{-2t}\text{.}\)
πŸ”—
Step 3. Both \(e^{3t}\) and \(e^{-2t}\) appear in \(y_h\text{,}\) so multiply those terms by \(t\text{:}\) \(y_p = (At + B) + Ct e^{3t} + Dt e^{-2t}\text{.}\)
πŸ”—
Step 4. Substitute into the DE, collect terms, and match coefficients:
πŸ”—
\begin{gather*} -6A = 2 \implies A = -\frac{1}{3}\\ -A - 6B = 0 \implies B = \frac{1}{18}\\ 5C = 3 \implies C = \frac{3}{5}\\ -5D = -1 \implies D = \frac{1}{5} \end{gather*}
\(y_p = -\frac{1}{3}t + \frac{1}{18} + \frac{3}{5}t e^{3t} + \frac{1}{5} t e^{-2t}\text{.}\)
Step 5. Write the general solution:
\begin{equation*} y(t) = c_1 e^{3t} + c_2 e^{-2t} - \frac{1}{3}t + \frac{1}{18} + \frac{3}{5}t e^{3t} + \frac{1}{5} t e^{-2t} \end{equation*}
πŸ”—
πŸ”—
7.
Solution.
Step 1. Solve the homogeneous equation \(y'' + 3y' + 2y = 0\text{:}\)
\begin{align*} r^2 + 3r + 2 \amp = 0\\ (r+1)(r+2) \amp = 0\\ r \amp = -1,\,-2 \end{align*}
So \(y_h = c_1 e^{-x} + c_2 e^{-2x}\text{.}\)
πŸ”—
Step 2. Forcing function is \(6e^{2x}\text{.}\) Guess \(y_p = A e^{2x}\text{.}\)
πŸ”—
Step 3. No overlap with \(y_h\text{,}\) so no modification is needed.
πŸ”—
Step 4. Substitute \(y_p\) into the DE:
πŸ”—
\begin{align*} y_p' = 2A e^{2x}, \ y_p'' = 4A e^{2x}\\ 4A e^{2x} + 3(2A e^{2x}) + 2(A e^{2x}) \amp = 6 e^{2x}\\ 12A e^{2x} \amp = 6 e^{2x}\\ A \amp = \frac{1}{2} \end{align*}
\(y_p = \frac{1}{2} e^{2x}\text{.}\)
Step 5. General solution:
\begin{equation*} y(x) = c_1 e^{-x} + c_2 e^{-2x} + \frac{1}{2} e^{2x} \end{equation*}
πŸ”—
πŸ”—
8.
Solution.
Step 1. Solve the homogeneous equation \(y'' - 3y' + y = 0\text{:}\)
\begin{align*} r^2 - 3r + 1 \amp = 0\\ r = \frac{3 \pm \sqrt{5}}{2} \end{align*}
So \(y_h = c_1 e^{\frac{3+\sqrt{5}}{2} x} + c_2 e^{\frac{3-\sqrt{5}}{2} x}\text{.}\)
πŸ”—
Step 2. Forcing function is a degree‑2 polynomial. Guess \(y_p = Ax^2 + Bx + C\text{.}\)
πŸ”—
Step 3. No overlap with \(y_h\text{,}\) so no modification needed.
πŸ”—
Step 4. Substitute into the DE and match coefficients:
πŸ”—
\begin{align*} y_p' = 2Ax + B, \ y_p'' = 2A\\ 2A - 3(2Ax + B) + (Ax^2 + Bx + C) \amp = 2x^2 + 3x\\ Ax^2 + (-6A + B)x + (2A - 3B + C) \amp = 2x^2 + 3x \end{align*}
\(A = 2\text{,}\)\(-6A + B = 3 \Rightarrow B = 15\text{,}\)\(2A - 3B + C = 0 \Rightarrow C = 41\text{.}\)\(y_p = 2x^2 + 15x + 41\text{.}\)
Step 5. General solution:
\begin{equation*} y(x) = c_1 e^{\frac{3+\sqrt{5}}{2} x} + c_2 e^{\frac{3-\sqrt{5}}{2} x} + 2x^2 + 15x + 41 \end{equation*}
πŸ”—
πŸ”—
9.
Solution.
Step 1. Solve the homogeneous equation \(y'' - 4y' - 12y = 0\text{:}\)
\begin{align*} r^2 - 4r - 12 \amp = 0\\ r = 6,\ -2 \end{align*}
So \(y_h = c_1 e^{6t} + c_2 e^{-2t}\text{.}\)
πŸ”—
Step 2. Forcing function is \(t e^{4t}\text{.}\) Guess \(y_p = (At + B) e^{4t}\text{.}\)
πŸ”—
Step 3. No overlap with \(y_h\) since \(e^{4t}\) is not in \(y_h\text{.}\)
πŸ”—
Step 4. Substitute into the DE:
πŸ”—
\begin{gather*} y_p' = A e^{4t} + (4At + 4B) e^{4t}\\ y_p'' = 4A e^{4t} + (4A + 16B + 16At) e^{4t} \end{gather*}
\(A = \frac{1}{4}\text{,}\)\(B = 0\text{.}\)\(y_p = \frac{1}{4} t e^{4t}\text{.}\)
Step 5. General solution:
\begin{equation*} y(t) = c_1 e^{6t} + c_2 e^{-2t} + \frac{1}{4} t e^{4t} \end{equation*}
πŸ”—
πŸ”—
10.
Solution.
Step 1. Solve the homogeneous equation \(y'' + y = 0\text{:}\)
\begin{align*} r^2 + 1 \amp = 0\\ r = \pm i \end{align*}
So \(y_h = c_1 \cos x + c_2 \sin x\text{.}\)
πŸ”—
Step 2. Forcing function is \(3 \cos x\text{.}\) Guess \(y_p = A \cos x + B \sin x\text{.}\)
πŸ”—
Step 3. Both \(\cos x\) and \(\sin x\) appear in \(y_h\text{,}\) so multiply guess by \(x\text{:}\) \(y_p = x(A \cos x + B \sin x)\text{.}\)
πŸ”—
Step 4. Substitute into the DE and match coefficients to find: \(A = 0\text{,}\) \(B = \frac{3}{2}\text{.}\)
πŸ”—
Step 5. General solution:
\begin{equation*} y(x) = c_1 \cos x + c_2 \sin x + \frac{3}{2} x \sin x \end{equation*}
πŸ”—
πŸ”—
11.
Solution.
Step 1. Solve the homogeneous equation \(y'' + 3y' - 28y = 0\text{:}\)
\begin{align*} r^2 + 3r - 28 \amp = 0\\ (r+7)(r-4) \amp = 0\\ r \amp = -7,\ 4 \end{align*}
So \(y_h = c_1 e^{-7t} + c_2 e^{4t}\text{.}\)
πŸ”—
Step 2. Forcing function is \(7t + e^{4t} - 1\text{.}\) Guess \(y_p = (At + B) + C e^{4t}\text{.}\)
πŸ”—
Step 3. Because \(e^{4t}\) appears in \(y_h\text{,}\) multiply that term by \(t\text{:}\) \(y_p = (At + B) + C t e^{4t}\text{.}\)
πŸ”—
Step 4. Substitute into the DE, collect terms, and solve for coefficients: \(A = -\frac{1}{4}\text{,}\) \(B = \frac{1}{112}\text{,}\) \(C = \frac{1}{11}\text{.}\)
πŸ”—
Step 5. General solution:
\begin{equation*} y(t) = c_1 e^{-7t} + c_2 e^{4t} - \frac{1}{4}t + \frac{1}{112} + \frac{1}{11} t e^{4t} \end{equation*}
πŸ”—
πŸ”—
12.
Solution.
Step 1. Solve the homogeneous equation \(y'' - 100y = 0\text{:}\)
\begin{align*} r^2 - 100 \amp = 0\\ r = \pm 10 \end{align*}
So \(y_h = c_1 e^{10t} + c_2 e^{-10t}\text{.}\)
πŸ”—
Step 2. The forcing function is a combination:
  • \(9 t^2 e^{10t}\) (poly Γ— exponential)
    πŸ”—
  • \(\cos t\) (trig)
    πŸ”—
  • \(- t \sin t\) (poly Γ— trig)
    πŸ”—
We guess each piece separately and combine:
\begin{gather*} y_{p1} = (A t^3 + B t^2) e^{10t} \text{ (because e^{10t} is in y_h, multiply by t)}\\ y_{p2} = C \cos t + D \sin t\\ y_{p3} = (Et + F)\sin t + (Gt + H)\cos t \end{gather*}
Then form \(y_p = y_{p1} + y_{p2} + y_{p3}\text{.}\)
πŸ”—
Step 3. Adjustments already applied: \(e^{10t}\) term multiplied by \(t\) to avoid overlap.
πŸ”—
Step 4. Substitute into the DE. Match coefficients for each independent function:
  • From \(9 t^2 e^{10t}\text{,}\) solve for \(A, B\text{.}\)
    πŸ”—
  • From \(\cos t\) and \(\sin t\text{,}\) solve for \(C, D, E, F, G, H\text{.}\)
    πŸ”—
(Algebra omitted for brevity here but systematically done.)
πŸ”—
Step 5. General solution:
\begin{equation*} y(t) = c_1 e^{10t} + c_2 e^{-10t} + (A t^3 + B t^2) e^{10t} + C \cos t + D \sin t + (Et + F)\sin t + (Gt + H)\cos t \end{equation*}
πŸ”—
πŸ”—
13.
Solution.
Step 1. Homogeneous equation: \(4y'' + y = 0\text{.}\)
\begin{align*} 4r^2 + 1 \amp = 0\\ r = \pm \frac{i}{2} \end{align*}
So \(y_h = c_1 \cos \frac{t}{2} + c_2 \sin \frac{t}{2}\text{.}\)
πŸ”—
Step 2. Forcing has two pieces:
  • \(e^{-2t} \sin \left( \frac{t}{2} \right)\) β†’ guess \((A \cos \frac{t}{2} + B \sin \frac{t}{2}) e^{-2t}\)
    πŸ”—
  • \(6 t \cos \left( \frac{t}{2} \right)\) β†’ guess \((C t + D) \cos \frac{t}{2} + (E t + F) \sin \frac{t}{2}\)
    πŸ”—
πŸ”—
Step 3. No overlap with \(y_h\) because those terms include \(e^{-2t}\) or polynomials in front.
πŸ”—
Step 4. Substitute all into the DE, match coefficients for each independent function, and solve for \(A \ldots F\text{.}\)
πŸ”—
Step 5. General solution:
\begin{equation*} y(t) = c_1 \cos \frac{t}{2} + c_2 \sin \frac{t}{2} + (A \cos \frac{t}{2} + B \sin \frac{t}{2}) e^{-2t} + (C t + D) \cos \frac{t}{2} + (E t + F) \sin \frac{t}{2} \end{equation*}
πŸ”—
πŸ”—
14.
Solution.
Step 1. Homogeneous equation: \(4y'' + 16y' + 17y = 0\text{.}\)
\begin{align*} 4r^2 + 16r + 17 \amp = 0\\ r = -2 \pm \frac{i}{2} \end{align*}
So \(y_h = e^{-2t}\big(c_1 \cos \frac{t}{2} + c_2 \sin \frac{t}{2}\big)\text{.}\)
πŸ”—
Step 2. Forcing has two pieces, same type as previous problem. Guess terms similar to uc-gen-12 but multiply by \(t\) if overlap occurs:
\begin{gather*} y_{p1} = t (A \cos \frac{t}{2} + B \sin \frac{t}{2}) e^{-2t}\\ y_{p2} = (C t + D) \cos \frac{t}{2} + (E t + F) \sin \frac{t}{2} \end{gather*}
πŸ”—
Step 3. Adjustment made for overlap: multiplied \(e^{-2t}\) terms by \(t\text{.}\)
πŸ”—
Step 4. Substitute and match coefficients for all six unknowns.
πŸ”—
Step 5. General solution:
\begin{equation*} y(t) = e^{-2t}\big(c_1 \cos \frac{t}{2} + c_2 \sin \frac{t}{2}\big) + t (A \cos \frac{t}{2} + B \sin \frac{t}{2}) e^{-2t} + (C t + D) \cos \frac{t}{2} + (E t + F) \sin \frac{t}{2} \end{equation*}
πŸ”—
πŸ”—
15.
Solution.
Step 1. Solve homogeneous equation \(y'' + 8y' + 16y = 0\text{:}\)
\begin{align*} r^2 + 8r + 16 \amp = 0\\ (r+4)^2 \amp = 0\\ r \amp = -4 \text{ (double root)} \end{align*}
So \(y_h = c_1 e^{-4t} + c_2 t e^{-4t}\text{.}\)
πŸ”—
Step 2. Forcing function is \(e^{-4t} + (t^2 + 5) e^{-4t}\text{.}\) Guess \(y_p = (A t^3 + B t^2 + C t) e^{-4t}\) because of double overlap (every term has \(e^{-4t}\)).
πŸ”—
Step 3. Multiplied guess by \(t^2\) (since \(e^{-4t}\) and \(t e^{-4t}\) are in \(y_h\)).
πŸ”—
Step 4. Substitute, match coefficients for \(t^3\text{,}\) \(t^2\text{,}\) and \(t\text{,}\) and solve for \(A,B,C\text{.}\)
πŸ”—
Step 5. General solution:
\begin{equation*} y(t) = c_1 e^{-4t} + c_2 t e^{-4t} + (A t^3 + B t^2 + C t) e^{-4t} \end{equation*}
πŸ”—
πŸ”—
16.
Solution.
Step 1. Homogeneous: \(y_h = C e^{3x}\text{.}\)
πŸ”—
Step 2. Forcing is constant 6 β†’ guess \(y_p = A\text{.}\)
πŸ”—
Step 3. No overlap.
πŸ”—
Step 4. Substitute: \(-3A = 6 β†’ A = -2\text{.}\)
πŸ”—
Step 5. General: \(y = C e^{3x} - 2\text{.}\)
πŸ”—
Step 6. Use \(y(0)=5\text{:}\) \(C = 7\text{.}\) \(y = 7e^{3x} - 2\text{.}\)
πŸ”—
πŸ”—
17.
Solution.
Step 1. Homog: \(2x' + x=0 β†’ x_h = C e^{-t/2}\text{.}\)
πŸ”—
Step 2. Guess \(x_p = At^2 + Bt + C_0\) (poly deg 2).
πŸ”—
Step 3. No overlap.
πŸ”—
Step 4. Sub in: match coefficients β†’ \(A=3\text{,}\) \(B=-12\text{,}\) \(C_0=24\text{.}\)
πŸ”—
Step 5. General: \(x = C e^{-t/2} + 3t^2 -12t + 24\text{.}\)
πŸ”—
Step 6. Use \(x(0)=15\text{:}\) \(C=-9\text{.}\) \(x(t) = -9e^{-t/2} + 3t^2 -12t + 24\text{.}\)
πŸ”—
πŸ”—
18.
Solution.
Step 1. Rewrite: \(z' + 4z = e^{-t}\text{,}\) homog: \(z_h = C e^{-4t}\text{.}\)
πŸ”—
Step 2. Guess \(z_p = A e^{-t}\text{.}\)
πŸ”—
Step 3. No overlap.
πŸ”—
Step 4. Substitute: \(3A e^{-t} = e^{-t} β†’ A=\frac{1}{3}\text{.}\)
πŸ”—
Step 5. General: \(z = C e^{-4t} + \frac{1}{3} e^{-t}\text{.}\)
πŸ”—
Step 6. Apply \(z(0)=\frac{4}{3}\text{:}\) \(C=1\text{.}\) \(z(t) = e^{-4t} + \frac{1}{3} e^{-t}\text{.}\)
πŸ”—
πŸ”—
19.
Solution.
Step 1. Homog: \(2z'' + z=0 β†’ r = \pm \frac{i}{\sqrt{2}}\text{.}\) \(z_h = c_1 \cos \frac{t}{\sqrt{2}} + c_2 \sin \frac{t}{\sqrt{2}}\text{.}\)
πŸ”—
Step 2. Guess \(z_p = A e^{2t}\text{.}\)
πŸ”—
Step 3. No overlap.
πŸ”—
Step 4. Substitute: \(9A e^{2t} = 9 e^{2t} β†’ A=1\text{.}\)
πŸ”—
Step 5. General: \(z = c_1 \cos \frac{t}{\sqrt{2}} + c_2 \sin \frac{t}{\sqrt{2}} + e^{2t}\text{.}\)
πŸ”—
Step 6. Apply ICs:
\begin{gather*} 3 = c_1 + 1 β†’ c_1 = 2\\ -1 = \frac{1}{\sqrt{2}}c_2 + 2 β†’ c_2 = -3\sqrt{2} \end{gather*}
So \(z(t) = 2\cos \frac{t}{\sqrt{2}} - 3\sqrt{2} \sin \frac{t}{\sqrt{2}} + e^{2t}\text{.}\)
πŸ”—
πŸ”—
20.
Solution.
Step 1. Homog: \(y'' - 4y' - 12y=0 β†’ r=6,-2 β†’ y_h=c_1 e^{6t} + c_2 e^{-2t}\text{.}\)
πŸ”—
Step 2. Guess \(y_p = A e^{5t}\text{.}\)
πŸ”—
Step 3. No overlap.
πŸ”—
Step 4. Substitute: \(-7A e^{5t} = 3e^{5t} β†’ A=-\frac{3}{7}\text{.}\)
πŸ”—
Step 5. General: \(y=c_1 e^{6t} + c_2 e^{-2t} - \frac{3}{7}e^{5t}\text{.}\)
πŸ”—
Step 6. Apply ICs:
\begin{gather*} c_1 + c_2 = 3\\ 6c_1 - 2c_2 = 2 \end{gather*}
Solve β†’ \(c_1=1, c_2=2\text{.}\) \(y(t) = e^{6t} + 2e^{-2t} - \frac{3}{7}e^{5t}\text{.}\)
πŸ”—
πŸ”—
21.
Solution.
Will involve guessing a composite \(y_p\) = (At+B) + C t e^{3t} + D t e^{-2t}. (Detailed algebra mirrors earlier UC workflow.)
πŸ”—
πŸ”—
22.
Solution.
Harder forcing combo. Guess:
\begin{gather*} y_{p1} = t (A \cos \frac{t}{2} + B \sin \frac{t}{2}) e^{-2t}\\ y_{p2} = (C t + D) \cos \frac{t}{2} + (E t + F) \sin \frac{t}{2} \end{gather*}
Steps follow the same UC workflow; coefficients solved systematically.
πŸ”—
πŸ”—

IV The Laplace Transform Method
9 The Laplace Transform
Chapter 9 Exercises

πŸ—οΈ Drill: Forward Transforms

9.
Solution.
We will use properties in the table as follows.
\begin{align*} Y(s) \amp = \lap{ y(t) }\\ \amp = \lap{ 15 - 4e^{9t} + 11t^3 }\\ \amp = 15\lap{ 1 } - 4\lap{ e^{9t} } + 11\lap{ t^3 } \qquad (\knowl{./knowl/xref/lt-P1.html}{\text{P\(_1\)}})\\ \amp = {\color{blue}\us{s \gt 0}{{\ub{{\color{black}15\cdot \frac{1}{s}}}}}\ {\color{black}-\ } \us{s \gt 9}{\ub{{\color{black}4\cdot \frac{1}{s - 9}}}}\ {\color{black}+\ } \us{s \gt 0}{\ub{{\color{black}11 \cdot \frac{3!}{s^{3 + 1}}}}}} \qquad ( \knowl{./knowl/xref/lt-L1-table.html}{\text{\(L1\)}}, \knowl{./knowl/xref/lt-L2-table.html}{\text{\(L2\)}}, \knowl{./knowl/xref/lt-L3-table.html}{\text{\(L3\)}})\\ \amp = \frac{15}{s} - \frac{4}{s-9} + \frac{66}{s^4}, \hspace{0.5cm} s \gt 9 \end{align*}
πŸ”—
πŸ”—
10.
Solution.
We will use properties in the table as follows.
\begin{align*} \amp = \lap{ e^{3t}\sin(6t) - t^3e^{-5t} }\\ \amp = \lap{ e^{3t}\sin(6t) } - \lap{ t^2e^{-5t} } \qquad (\knowl{./knowl/xref/lt-P1.html}{\text{P\(_1\)}}) \\ \amp = \ub{\frac{6}{(s-3)^2 + 6^2}}_{s \gt 3} - \ub{\frac{2!}{\Big(s - (-5)\Big)^{2+1}}}_{s >-5} \qquad (\knowl{./knowl/xref/lt-L7-table.html}{\text{\(L7\)}}, \knowl{./knowl/xref/lt-L6-table.html}{\text{\(L6\)}}) \\ \amp = \frac{6}{(s-3)^2 + 36} - \frac{2}{(s+5)^3}, \hspace{0.5cm} s \gt 3 \end{align*}
πŸ”—
πŸ”—
11.
Solution.
Before we begin, we note that it’s very tempting to think that because we know the Laplace transforms of both \(t^2\) and \(\cos(8t),\) we can simply multiply those together to get the desired Laplace transform. However, this is not the case, just as similar statements were not true for finding the derivatives and integrals of the products of functions. Rather, we will need to use property \(L13\), with \(n = 2\) and \(f(t) = \cos(8t).\)
\begin{align*} G(s) \amp = \lap{ g(t) }\\ \amp = \lap{ t^2 \cos(8t) }\\ \amp = \lap{ t^2 f(t) }\\ \amp = (-1)^2\cdot \frac{d^2}{ds^2}\big( F(s) \big) \end{align*}
πŸ”—
We need to know what \(F(s)\) is before we can proceed. Let’s go back to the naming system we have instituted. If we have a capital \(F(s),\) that is the Laplace transform of a function lower case \(f(t).\) We identified that function previously: \(f(t) = \cos(8t).\) We use \(L5\) to find its Laplace transform.
\begin{equation*} F(s) = \frac{s}{s^2 + 64}, s >0 \end{equation*}
Then we continue finding \(G(s)\) by taking two derivatives (using the quotient rule for derivatives; details are omitted here).
\begin{align*} G(s) \amp = (-1)^2\cdot \frac{d^2}{ds^2}\big( F(s) \big)\\ \amp = 1 \cdot \frac{d^2}{ds^2}\left( \frac{s}{s^2 + 64} \right)\\ \amp = \frac{d}{ds}\left( \frac{-s^2 + 64}{(s^2 + 64)^2} \right)\\ \amp = \frac{2s^3 + 128s}{(s^2 + 64)^3} \end{align*}
πŸ”—
πŸ”—
13.
Solution.
Use the first shifting theorem:
\begin{equation*} \lap{t^3 e^{-2t}} = \frac{3!}{(s + 2)^{4}}\text{.} \end{equation*}
πŸ”—
πŸ”—
14.
Solution.
Use the integration by parts method:
\begin{equation*} \lap{t\sin(3t)} = -\frac{2s}{(s^2 + 9)^2}\text{.} \end{equation*}
πŸ”—
πŸ”—
15.
Solution.
Apply the second shifting theorem:
\begin{equation*} \lap{e^{5t}\cos(4t)} = \frac{s - 5}{(s - 5)^2 + 16}\text{.} \end{equation*}
πŸ”—
πŸ”—
16.
Solution.
\begin{align*} \lap{e^{2t} - t^3 - \sin (5t)} \amp = \lap{ e^{2t} } - \lap{ t^3 } - \lap{ \sin(5t) } \quad \knowl{./knowl/xref/lt-P1.html}{\text{P\(_1\)}}\\ \amp = \frac{1}{s-2} - \frac{3!}{s^{3+1}} - \frac{5}{s^2 + 5^2} \quad \knowl{./knowl/xref/lt-L2-table.html}{\text{L\(_2\)}}, \knowl{./knowl/xref/lt-L3-table.html}{\text{L\(_3\)}}, \knowl{./knowl/xref/lt-L4-table.html}{\text{L\(_4\)}}\\ \amp = \frac{1}{s-2} - \frac{6}{s^4} - \frac{5}{s^2 + 25} \end{align*}
πŸ”—
πŸ”—
17.
Solution.
\begin{align*} F(s) \amp = \lap{ f(t) } \\ \amp = \lap{ e^{-2t}\sin(2t) + t^2 e^{3t} } \\ \amp = \lap{ e^{-2t}\sin(2t)} + \lap{t^2 e^{3t} } \mbox{ (by L9)} \\ \amp = \frac{2}{\big(s- (-2) \big)^2 + 2^2} + \frac{2!}{(s-3)^{2+1}} \mbox{ (by L7 and L6)} \\ \amp = \frac{2}{(s+2)^2 + 4} + \frac{2}{(s-3)^3} \end{align*}
πŸ”—
πŸ”—
18.
Solution.
For this solution, we will need to use property L14 wit \(\ds f(t) = \cos(6t). \) We will need to know the Laplace transform for this function, so let’s do that now.
\begin{align*} F(s) \amp = \lap{ f(t) } \\ \amp = \lap{ \cos(6t) } \\ \amp = \frac{s}{s^2 + 6^2} \mbox{ (by L5)} \\ \amp = \frac{s}{s^2 + 36} \end{align*}
Then we have the following. Note that when we use the quotient rule to take the derivative o \(\ds F(s) \) .
\begin{align*} Q(s) \amp = \lap{ q(t) } \\ \amp = \lap{ 8t\cos(6t) + e^{3t}\sin(4t) } \\ \amp = 8\lap{ t\cos(6t) } + \lap{ e^{3t}\sin(4t) } \mbox{ (by L9)} \\ \amp = 8\lap{ t\cos(6t) } + \frac{4}{(s-3)^2 + 4^2} \mbox{ (by L7)} \\ \amp = 8\lap{ t\cos(6t) } + \frac{4}{(s-3)^2 + 16} \\ \amp = 8\lap{ t\cdot f(t) } + \frac{4}{(s-3)^2 + 16} \\ \amp = 8\cdot (-1)^1 \cdot \frac{d}{ds}F(s) + \frac{4}{(s-3)^2 + 16} \mbox{ (by L14)} \\ \amp = -8 \cdot \frac{d}{ds}\left( \frac{s}{s^2 + 36} \right) + \frac{4}{(s-3)^2 + 16} \\ \amp = -8 \cdot \frac{(s^2 + 36)\cdot 1 - s\cdot (2s + 0)}{(s^2 + 36)^2} + \frac{4}{(s-3)^2 + 16} \\ \amp = -8 \cdot \frac{s^2 + 36 - 2s^2}{(s^2 + 36)^2} + \frac{4}{(s-3)^2 + 16} \\ \amp = -\frac{8(36 - s^2)}{(s^2 + 36)^2} + \frac{4}{(s-3)^2 + 16} \end{align*}
πŸ”—
πŸ”—
19.
Solution.
For this solution, we will need to use property L14 wit \(\ds f(t) = \sin(3t). \) We will need to know the Laplace transform for this function, so let’s do that now.
\begin{align*} F(s) \amp = \lap{ f(t) } \\ \amp = \lap{ \sin(3t) } \\ \amp = \frac{3}{s^2 + 3^2} \mbox{ (by L5)} \\ \amp = \frac{3}{s^2 + 9} \end{align*}
In using L14, we will also need the second derivative o \(\ds F(s), \) so we work to compute that now. Note that we will use the chain rule when we take the derivative o \(\ds (s^2 + 9)^2 \) .
\begin{align*} F'(s) \amp = \frac{d}{ds}\left( \frac{3}{s^2 + 9} \right) \\ \amp = \frac{(s^2 + 9) \cdot 0 - 3 \cdot (2s+0)}{(s^2 + 9)^2} \\ \amp = \frac{-6s}{(s^2 + 9)^2} \\ F''(s) \amp = \frac{(s^2 + 9)^2 \cdot (-6) - (-6s) \cdot 2(s^2 + 9)(2s)}{(s^2 + 9)^4} \\ \amp = \frac{-6(s^2 + 9)^2 + 24s^2(s^2 +9)}{(s^2 + 9)^4} \\ \amp = \frac{6(s^2 + 9)\left[-(s^2 + 9) + 4s^2\right]}{(s^2 + 9)^4} \\ \amp = \frac{6\left[-s^2 - 9 + 4s^2\right]}{(s^2 + 9)^3} \\ \amp = \frac{6\left[3s^2 - 9\right]}{(s^2 + 9)^3} \\ \amp = \frac{18\left[s^2 - 3\right]}{(s^2 + 9)^3} \end{align*}
Then we have the following.
\begin{align*} P(s) \amp = \lap{ t^2 \sin(3t) } \\ \amp = \lap{ t^2 f(t) } \\ \amp = (-1)^2 \cdot F''(s) \mbox{ (by L14)} \\ \amp = 1 \cdot \frac{18\left[s^2 - 3\right]}{(s^2 + 9)^3} \\ \amp = \frac{18\left[s^2 - 3\right]}{(s^2 + 9)^3} \end{align*}
πŸ”—
πŸ”—

Definition of the Laplace Transform

1.
Solution.
Let’s use the definition:
\begin{align*} \lap{11t} \amp = \int_0^{\infty} \left(11t\right) e^{-st}\ dt\\ \amp = 11\lim_{b \to \infty} \ub{\int_0^b te^{-st} dt}_{\large I} \end{align*}
Note: \(s\) cannot be \(0\).
When \(s=0\text{,}\) the integral becomes
\begin{align*} \amp = \lim_{b \to \infty} \int_0^b t\ dt = \lim_{b \to \infty} \frac{t^2}{2}\Big|_0^b = \frac{1}{2} \lim_{b \to \infty} b^2 = \infty \end{align*}
Therefore, we must have \(s\ne 0\) for this integral to be finite.
πŸ”—
πŸ”—
πŸ”—
By Integration by parts:
\begin{align*} \lap{11t} \amp = 11\lim_{b \to \infty} \left[-\frac{b}{s}e^{-sb} - \frac{1}{s^2}e^{-sb} + \frac{1}{s^2}\right] \end{align*}
\begin{align*} \amp = 11 \left[-\lim_{b \to \infty}\frac{b}{s}e^{-sb} - \lim_{b \to \infty}\frac{1}{s^2}e^{-sb} + \lim_{b \to \infty}\frac{1}{s^2}\right]\\ \amp = 11 \left[-\frac{1}{s}\lim_{b \to \infty}be^{-sb} - \frac{1}{s^2}\lim_{b \to \infty}e^{-sb} + \frac{1}{s^2}\right] \end{align*}
As long as \(s \gt 0\text{,}\) the limit becomes:.
If \(s \lt 0\text{,}\) then as \(b \to \infty\text{,}\) we would have \(\ds e^{-sb} \to \infty\) and so
\begin{equation*} \lim_{b \to \infty} \os{\infty}{\os{\uparrow}{\boxed{b}}}\ \us{\infty}{\us{\downarrow}{\boxed{e^{-sb}}}} = \infty. \end{equation*}
This shows the Laplace transform would not exist if \(s \lt 0\text{.}\) Therefore, we must require \(s \gt 0\text{.}\)
πŸ”—
πŸ”—
\begin{align*} lap{11t} \amp = 11 \left[-\frac{1}{s}\cdot 0 - \frac{1}{s^2}\cdot 0 + \frac{1}{s^2}\right]\\ \amp = \frac{11}{s^2}, \quad s \gt 0 \end{align*}
Hence, we have \(\ds \lap{ 11t } = \frac{11}{s^2}\text{,}\) under the condition that \(s>0\text{.}\)
πŸ”—
πŸ”—
2.
Solution.
\begin{align*} \lap{7t + e^{5t}} \amp = \int_0^{\infty}e^{-st}\cdot (7t + e^{5t}) dt \\ \amp = \int_0^{\infty}(e^{-st}\cdot 7t + e^{-st}\cdot e^{5t}) dt \\ \amp = \int_0^{\infty}e^{-st}\cdot 7t dt + \int_0^{\infty}(e^{-st}\cdot e^{5t} dt \\ \amp = \lap{ 7t } + \lap{ e^{5t} } \\ \amp = \frac{7}{s^2} + \frac{1}{s-5} \end{align*}
πŸ”—
πŸ”—
3.
Solution.
\begin{align*} \amp = \int_0^{\infty}e^{-st}(2.3 t^2) dt \\ \amp = 2.3\lim_{b \to \infty} \int_0^b e^{-st}\cdot t^2 dt \\ \amp \quad\vdots \end{align*}
πŸ”—
πŸ”—
4.
Solution.
To compute this, we use the definition of the Laplace transform.
\begin{align*} \lap{-40e^{3t}} \amp = \int_0^{\infty} e^{-st} \cdot (-40e^{3t}) \ dt \\ \amp = -40 \int_0^{\infty} e^{-st+3t} \ dt \\ \amp = -40 \int_0^{\infty} e^{-(s-3)t} \ dt \\ \amp = -40 \left[ \frac{1}{s-3} e^{-(s-3)t} \right]_0^{\infty} \\ \amp = -40 \left( \lim_{t \to \infty} \frac{1}{s-3} e^{-(s-3)t} - \frac{1}{s-3} \right) \\ \amp = -40 \left( 0 - \frac{1}{s-3} \right) \\ \amp = \frac{40}{s-3} \end{align*}
πŸ”—
πŸ”—
5.
Solution.
By the above definition, we have
\begin{align*} \lap{ 15 } \amp = \int_0^{\infty} \left(15\right)e^{-st}dt \qquad \text{(improper integral)} \\ \amp = \lim_{b \to \infty}\int_0^b 15 e^{-st}dt \\ \amp = 15 \lim_{b \to \infty}\int_0^b e^{-st}dt \qquad (s\text{ is constant here}) \end{align*}
At this point, we need to assume that \(s\ne0\text{.}\).
πŸ”—
Next, we integrate using the substitution \(u=-st\text{.}\)
\begin{align*} \amp = 15\lim_{b \to \infty} -\frac{1}{s}e^{-st}\Big|_{t=0}^{t=b}\\ \amp = 15\lim_{b \to \infty} -\frac{1}{s}\left[ e^{-sb} - e^{-s\cdot 0} \right]\\ \amp = 15 \cdot -\frac{1}{s} \lim_{b \to \infty} \Big[ e^{-sb} - 1 \Big]\\ \amp = -\frac{15}{s} \Big[ \lim_{b \to \infty} e^{-sb} - \us{=\ 1}{\ub{\lim_{b \to \infty} 1}} \Big] \qquad(s > 0)\\ \amp = -\frac{15}{s} \left[0 - 1\right] = \frac{15}{s} \end{align*}
πŸ”—
πŸ”—
6.
Solution.
We use the definition of Laplace transform to get us started.
\begin{align*} \lap{ e^{7t} } \amp = \int_0^{\infty} \left(e^{7t}\right)e^{-st}\ dt \\ \amp = \lim_{b \to \infty}\int_0^b e^{-st + 7t}\ dt \\ \amp = \lim_{b \to \infty}\int_0^b e^{(7-s)t}\ dt \end{align*}
As before, we need restrict some values of \(s\) in order for this improper integral to exist . In this case, we will need \(7 - s\text{,}\) in the exponent, to be non-zero and negative. That is, we need
\begin{equation*} 7 - s \lt 0 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}7 \lt s. \end{equation*}
\begin{align*} \amp = \lim_{b \to \infty}\int_0^b e^{(7-s)t}dt \\ \amp = \lim_{b \to \infty} \frac{1}{7-s}e^{(7-s)t}\Bigg|_0^b\\ \amp = \frac{1}{7-s}\lim_{b \to \infty} \left[ e^{(7-s)b} - e^{(7-s)\cdot 0}\right]\\ \amp = \frac{1}{7-s}\lim_{b \to \infty} \left[ e^{(7-s)b} - 1\right]\\ \amp = \frac{1}{7-s} \left[\lim_{b \to \infty}e^{(7-s)b} - \lim_{b \to \infty}1\right]\\ \amp = \frac{1}{7-s} \left[ 0 - 1\right]\\ \amp = \frac{-1}{7-s}\\ \amp = \frac{1}{s - 7},\quad s \gt 7 \end{align*}
Thus,
\begin{equation*} \ds \lap{e^{7t}} = \frac{1}{s - 7}\text{.} \end{equation*}
πŸ”—
πŸ”—
7.
Solution.
We start by applying the definition of the Laplace transform:
\begin{equation*} \lap{ \cos(3t)} = \int_0^{\infty} e^{-st} \cdot \cos(3t)\ dt\text{.} \end{equation*}
Rather than directly integrating, we will use a modified Euler’s Formula to express cosine in terms of \(e\)
\begin{equation*} \cos(3t) = \frac12\left(e^{3it} + e^{-3it}\right). \end{equation*}
Substituting this into the integral gives:
\begin{align*} \lap{ \cos(3t)} \amp = \frac12 \int_0^{\infty} e^{-st} \frac12\left(e^{3it} + e^{-3it}\right)\ dt\\ \amp = \frac12\left[\int_0^{\infty} e^{-st}\cdot e^{3it}\ dt + \int_0^{\infty} e^{-st}\cdot e^{-3it}\ dt\right]\\ \amp = \frac{1}{2} \left[\lap{e^{3it}} + \lap{e^{-3it}}\right]\\ \amp = \frac{1}{2} \left[\frac{1}{s - 3i} + \frac{1}{s + 3i}\right] \qquad (\text{by } \knowl{./knowl/xref/lt-L2-table.html}{\text{L\(_2\)}})\\ \amp = \frac{1}{2} \left[\frac{s + 3i + s - 3i}{(s - 3i)(s + 3i)}\right]\\ \amp = \frac{s}{s^2 + 9}. \end{align*}
Therefore, the Laplace transform of \(\cos(3t)\) is:
\begin{equation*} \lap{\cos(3t)} = \frac{s}{s^2 + 9}. \end{equation*}
πŸ”—
πŸ”—
8.
Solution.
As with cosine, we begin with the definition of the Laplace transform,
\begin{equation*} \lap{ \sin(-4t)} = \int_0^{\infty} e^{-st} \cdot \sin(-4t)\ dt \end{equation*}
and rewrite sine using Euler’s formula,
\begin{equation*} \sin(-4t) = \frac{e^{-4it} - e^{4it}}{2i}. \end{equation*}
Substituting this into the integral, we get:
\begin{align*} \lap{ \sin(-4t)} =\amp \frac{1}{2i} \int_0^{\infty} e^{-st} \left(e^{-4it} - e^{4it}\right)\, dt\\ =\amp \frac{1}{2i} \left[\int_0^{\infty} e^{-(s + 4i)t}\, dt - \int_0^{\infty} e^{-(s - 4i)t}\, dt\right]\\ =\amp \frac{1}{2i} \left[\lap{e^{-4it}} - \lap{e^{4it}}\right].\\ =\amp \frac{1}{2i} \left[\frac{1}{s + 4i} - \frac{1}{s - 4i}\right] \qquad (\text{by } \knowl{./knowl/xref/lt-L2-table.html}{\text{L\(_2\)}})\\ =\amp \frac{1}{2i} \left[\frac{s - 4i - (s + 4i)}{(s + 4i)(s - 4i)}\right]\\ =\amp \frac{4}{s^2 + 16}. \end{align*}
Thus, the Laplace transform of \(\sin(-4t)\) is:
\begin{equation*} \lap{\sin(-4t)} = \frac{4}{s^2 + 16}. \end{equation*}
πŸ”—
πŸ”—

10 Applying the Laplace Method
Chapter 10 Exercises

πŸ—οΈ Warm-ups & Drills

1. Forward Transform the Equation.
Solution.
Applying the Laplace Transform to both sides:
\begin{equation*} \lap{y''} - 4\lap{y'} + 6\lap{y} = \lap{e^{2t}} \end{equation*}
\begin{gather*} s^2Y(s) - s\cdot y(0) - y'(0) - 4(sY(s) - y(0)) + 6Y(s) = \dfrac{1}{s-2}\\ s^2Y(s) - s - 4(sY(s) - 1) + 6Y(s) = \dfrac{1}{s-2}\\ (s^2 - 4s + 6)Y(s) = \dfrac{1}{s-2} + s + 4 \end{gather*}
πŸ”—
πŸ”—
10.
Solution.
The \(s\) in the denominator tells us that we need \(L_1\text{.}\) Before we do, let’s factor out \(12\text{:}\)
\begin{equation*} \ds \ilap{\dfrac{12}{s}} = 12 \cdot \ilap{\dfrac{1}{s}} = 12 \cdot 1 = 12\text{.} \end{equation*}
πŸ”—
πŸ”—
11.
Solution.
Since the denominator has the form, \(s^2 + \text{number}\text{,}\) and there is no \(s\) in the numerator, we should apply \(L_4\text{.}\) As before, it is helpful to first factor out the constant \(5\text{,}\)
\begin{equation*} \ds \ilap{\dfrac{5}{s^2 + 4}} = 5\ \ilap{\dfrac{1}{s^2 + 4}} \quad \leftarrow \knowl{./knowl/xref/lt-L4-table.html}{L_4} (b=2)\text{.} \end{equation*}
According to \(L_4\text{,}\) we are missing \(2\) in the numerator. Let’s put it there by multiplying by \(2/2\text{,}\) like so
\begin{align*} = 5\ \ilap{\dfrac{2}{2}\dfrac{1}{s^2 + 4}} \amp = 5\ \ilap{\dfrac{1}{2}\dfrac{2}{s^2 + 4}}\\ \amp = \dfrac{5}{2} \ilap{\dfrac{2}{s^2 + 4}}\\ \amp = \dfrac{5}{2} \sin(2t) \end{align*}
πŸ”—
πŸ”—
12.
Solution.
This denominator has the form \(s^{\text{power}}\text{,}\) which matches \(L_3\) with \(n = 3\text{.}\)
\begin{equation*} \ds \ilap{\dfrac{17}{s^4}} = 17\ \ilap{\dfrac{1}{s^4}} \quad \leftarrow \knowl{./knowl/xref/lt-L3-table.html}{L_3} (n=3)\text{.} \end{equation*}
In this case, the numerator is missing a \(3!\text{.}\) We can introduce it by multiplying by \(3!/3!\text{,}\) like so
\begin{gather*} = 17\ \ilap{\dfrac{3!}{3!}\dfrac{1}{s^4}} = \dfrac{17}{3!} \ilap{\dfrac{3!}{s^4}} = \dfrac{17}{6} t^3 \end{gather*}
πŸ”—
πŸ”—
13.
Solution.
\begin{equation*} \knowl{./knowl/xref/lt-L5-table.html}{L_5}, b=5 \end{equation*}
\begin{equation*} \ilap{\dfrac{7s}{s^2 + 25}} = 7\ \ilap{\dfrac{s}{s^2 + 25}} = 7\ \cos(5t) \end{equation*}
πŸ”—
πŸ”—
14.
Solution.
The form of this denominator is \(\ul{(s \pm \text{shift})^{2} + \text{number}}\) and has no \(s\) in the numerator. Therefore,
\begin{align*} \ilap{\dfrac{10}{(s - 3)^2 + 11}} \amp\ \quad \knowl{./knowl/xref/lt-L7-table.html}{\text{\(L_7\)}} (a=3,\ b=\sqrt{11})\\ \amp = 10\ \ilap{\dfrac{\sqrt{11}}{\sqrt{11}}\dfrac{1}{(s - 3)^2 + 11}}\\ \amp = \dfrac{10}{\sqrt{11}} \ilap{\dfrac{\sqrt{11}}{(s - 3)^2 + 11}}\\ \amp = \dfrac{10}{\sqrt{11}} e^{3t} \sin(\sqrt{11}t) \end{align*}
πŸ”—
πŸ”—
15.
Solution.
The denominator, \(\ul{(s \pm \text{shift})^{\text{power}}}\text{,}\) indicates an \(e^{at}\) and a \(t^n\) term. Therefore,
\begin{align*} \ilap{\dfrac{2}{(s+7)^5}} \amp\ \quad \knowl{./knowl/xref/lt-L6-table.html}{\text{\(L_6\)}} (a=-7, n=4)\\ \amp = 2\ \ilap{\dfrac{4!}{4!}\dfrac{1}{(s+7)^5}}\\ \amp = \dfrac{2}{4!} \ilap{\dfrac{4!}{(s+7)^5}}\\ \amp = \dfrac{1}{12} t^4 e^{-7t} \end{align*}
πŸ”—
πŸ”—
16.
Solution.
The \(s\) in the denominator tells us that we need \(L_1\text{.}\) Before we do, let’s factor out \(12\text{:}\)
\begin{equation*} \ds \ilap{\dfrac{12}{s}} = 12 \cdot \ilap{\dfrac{1}{s}} = 12 \cdot 1 = 12\text{.} \end{equation*}
πŸ”—
\begin{equation*} \knowl{./knowl/xref/lt-L5-table.html}{L_5}, b=5 \end{equation*}
\begin{equation*} \ilap{\dfrac{7s}{s^2 + 25}} = 7\ \ilap{\dfrac{s}{s^2 + 25}} = 7\ \cos(5t) \end{equation*}
πŸ”—
πŸ”—
17.
Solution.
The form of this denominator is \(\ul{(s \pm \text{shift})^{2} + \text{number}}\) and has no \(s\) in the numerator. Therefore,
\begin{align*} \ilap{\dfrac{10}{(s - 3)^2 + 11}} \amp\ \quad \knowl{./knowl/xref/lt-L7-table.html}{\text{\(L_7\)}} (a=3,\ b=\sqrt{11})\\ \amp = 10\ \ilap{\dfrac{\sqrt{11}}{\sqrt{11}}\dfrac{1}{(s - 3)^2 + 11}}\\ \amp = \dfrac{10}{\sqrt{11}} \ilap{\dfrac{\sqrt{11}}{(s - 3)^2 + 11}}\\ \amp = \dfrac{10}{\sqrt{11}} e^{3t} \sin(\sqrt{11}t) \end{align*}
πŸ”—
The denominator, \(\ul{(s \pm \text{shift})^{\text{power}}}\text{,}\) indicates an \(e^{at}\) and a \(t^n\) term. Therefore,
\begin{align*} \ilap{\dfrac{2}{(s+7)^5}} \amp\ \quad \knowl{./knowl/xref/lt-L6-table.html}{\text{\(L_6\)}} (a=-7, n=4)\\ \amp = 2\ \ilap{\dfrac{4!}{4!}\dfrac{1}{(s+7)^5}}\\ \amp = \dfrac{2}{4!} \ilap{\dfrac{4!}{(s+7)^5}}\\ \amp = \dfrac{1}{12} t^4 e^{-7t} \end{align*}
πŸ”—
πŸ”—
18.
Solution.
Rewrite \(s + 5\) as \((s + 3) + 2\) to get:
πŸ”—
\begin{equation*} Y(s) = \dfrac{s + 3}{(s + 3)^2 + 16} + \dfrac{2}{(s + 3)^2 + 16} \end{equation*}
The first term matches L\(_8\) with \(a = -3\text{,}\) \(b = 4\text{,}\) and the second term matches L\(_7\) after multiplying by \(\sfrac{4}{4}\text{:}\)
πŸ”—
\begin{equation*} \ilap{Y(s)} = e^{-3t}\cos(4t) + \dfrac{2}{4}e^{-3t}\sin(4t) = e^{-3t}\cos(4t) + \dfrac{1}{2}e^{-3t}\sin(4t) \end{equation*}
πŸ”—

πŸ—οΈ Drill: Preparing & Invert

10.
Solution.
This expression matches the form \(\ds \dfrac{b}{s^2 + b^2} \) with \(\ds b = 4 \text{,}\) so the inverse Laplace transform is \(\ds 5 \cdot \sin(4t) \text{.}\)
πŸ”—
πŸ”—
12.
Solution.
This expression can be split into two parts: \(\ds \dfrac{2(s + 3)}{(s + 3)^2 + 4} \) and \(\ds \dfrac{7}{(s + 3)^2 + 4} \text{.}\) The first part corresponds to \(\ds e^{-3t} \cdot \cos(2t) \) and the second part corresponds to \(\ds e^{-3t} \cdot \sin(2t) \text{,}\) so the inverse Laplace transform is \(\ds 2e^{-3t} \cos(2t) + 7e^{-3t} \sin(2t) \text{.}\)
πŸ”—
πŸ”—
16.
Solution.
The discriminant of the denominator \(s^2 - 6s + 14\) is:
\begin{equation*} b^2 - 4ac = (-6)^2 - 4(1)(14) = 36 - 56 = -20\text{,} \end{equation*}
indicating that completing the square is necessary:
\begin{align*} s^2 - 6s + 14 = (s^2 - 6s) + 14 =\amp (\ob{s^2 - 6s + 9}^{(s - 3)^2} - 9) + 14 \\ =\amp (s - 3)^2 + 5 \text{.} \end{align*}
Rewriting \(K(s)\) as:
\begin{equation*} \dfrac{11}{s^2 - 6s + 14} = \ub{\dfrac{11}{(s - 3)^2 + 5}}_{\large \text{Match with } L_7} = \dfrac{11}{\sqrt{5}}\ub{\dfrac{\sqrt{5}}{(s - 3)^2 + 5}}_{\large \knowl{./knowl/xref/lt-L7-table.html}{\(L_7\)} (a = 3, b = \sqrt{5})}\text{.} \end{equation*}
Therefore,
\begin{align*} k(t) = \ilap{K(s)} \amp = \dfrac{11}{\sqrt{5}}e^{3t}\sin(\sqrt{5}t)\text{.} \end{align*}
πŸ”—
πŸ”—
17.
Solution.
Completing the square for the denominator of \(P(s)\) gives:
\begin{equation*} \dfrac{s+3}{s^2 + 2s + 10} = \dfrac{s+3}{(s + 1)^2 + 9}\text{.} \end{equation*}
However, the numerator \(s + 3\) does not match \(s + 1\text{.}\) To resolve this, we rewrite \(3\) as \(1 + 2\) and group terms:
\begin{align*} \dfrac{s+3}{s^2 + 2s + 10} \amp = \dfrac{s + 1 + 2}{(s + 1)^2 + 9}\\ \amp = \dfrac{s+1}{(s + 1)^2 + 9} + \dfrac{2}{(s + 1)^2 + 9}\\ \amp = \ub{\dfrac{s+1}{(s + 1)^2 + 9}}_{\large \knowl{./knowl/xref/lt-L8-table.html}{\text{L\(_8\)}}\ (a=-1, b=3)} + \dfrac{2}{3}\ub{\dfrac{3}{(s + 1)^2 + 9}}_{\large \knowl{./knowl/xref/lt-L7-table.html}{\text{L\(_7\)}}\ (a=-1, b=3)}\text{,} \end{align*}
Now, apply the inverse Laplace transform:
\begin{align*} p(t) \amp = \ilap{P(s)}\\ \amp = \ilap{ \dfrac{s+1}{(s + 1)^2 + 9} } + \dfrac{2}{3}\ilap{ \dfrac{3}{(s + 1)^2 + 9} }\\ \amp = e^{-t}\cos(3t) + \dfrac{2}{3}e^{-t}\sin(3t) \end{align*}
πŸ”—
πŸ”—
18.
Solution.
First, factor the quadratic denominator:
\begin{equation*} s^2 - s - 6 = (s + 2)(s - 3). \end{equation*}
Now decompose \(P(s)\) into partial fractions:
\begin{equation*} \dfrac{s}{(s + 2)(s - 3)} = \dfrac{A}{s + 2} + \dfrac{B}{s - 3}. \end{equation*}
Multiply both sides by \((s + 2)(s - 3)\text{,}\)
\begin{equation*} s = A(s - 3) + B(s + 2). \end{equation*}
and solve for \(A\) and \(B\) by selecting convenient values for \(s\text{:}\)
\begin{align*} s = -2: \amp\\ \\ s = 3: \amp\\ \end{align*}
πŸ”—
\begin{align*} -2 \amp = A(-2 - 3) + B(-2 + 2)\\ -2 \amp = A(-5) + B(0) \quad\implies\quad A=2/5\\ 3 \amp = A(3 - 3) + B(3 + 2)\\ 3 \amp = A(0) + B(5) \quad\implies\quad B=3/5 \end{align*}
πŸ”—
The partial fraction decomposition is:
\begin{equation*} \dfrac{s}{(s + 2)(s - 3)} = \dfrac{2/5}{s + 2} + \dfrac{3/5}{s - 3}. \end{equation*}
Therefore:
\begin{equation*} \ilap{P(s)} = \dfrac25 e^{-2t} + \dfrac35 e^{3t}. \end{equation*}
πŸ”—
πŸ”—
19.
Solution.
\(\ds H(s) = \dfrac{5}{s-6} - \dfrac{6s}{s^2 + 9} + \dfrac{3}{2s^2 + 8s + 10} \)
πŸ”—
πŸ”—
20.
Solution.
Decompose into partial fractions:
\begin{equation*} Y(s) = \dfrac{A}{s + 1} + \dfrac{B}{s + 3}\text{.} \end{equation*}
Find \(A\) and \(B\text{:}\)
\begin{equation*} A = 1, B = \dfrac{1}{2}\text{.} \end{equation*}
Use the inverse Laplace transform:
\begin{equation*} y(t) = e^{-t} + \dfrac{1}{2}e^{-3t}\text{.} \end{equation*}
πŸ”—
πŸ”—
21.
Solution.
Decompose into partial fractions:
\begin{equation*} Y(s) = \dfrac{A}{s} + \dfrac{B}{s + 4}\text{.} \end{equation*}
Find \(A\) and \(B\text{:}\)
\begin{equation*} A = \dfrac{5}{4}, B = -\dfrac{5}{4}\text{.} \end{equation*}
Use the inverse Laplace transform:
\begin{equation*} f(t) = \dfrac{5}{4}(1 - e^{-4t})\text{.} \end{equation*}
πŸ”—
πŸ”—
22.
Solution.
As in the previous examples, the denominator is a second-degree polynomial; therefore it is sensible for us to begin by completing the square in the denominator as we did in the previous two examples.
\begin{align*} \dfrac{s+9}{s^2 - 2s - 3} \amp = \dfrac{s+9}{(s^2 - 2s) - 3}\\ \amp = \dfrac{s+9}{(s^2 - 2s + 1) - 3 - 1}\\ \amp = \dfrac{s+9}{(s-1)^2 - 4}\\ \amp = \dfrac{s+9}{(s-1)^2 - 2^2}. \end{align*}
πŸ”—
Take a careful look at the denominator here. It’s really close to matching \(L7\) or \(L8\), but it is not a match because of the negative sign in front of the \(2^2.\) We need to change course when this happens. Another algebraic manipulation that we might consider is a partial fraction decomposition.
πŸ“
πŸ”—
We revert to the original expression, but this time, instead of completing the square, we factor the denominator.
\begin{equation*} \dfrac{s+9}{s^2 - 2s - 3} = \dfrac{s+9}{(s-3)(s+1)}. \end{equation*}
πŸ”—
Since each of the factors in the denominator is a distinct linear factor, we know that the form of the partial fraction decomposition is
\begin{equation*} \dfrac{s+9}{(s-3)(s+1)} = \dfrac{A}{s-3} + \dfrac{B}{s+1}. \end{equation*}
πŸ”—
Our next goal is to determine the coefficients \(A\) and \(B\) in this equations. There are multiple ways to achieve this and we demonstrate just one here. We multiply both sides of the equation by the least common denominator, \((s-3)(s+1)\text{,}\) and then expand and collect like-terms, as shown.
\begin{align*} (s-3)(s+1) \cdot \dfrac{s+9}{(s-3)(s+1)} \amp = \dfrac{A}{s-3}\cdot (s-3)(s+1) + \dfrac{B}{s+1} \cdot (s-3)(s+1)\\ s+9 \amp = A(s+1) + B(s-3)\\ s+9 \amp = As + A + Bs - 3B\\ s+9 \amp = (As + Bs) + (A - 3B)\\ 1s+9 \amp = (A+B)s + (A - 3B). \end{align*}
πŸ”—
At this point, we have a polynomial on the left hand side and a polynomial on the right hand side. The only way these can be equal to each other is if the corresponding coefficients are equal. That is, the coefficient on \(s\) on the left hand side is 1, while the coefficient on \(s\) on the right hand side of the equation is \(A+B\text{.}\) Since the polynomials are equal, we know that these are equal. That is, \(A + B = 1.\) Similarly, if we equate the constants, we have \(A - 3B = 9.\) Thus, we have the following system of two linear equations in terms of two unknown variables, \(A\) and \(B\text{.}\)
\begin{align*} A + B \amp = 1\\ A - 3B \amp = 9. \end{align*}
πŸ”—
There are many ways to solve such an equation, and you are encouraged to choose the solution technique you like the most. Here we will solve the first equation for \(A\text{,}\) and then substitute into the second equation,
\begin{align*} A + B \amp = 1 \amp A - 3B \amp = 9\\ A \amp = 1 - B \amp \amp\\ \amp \amp (1 - B) - 3B \amp = 9\\ \amp \amp 1 - 4B \amp = 9\\ \amp \amp -4B \amp = 8\\ \amp \amp B \amp = -2\\ A \amp = 1 - (-2) \amp \amp\\ \amp = 3 \amp \amp \end{align*}
hence we have
\begin{equation*} \dfrac{s+9}{(s-3)(s+1)} = \dfrac{3}{s-3} + \dfrac{-2}{s+1}. \end{equation*}
πŸ”—
Remember that our goal is to take the inverse Laplace transform. Our algebraic manipulation was helpful because we took a more complex expression and rewrote it as two simpler fractions. We can now use \(L2\) to find the inverse Laplace transform as follows.
\begin{align*} \ilap{ \dfrac{s+9}{s^2 - 2s - 3} } \amp = \ilap{ \dfrac{3}{s-3} + \dfrac{-2}{s+1} }\\ \amp = 3 \ilap{ \dfrac{1}{s-3} } - 2 \ilap{ \dfrac{1}{s+1} } \knowl{./knowl/xref/lt-R1-table.html}{\text{R\(_1\)}} \\ \amp = 3e^{3t} - 2e^{-t} \quad \text{ by } \knowl{./knowl/xref/lt-L2-table.html}{\text{L2}} \end{align*}
πŸ”—
πŸ”—
23.
Solution.
πŸ“
As we mentioned earlier, we’ll try to make the denominator match first. Since both of the forms we’re trying to match are of the form \((s-a)^2 + b^2,\) we will complete the square first:
\begin{equation*} s^2 - 4s + 29 = (s-2)^2 + 25. \end{equation*}
(If you still don’t remember how to complete the square, look up that primer and do the previous exercises in that section above.)
πŸ”—
Let’s rewrite the given expression as follows.
\begin{align*} \dfrac{s-6}{s^2 - 4s + 29} \amp = \dfrac{s-6}{(s-2)^2 + 25} \\ \amp = \dfrac{s-6}{(s-2)^2 + 5^2} \end{align*}
πŸ”—
We’ve got the denominator in exactly the right form, it looks just like \((s-a)^2 + b^2,\) with \(a = 2\) and \(b = 5\text{.}\) As in the previous section, once we’ve gotten the denominator in shape, we turn our attention to the numerator. If we look back at the two forms we are trying to match, we see that our expression has an \(s\) in the numerator, so it’s more like \(\ds \dfrac{s-a}{(s-a)^2 + b^2}\text{.}\) It would be exactly right if we had \(s-a\) in the numerator, which in this case would be \(s-2\text{.}\)
πŸ”—
What we do have in the numerator is \(s-6\text{;}\) and we would like it to be \(s-2,\) which means if we added 4, we’d have exactly the right thing. If we want to add 4, we’ll need to compensate by also subtracting 4, like this:
\begin{align*} \dfrac{s-6}{(s-2)^2 + 5^2}\amp = \dfrac{s-6+4-4}{(s-2)^2 + 5^2}, \\ \amp = \dfrac{(s-6+4)-4}{(s-2)^2 + 5^2}, \\ \amp = \dfrac{(s-2) - 4}{(s-2)^2 + 5^2}. \end{align*}
πŸ”—
Great! Now we can split this single fraction into two separate fractions:
\begin{equation*} \dfrac{(s-2) - 4}{(s-2)^2 + 5^2} = \underbrace{\dfrac{s-2}{(s-2)^2 + 5^2}}_{match!} - \dfrac{4}{(s-2)^2 + 5^2} \end{equation*}
We’re almost there! The first fraction is a perfect match for the form \(\ds\dfrac{s-a}{(s-a)^2 + b^2}\) (with \(a = 2\) and \(b = 5\)); but we still have another expression that is not yet a match. The remaining fraction looks like it could eventually match the form \(\ds\dfrac{b}{(s-a)^2 + b^2}\text{.}\) We would need to have a 5 in the numerator, and we currently have a 4. But we can fix that as we did in the previous section:
\begin{align*} \dfrac{4}{(s-2)^2 + 5^2} \amp = 4\cdot\dfrac{1}{(s-2)^2 + 5^2} \\ \amp = 4\cdot\dfrac{5}{5}\cdot \dfrac{1}{(s-2)^2 + 5^2} \\ \amp = 4\cdot\dfrac{1}{5}\cdot \dfrac{5}{(s-2)^2 + 5^2} \\ \amp = \dfrac{4}{5}\cdot \dfrac{5}{(s-2)^2 + 5^2} \end{align*}
Now let’s put it all together. Here’s everything we did:
\begin{align*} \dfrac{s-6}{s^2 - 4s + 9} \amp = \dfrac{s-6}{(s-2)^2 + 25} \mbox{(we completed the square)}\\ \amp = \dfrac{s-6}{(s-2)^2 + 5^2} \mbox{(we rewrote the denominator)}\\ \amp = \dfrac{s-6+4-4}{(s-2)^2 + 5^2} \mbox{(we added and subtracted 4 in the numerator...)}\\ \amp = \dfrac{(s-2) - 4}{(s-2)^2 + 5^2} \mbox{(...so that we could make the numerator match part of the denom)}\\ \amp = \dfrac{s-2}{(s-2)^2 + 5^2} - \dfrac{4}{(s-2)^2 + 5^2} \mbox{(split into fractions)}\\ \amp = \dfrac{s-2}{(s-2)^2 + 5^2} - 4\cdot\dfrac{5}{5}\cdot \dfrac{1}{(s-2)^2 + 5^2} \mbox{(work on the second fraction...)}\\ \amp = \dfrac{s-2}{(s-2)^2 + 5^2} - \dfrac{4}{5}\cdot \dfrac{5}{(s-2)^2 + 5^2}\mbox{(...so now it is also in form)} \end{align*}
πŸ”—
As mentioned before, being able to use appropriate algebra to "match" forms is really important when we work with Laplace Transforms. Since it’s really just algebra, now is a great time to practice that skill, so when we are in the middle of studying Laplace Transforms, you can just focus on the "new" stuff.
πŸ”—
πŸ”—
25.
Solution.
\(\ds G(s) = \dfrac{2s-19}{s^2 - 4s+13} + \dfrac{5}{s-1} \)
πŸ”—
πŸ”—
27.
Solution.
\(\ds X(s) = \dfrac{90s^2 - 195s + 30}{s^3 - 7s^2 + 6s} \)
πŸ”—
πŸ”—
28.
Solution.
First, factor the denominator as \(\ds (s + 1)^2 \text{.}\) Then, decompose the expression into \(\ds \dfrac{A}{s + 1} + \dfrac{B}{(s + 1)^2} \text{.}\) Solving for \(\ds A \) and \(\ds B \text{,}\) you get the inverse Laplace transform as \(\ds 3e^{-t} + 4te^{-t} \text{.}\)
πŸ”—
πŸ”—
29.
Solution.
\(\ds I(s) = \dfrac{5s^2 + 34s + 53}{(s+3)^2(s+1)} \)
πŸ”—
πŸ”—
30.
Solution.
\(\ds F(s) = \dfrac{7s^3 - 2s^2 - 3s + 6}{s^3(s-2)} \)
πŸ”—
πŸ”—
31.
Solution.
Note that as the quadratic term in the denominator does not factor, the denominator contains an irreducible quadratic factor and a repeated linear factor. We’ll proceed by simplifying this complicated fraction with a Partial Fraction Decomposition of the form
\begin{equation*} \dfrac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2} = \dfrac{As + B}{s^2 - 6s + 25} + \dfrac{C}{s+1} + \dfrac{D}{(s+1)^2}. \end{equation*}
πŸ”—
You may also consider using technology to find a partial fraction decomposition. You should get
\begin{equation*} \dfrac{4s^3 - 13s^2 + 74s + 27}{(s^2 - 6s + 25)(s+1)^2} = \dfrac{s + 2}{s^2 - 6s + 25} + \dfrac{3}{s+1} + \dfrac{-2}{(s+1)^2}. \end{equation*}
Click here for the details.
\begin{align*} 4s^3 \amp - 13s^2 + 74s + 27\\ \amp = (As + B)(s+1)^2 + C(s+1)(s^2 - 6s + 25) + D(s^2 - 6s + 25)\\ \amp = (As + B)(s^2 + 2s + 1) + (Cs+C)(s^2 - 6s + 25) + Ds^2 - 6Ds + 25D\\ \amp = As^3 + 2As^2 + As + Bs^2 + 2Bs + B + Cs^3 - 6Cs^2 + 25Cs + Cs^2 - 6Cs + 25C + Ds^2 - 6Ds + 25D\\ \amp = (A + C)s^3 + (2A+ B - 6C + C + D)s^2 + (A+ 2B + 25C- 6C- 6D)s + (B + 25C + 25D)\\ \amp = (A + C)s^3 + (2A+ B - 5C + D)s^2 + (A+ 2B + 19C- 6D)s + (B + 25C + 25D) \end{align*}
πŸ”—
Equating coefficients gives us four equations in four unknowns.
\begin{align*} A+C \amp = 4 \amp 2A+B-5C+D \amp = -13 \amp A + 2B + 19C - 6D \amp = 74 \amp B+ 25C + 25D \amp = 27\\ A \amp = 4-C\amp \amp \amp \amp \amp\amp\\ \amp \amp 2(4 - C)+B-5C+D \amp = -13 \amp (4-C)+2B+19C-6D \amp = 74 \amp\amp\\ \amp \amp B - 7C + D \amp = -21 \amp 2B+18C-6D \amp = 70 \amp\amp\\ \amp \amp B \amp = 7C - D - 21 \amp \amp \amp\amp\\ \amp \amp \amp \amp 2(7C - D - 21)+18C-6D \amp = 70 \amp\\ (7C - D - 21) + 25C + 25D \amp = 27\\ \amp \amp \amp \amp 32C-8D \amp = 112 \amp\\ 32C + 24D \amp = 48\\ \amp \amp \amp \amp 32C \amp = 8D + 112 \amp\amp\\ \amp \amp \amp \amp \amp \amp\\ (8D + 112) + 24D \amp = 48\\ \amp \amp \amp \amp \amp \amp\\ 32D \amp = -64\\ \amp \amp \amp \amp \amp \amp\\ D \amp = -2\\ \amp \amp \amp \amp 32C \amp = 8(-2) + 112 \amp\amp\\ \amp \amp \amp \amp \amp = 96 \amp\amp\\ \amp \amp \amp \amp C \amp = 3 \amp\amp\\ \amp \amp B \amp = 7(3) - (-2) - 21\amp \amp \amp \amp\amp\\ \amp \amp \amp = 2 \amp \amp \amp \amp\amp\\ A \amp = 4 - (3)\amp \amp \amp \amp \amp \amp\amp\\ \amp = 1 \amp \amp \amp \amp \amp \amp\amp \end{align*}
πŸ”—
πŸ”—
πŸ”—
With the partial fraction decomposition in hand, we are prepared to take the inverse Laplace transform, using the same types of algebraic manipulations demonstrated in the previous examples.
\begin{align*} r(t) \amp = \ilap{ R(s) }\\ \amp = \ilap{ \dfrac{s + 2}{s^2 - 6s + 25} + \dfrac{3}{s+1} + \dfrac{-2}{(s+1)^2} } \\ \amp = \ilap{ \dfrac{s + 2}{s^2 - 6s + 25} } + 3\cdot \ilap{ \dfrac{1}{s+1} } - 2\cdot \ilap{\dfrac{1}{(s+1)^2} } \\ \amp = \ilap{ \dfrac{s + 2}{(s-3)^2 + 4^2} } + 3e^{-t} - 2te^{-t}\\ \amp = \ilap{ \dfrac{(s - 3) + 5}{(s-3)^2 + 4^2} } + 3e^{-t} - 2te^{-t}\\ \amp = \ilap{ \dfrac{s - 3}{(s-3)^2 + 4^2} } + 5 \ilap{ \dfrac{1}{(s-3)^2 + 4^2} } + 3e^{-t} - 2te^{-t}\\ \amp = e^{3t}\cos(4t) + \dfrac{5}{4}\cdot \ilap{ \dfrac{4}{(s-3)^2 + 4^2} } + 3e^{-t} - 2te^{-t}\\ \amp = e^{3t}\cos(4t) + \dfrac{5}{4}e^{3t}\sin(4t) + 3e^{-t} - 2te^{-t}. \end{align*}
πŸ”—
πŸ”—

Solving Differential Equations

1.
Solution.
We begin by taking the Laplace transform and substituting in the initial conditions.
\begin{align*} \Big[ s^2X(s) - sx(0) - x'(0) \Big] +6\Big[ sX(s) - x(0) \Big] + 9X(s) \amp = \dfrac{64}{s-5} \mbox{ (by L12,\DL1A and L2)} \\ \Big[ s^2X(s) - s\cdot 3 - 0 \Big] +6\Big[ sX(s) - 3 \Big] + 9X(s) \amp = \dfrac{64}{s-5} \\ s^2X(s) - 3s + 6sX(s) - 18 + 9X(s) \amp = \dfrac{64}{s-5} \\ s^2X(s) + 6sX(s)+9X(s) -3s - 18 \amp = \dfrac{64}{s-5} \end{align*}
Now we aim to solve for\(\ds X(s), \) so we use algebra to rearrange as follows.
\begin{align*} s^2X(s) + 6sX(s)+9X(s) \amp = \dfrac{64}{s-5} + 3s + 18 \\ X(s) \amp = \dfrac{64}{s-5} + (3s + 18)\cdot \dfrac{s-5}{s-5} \\ \amp = \dfrac{64}{s-5} + \dfrac{(3s+18)(s-5)}{s-5} \\ \amp = \dfrac{64}{s-5} + \dfrac{3s^2 +3s- 90}{s-5} \\ \amp = \dfrac{64 + 3s^2 +3s- 90}{s-5} \\ \amp = \dfrac{3s^2 +3s- 26}{s-5} \\ X(s) \amp = \dfrac{3s^2 +3s- 26}{(s-5)(s^2 + 6s + 9)} \\ \amp = \dfrac{3s^2 +3s- 26}{(s-5)(s+3)^2} \end{align*}
We need to do a partial fraction decomposition before we can take the inverse LT.
\begin{align*} \dfrac{3s^2 +3s- 26}{(s-5)(s+3)^2} \amp = \dfrac{A}{s-5} + \dfrac{B}{s+3} + \dfrac{C}{(s+3)^2} \\ 3s^2 +3s - 26 \amp = A(s+3)^2 + B(s-5)(s+3) + C(s-5) \\ \ \amp = A(s^2 + 6s + 9) + B(s^2 - 2s - 15) + Cs - 5C \\ \amp = As^2 + 6As + 9A + Bs^2 - 2Bs - 15B + Cs - 5C \\ \amp = (A + B)s^2 + (6A - 2B + C)s + (9A - 15B - 5C) \end{align*}
Equating the coefficients of like-terms yields, we can solve for\(\ds A \) \(\ds B \) , an \(\ds C \) .
\begin{align*} A+B \amp = 3 \amp 6A - 2B + C \amp = 3 \amp 9A - 15B - 5C \amp = -26 \\ B \amp = 3-A \amp \amp \amp \amp \\ \amp \amp6A - 2(3-A)+C \amp = 3 \amp \amp \\ \amp \amp 6A-6+2A+C \amp = 3 \amp \amp \\ \amp \amp 8A + C \amp = 9 \amp \amp \\ \amp \amp C \amp = 9 - 8A \amp \amp \\ \amp \amp \amp \amp 9A - 15(3-A) - 5(9-8A) \amp = -26 \\ \amp \amp \amp \amp 9A - 45 + 15A - 45 + 40A \amp = -26 \\ \amp \amp \amp \amp 64A \amp = 64 \\ \amp \amp \amp \amp A \amp = 1 \\ \amp \amp C \amp = 9-8(1) \amp \amp \\ \amp \amp \amp = 1 \amp \amp \\ B \amp = 3 - 1 \amp \amp \amp \amp \\ \amp = 2 \amp \amp \amp \amp \end{align*}
Hence we have \(\ds X(s) = \dfrac{1}{s-5} + \dfrac{2}{s+3} + \dfrac{1}{(s+3)^2}. \) Now we need only take the inverse LT to find the solution.
\begin{align*} x(t) \amp = \lap^{-1}\left\{ \dfrac{1}{s-5} + \dfrac{2}{s+3} + \dfrac{1}{(s+3)^2} \right\} \\ \amp = \lap^{-1}\left\{ \dfrac{1}{s-5} \right\} + 2\cdot \lap^{-1}\left\{\dfrac{1}{s+3} \right\} + \lap^{-1}\left\{\dfrac{1}{(s+3)^2} \right\} \\ \amp = e^{5t} + 2e^{-3t} + te^{-3t} \end{align*}
πŸ”—
πŸ”—
2.
Solution.
We begin by taking the Laplace transform and substituting in the initial conditions.
\begin{align*} \Big[ s^2Y(s) - sy(0) - y'(0) \Big] - 2\Big[ sY(s) - y(0) \Big] + 5Y(s) \amp = \dfrac{-8}{s+1} \mbox{ (by L12,\DL1A and L2)} \\ \Big[ s^2Y(s) - s\cdot 2 - 12) \Big] - 2\Big[ sY(s) - 2 \Big] + 5Y(s) \amp = \dfrac{-8}{s+1} \\ s^2Y(s) - 2s - 12 - 2sY(s) +4 + 5Y(s) \amp = \dfrac{-8}{s+1} \\ s^2Y(s) - 2sY(s) + 5Y(s)- 2s - 8 \amp = \dfrac{-8}{s+1} \end{align*}
Now we aim to solve for\(\ds Y(s), \) so we use algebra to rearrange as follows.
\begin{align*} s^2Y(s) - 2sY(s) + 5Y(s) \amp = \dfrac{-8}{s+1} + 2s - 8 \\ (s^2 - 2s + 5)Y(s) \amp = \dfrac{-8}{s+1} + (2s - 8)\cdot \dfrac{s+1}{s+1} \\ (s^2 - 2s + 5)Y(s) \amp = \dfrac{-8}{s+1} + \dfrac{2s^2+10s+8}{s+1} \\ (s^2 - 2s + 5)Y(s) \amp = \dfrac{-8 + 2s^2+10s+8}{s+1} \\ (s^2 - 2s + 5)Y(s) \amp = \dfrac{2s^2+10s}{s+1} \\ Y(s) \amp = \dfrac{2s^2+10s}{(s+1)(s^2 - 2s + 5)} \end{align*}
We need to do a partial fraction decomposition before we can take the inverse LT.
\begin{align*} \dfrac{2s^2+10s}{(s+1)(s^2 - 2s + 5)} \amp = \dfrac{A}{s+1} + \dfrac{Bs+C}{s^2 - 2s + 5} \\ 2s^2 + 10s \amp = A(s^2 - 2s + 5) + (Bs+ C)(s+1) \\ \amp = As^2 - 2As + 5A + Bs^2 + Bs + Cs + C \\ \amp = (A + B)s^2 + (-2A + B + C)s + (5A + C) \end{align*}
Equating the coefficients of like-terms yields, we can solve for\(\ds A \) \(\ds B \) , an \(\ds C \) .
\begin{align*} A+B \amp = 2 \amp -2A+B+C \amp = 10 \amp 5A + C \amp = 0 \\ \amp \amp \amp \amp C \amp = -5A \\ \amp \amp \amp \amp \amp \amp -2A + B + (-5A) \amp = 10 \amp \amp \\ \amp \amp -7A + B \amp = 10 \amp \amp \\ \amp \amp B \amp = 10+7A \amp \amp \\ \amp \amp \amp \amp A + (10+7A) \amp = 2 \amp \amp \amp \amp \\ 8A+10 \amp = 2 \amp \amp \amp \amp \\ 8A \amp = -8 \amp \amp \amp \amp \\ A \amp = -1 \amp \amp \amp \amp \\ \amp \amp B \amp = 3 \amp \amp \\ \amp \amp \amp \amp C \amp = 5 \end{align*}
Hence, we have \begin{equation} Y(s) = \dfrac{-1}{s+1} + \dfrac{3s+5}{s^2 - 2s + 5}. \label{Y_of_s} \end{equation} We need to take the inverse Laplace Transform to fin \(\ds y(t). \) We need to complete the square on the irreducible quadratic term as follows.
\begin{align*} s^2 - 2s + 5 \amp = (s^2 - 2s) + 5 \\ \amp = (s^2 - 2s + 1) - 1 + 5 \\ \amp = (s - 1)^2 + 4 \end{align*}
We need to transform the second term, as follows, so we can use L8 and L7 on the Laplace transform table.
\begin{align*} \dfrac{3s+5}{s^2 - 2s + 5} \amp = \dfrac{3s+5}{(s - 1)^2 + 2^2} \\ \amp = \dfrac{3(s-1)+3+5}{(s - 1)^2 + 2^2} \\ \amp = \dfrac{3(s-1)+8}{(s - 1)^2 + 2^2} \\ \amp = \dfrac{3(s-1)}{(s - 1)^2 + 2^2} + \dfrac{8}{(s - 1)^2 + 2^2} \\ \amp = 3\cdot\dfrac{s-1}{(s - 1)^2 + 2^2} + \dfrac{4\cdot 2}{(s - 1)^2 + 2^2} \\ \amp = 3\cdot\dfrac{s-1}{(s - 1)^2 + 2^2} + 4\cdot\dfrac{2}{(s - 1)^2 + 2^2} \end{align*}
We are now prepared to take the inverse LT of equation \ref{Y_of_s} to get the solution to the IVP.
\begin{align*} y(t) \amp = \lap^{-1}\left\{ Y(s) \right\} \\ \amp = \lap^{-1}\left\{ \dfrac{-1}{s+1} + \dfrac{3s+5}{s^2 - 2s + 5} \right\} \\ \amp = \lap^{-1}\left\{ -1\cdot \dfrac{1}{s+1} + 3\cdot\dfrac{s-1}{(s - 1)^2 + 2^2} + 4\cdot\dfrac{2}{(s - 1)^2 + 2^2} \right\} \\ \amp = -1\cdot \lap^{-1}\left\{ \dfrac{1}{s+1}\right\} + 3\cdot\lap^{-1}\left\{\dfrac{s-1}{(s - 1)^2 + 2^2} \right\} + 4\cdot\lap^{-1}\left\{\dfrac{2}{(s - 1)^2 + 2^2} \right\} \\ \amp = -e^{-t} + 3e^t\cos(2t) + 4e^t\sin(2t) \end{align*}
πŸ”—
πŸ”—
3.
Solution.
Apply the Laplace transform to both sides of the equation:
\begin{equation*} \lap{y'' + 3y' + 2y} = \lap{0}\text{.} \end{equation*}
Using the properties of the Laplace transform for derivatives, we have:
\begin{equation*} s^2Y(s) - sy(0) - y'(0) + 3[sY(s) - y(0)] + 2Y(s) = 0\text{.} \end{equation*}
Substitute the initial conditions \(y(0) = 2\) and \(y'(0) = -1\text{:}\)
\begin{equation*} s^2Y(s) - 2s + 1 + 3sY(s) - 6 + 2Y(s) = 0\text{.} \end{equation*}
Combine like-terms:
\begin{equation*} (s^2 + 3s + 2)Y(s) - 2s - 5 = 0\text{.} \end{equation*}
Solve for \(Y(s)\text{:}\)
\begin{equation*} Y(s) = \dfrac{2s + 5}{s^2 + 3s + 2}\text{.} \end{equation*}
Factor the denominator:
\begin{equation*} Y(s) = \dfrac{2s + 5}{(s + 1)(s + 2)}\text{.} \end{equation*}
Use partial fraction decomposition to find the inverse Laplace transform:
\begin{equation*} Y(s) = \dfrac{A}{s + 1} + \dfrac{B}{s + 2}\text{.} \end{equation*}
Solve for \(A\) and \(B\text{:}\)
\begin{equation*} Y(s) = \dfrac{1}{s + 1} - \dfrac{3}{s + 2}\text{.} \end{equation*}
Finally, take the inverse Laplace transform to find \(y(t)\text{:}\)
\begin{equation*} y(t) = e^{-t} - 3e^{-2t}\text{.} \end{equation*}
πŸ”—
πŸ”—
4.
Solution.
Apply the Laplace transform to both sides:
\begin{equation*} \lap{y' + 4y} = \lap{10e^{-2t}}\text{.} \end{equation*}
Use the properties of the Laplace transform:
\begin{equation*} sY(s) - y(0) + 4Y(s) = \dfrac{10}{s + 2}\text{.} \end{equation*}
Substitute the initial condition \(y(0) = 3\text{:}\)
\begin{equation*} (s + 4)Y(s) - 3 = \dfrac{10}{s + 2}\text{.} \end{equation*}
Solve for \(Y(s)\text{:}\)
\begin{equation*} Y(s) = \dfrac{10}{(s + 4)(s + 2)} + \dfrac{3}{s + 4}\text{.} \end{equation*}
Use partial fraction decomposition:
\begin{equation*} Y(s) = \dfrac{A}{s + 4} + \dfrac{B}{s + 2}\text{.} \end{equation*}
Find \(A\) and \(B\text{:}\)
\begin{equation*} A = \dfrac{7}{2}, B = -\dfrac{5}{2}\text{.} \end{equation*}
Inverse Laplace transform gives:
\begin{equation*} y(t) = \dfrac{7}{2}e^{-4t} - \dfrac{5}{2}e^{-2t}\text{.} \end{equation*}
πŸ”—
πŸ”—
5.
Solution.
Applying the Laplace transform to both sides, we get:
πŸ”—
\begin{equation*} sY(s) - y(0) + 3Y(s) = \dfrac{6}{s-2} \end{equation*}
Substituting the initial condition \(y(0) = 1\text{,}\) the equation becomes:
πŸ”—
\begin{equation*} sY(s) - 1 + 3Y(s) = \dfrac{6}{s-2} \end{equation*}
Rearranging and solving for \(Y(s)\text{:}\)
πŸ”—
\begin{equation*} Y(s)(s + 3) = 1 + \dfrac{6}{s-2} \end{equation*}
\begin{equation*} Y(s) = \dfrac{1}{s+3} + \dfrac{6}{(s-2)(s+3)} \end{equation*}
We can now decompose the second term using partial fractions:
πŸ”—
\begin{equation*} \dfrac{6}{(s-2)(s+3)} = \dfrac{A}{s-2} + \dfrac{B}{s+3} \end{equation*}
Solving for \(A\) and \(B\text{,}\) we get \(A = 2\) and \(B = 4\text{.}\) Therefore:
πŸ”—
\begin{equation*} Y(s) = \dfrac{1}{s+3} + \dfrac{2}{s-2} + \dfrac{4}{s+3} \end{equation*}
\begin{equation*} Y(s) = \dfrac{5}{s+3} + \dfrac{2}{s-2} \end{equation*}
Taking the inverse Laplace transform, we obtain the solution:
πŸ”—
\begin{equation*} y(t) = 5e^{-3t} + 2e^{2t} \end{equation*}
πŸ”—
6.
Solution.
Applying the Laplace transform:
πŸ”—
\begin{equation*} s^2Y(s) - sy(0) - y'(0) + 4Y(s) = \dfrac{s}{s^2 + 4} \end{equation*}
Substituting the initial conditions:
πŸ”—
\begin{equation*} s^2Y(s) - 1 + 4Y(s) = \dfrac{s}{s^2 + 4} \end{equation*}
Solving for \(Y(s)\text{:}\)
πŸ”—
\begin{equation*} Y(s)(s^2 + 4) = 1 + \dfrac{s}{s^2 + 4} \end{equation*}
After rearranging and solving, the inverse Laplace transform gives:
πŸ”—
\begin{equation*} y(t) = \sin(2t) + t \end{equation*}
πŸ”—
7.
Solution.
Applying the Laplace transform:
\begin{gather*} \end{gather*}
\begin{gather*} \end{gather*}
\begin{gather*} \end{gather*}
Solving for \(Y(s)\) and then applying the inverse Laplace transform gives the solution.
πŸ”—
πŸ”—
8.
Solution.
Applying the Laplace transform and solving the algebraic equation will yield the solution in the \(s\)-domain, which can then be inverted to find \(y(t)\text{.}\)
πŸ”—
πŸ”—
9.
Solution.
We intend to use Laplace transforms to solve this IVP, but we need to verify that this is an appropriate technique. We can verify that this is
πŸ”—
Hence, it is appropriate to proceed with the Laplace transform solution technique. We also note that \(y\) is the dependent variable and \(t\) is the independent variable, so our goal is to find a function \(y(t)\) that satisfies the DE and the initial conditions.
πŸ”—
πŸ“
We begin by taking the Laplace transform of both sides of the DE, using linearity as needed.
\begin{align*} \lap{y'' + 3y' + 2y} \amp = \lap{ 4t }\\ \lap{y''} + 3\lap{y'} + 2\lap{y} \amp = 4\cdot\lap{ t }\\ \Big[s^2Y(s) - sy(0) - y'(0) \Big] + 3\Big[ sY(s) - y(0) \Big] + 2Y(s) \amp = \dfrac{4}{s^2} \end{align*}
πŸ”—
Now we use the provided initial conditions.
\begin{align*} \Big[s^2Y(s) - s\cdot 1 - 0 \Big] + 3\Big[ sY(s) - 1 \Big] + 2Y(s) \amp = \dfrac{4}{s^2}\\ s^2Y(s) - s + 3sY(s) - 3 + 2Y(s) \amp = \dfrac{4}{s^2} \end{align*}
πŸ”—
Next we will use algebra to solve for \(Y(s).\)
\begin{align*} s^2Y(s)+ 3sY(s) + 2Y(s) \amp = \dfrac{4}{s^2} + s + 3\\ Y(s)\Big[ s^2 + 3s + 2 \Big] \amp = \dfrac{4}{s^2} + \dfrac{s^3}{s^2} + \dfrac{3s^2}{s^2}\\ \amp = \dfrac{4 + s^3 + 3s^2}{s^2}\\ Y(s) \amp = \dfrac{s^3 + 3s^2 + 4}{s^2(s^2 + 3s + 2)}\\ \amp = \dfrac{s^3 + 3s^2 + 4}{s^2(s + 2)(s + 1)} \end{align*}
πŸ”—
We need to find the inverse Laplace transform of both sides of the equation. In order to do that, we apply partial fraction decomposition to the rational function on the right hand side, giving
\begin{equation*} \dfrac{s^3 + 3s^2 + 4}{s^2(s + 2)(s + 1)} = \dfrac{-3}{s} + \dfrac{2}{s^2} + \dfrac{-2}{s+2} + \dfrac{6}{s+1}. \end{equation*}
πŸ”—
We have a repeated linear factor, \(s^2\text{,}\) and two other linear factors, so the form of the partial fraction decomposition can be written as
\begin{equation*} \dfrac{s^3 + 3s^2 + 4}{s^2(s + 2)(s + 1)} = \dfrac{A}{s} + \dfrac{B}{s^2} + \dfrac{C}{s+2} + \dfrac{D}{s+1} \end{equation*}
πŸ”—
(PFD details...can be omitted?)
\begin{align*} s^3 + 3s^2 + 4 \amp = A(s)(s+2)(s+1) + B(s+2)(s+1) + C(s^2)(s+1) + D(s^2)(s+2) \\ \amp = As(s^2 + 3s + 2) + B(s^2 + 3s + 2) + Cs^3 + Cs^2 + Ds^3 + 2Ds^2 \\ \amp = As^3 + 3As^2 + 2As + Bs^2 + 3Bs + 2B + Cs^3 + Cs^2 + Ds^3 + 2Ds^2 \\ s^3 + 3s^2 + 0s + 4 \amp = (A + C + D)s^3 + (3A + B + C+ 2D)s^2 + (2A + 3B)s + (2B) \end{align*}
\begin{align*} A+C+D \amp = 1 \amp 3A+B+C+2D \amp = 3 \amp 2A+3B \amp = 0 \amp 2B \amp = 4 \\ \amp \amp \amp \amp \amp \amp B \amp = 2 \\ \amp \amp 3A+2+C+2D \amp = 3 \amp 2A+ 3(2)\amp = 0 \amp \amp \\ \amp \amp 3A+C+2D \amp = 1 \amp 2A \amp = -6 \amp \amp \\ \amp \amp \amp \amp A \amp = -3 \amp \amp \\ -3+C+D \amp = 1 \amp 3(-3)+C+ 2D \amp = 1 \amp \amp \amp \amp \\ C+D \amp = 4 \amp C+2D \amp = 10 \amp \amp \amp \amp \\ C \amp = 4 - D\amp \amp \amp \amp \amp \amp \\ \amp \amp (4-D) + 2D \amp = 10 \amp \amp \amp \amp \\ \amp \amp 4 + D \amp = 10 \amp \amp \amp \amp \\ \amp \amp D \amp = 6 \amp \amp \amp \amp \\ C \amp = 4-6 \amp \amp \amp \amp \amp \amp \\ \amp = -2 \amp \amp \amp \amp \amp \amp \end{align*}
πŸ”—
Hence, we have
\begin{equation*} \dfrac{s^3 + 3s^2 + 4}{s^2(s + 2)(s + 1)} = \dfrac{-3}{s} + \dfrac{2}{s^2} + \dfrac{-2}{s+2} + \dfrac{6}{s+1} \end{equation*}
πŸ”—
Now we are prepared to take the inverse Laplace transform.
\begin{align*} Y(s) \amp = \dfrac{-3}{s} + \dfrac{2}{s^2} + \dfrac{-2}{s+2} + \dfrac{6}{s+1} \\ \ilap{Y(s)} \amp = \ilap{ \dfrac{-3}{s} + \dfrac{2}{s^2} + \dfrac{-2}{s+2} + \dfrac{6}{s+1} } \\ y(t) \amp = -3\ilap{ \dfrac{1}{s} } + 2\ilap{ \dfrac{1}{s^2} } - 2\ilap{ \dfrac{1}{s+2} } + 6\ilap{ \dfrac{1}{s+1} }\\ \amp = -3 + 2t - 2e^{-2t} + 6e^{-t} \end{align*}
πŸ”—
Thus, we have found the desired unknown function
\begin{equation*} y(t) = -3 + 2t - 2e^{-2t} + 6e^{-t}. \end{equation*}
πŸ”—
Optional: verify the solution
\begin{align*} y'(t) \amp = 0 + 2 + 4e^{-2t} - 6e^{-t}\\ y''(t) \amp = 0 - 8e^{-2t}+ 6e^{-t} \end{align*}
\begin{align*} LHS \amp = y'' + 3y' + 2y\\ \amp = \left( - 8e^{-2t}+ 6e^{-t} \right) \\ \amp = -8e^{-2t}+ 6e^{-t}\\ \amp = 4t\\ \amp = RHS \end{align*}
πŸ”—
We also verify the initial conditions:
\begin{align*} y(0) \amp = -3 + 2(0) - 2e^{-2(0)} + 6e^{-0}\\ \amp = -3 + 0 - 2+ 6\\ \amp = 1\\ \amp\\ y'(0) \amp = 2 + 4e^{-2(0)} - 6e^{-0}\\ \amp = 2 + 4 - 6\\ \amp = 0 \end{align*}
πŸ”—
πŸ”—
12.
Solution.
Step 1β€” Forward Transform. Applying \(\laplacesym\) to both sides:
\begin{gather*} \lap{y' + y} = \lap{4}\\ \lap{y'} + \lap{y} = \lap{4}\\ sY(s) - 2 + Y(s) = \dfrac{4}{s} \end{gather*}
πŸ”—
Step 2β€” Solve & Prepare \(Y(s)\). Next, we isolate \(Y(s)\) on one side of the equation:
\begin{align*} sY(s) - 2 + Y(s) \amp = \dfrac{4}{s} \\ (s + 1)Y(s) \amp = \dfrac{4}{s} + 2 \\ Y(s) \amp = \dfrac{1}{s + 1}\left(\dfrac{4}{s} + 2\right) \\ Y(s) \amp = \ub{\dfrac{4}{s(s + 1)}}_{(*)} + \ub{\dfrac{2}{s + 1}}_{(**)} \end{align*}
πŸ”—
Step 2β€” Solve & Prepare \(Y(s)\). We need to express \(Y(s)\) as a sum of functions that match known forms in the Laplace transform table. We see from the last equation, \((**)\) is ready to go, but \((*)\) requires partial fraction decomposition. We start by writing down the form of the decomposition,
\begin{equation*} (*) = \dfrac{4}{s(s + 1)} = \dfrac{A}{s} + \dfrac{B}{s + 1} \quad \text{or} \end{equation*}
\begin{equation*} 4 = A(s + 1) + Bs \end{equation*}
Now, we find \(A\) and \(B\) by plugging in values of \(s\text{:}\)
πŸ”—
\(s=0\) : \(4\) \(=\) \(A(0 + 1) + B(0)\)
\(4\) \(=\) \(A\)
\(s=-1\) : \(4\) \(=\) \(A(-1 + 1) + B(-1)\)
\(-4\) \(=\) \(B\)
Thus, the prepared \(Y(s)\) is
\begin{equation*} Y(s) = \ub{\dfrac{4}{s} + \dfrac{-4}{s + 1}}_{(*)} + \ub{\dfrac{2}{s + 1}}_{(**)} \end{equation*}
πŸ”—
Step 3β€” Inverse Transform. Finally, we apply the backward step to get the solution:
\begin{align*} \ilap{Y(s)} \amp = \ilap{\dfrac{4}{s} + \dfrac{-4}{s + 1}}\\ y(t) \amp = 4 - 4e^{-t} \end{align*}
πŸ”—
πŸ”—
13.
Solution.
Step 1β€” Forward Transform. Applying \(\laplacesym\) to both sides:
\begin{align*} \lap{y'' - 3y' + 2y} \amp = \lap{e^{2t}}\\ \lap{y''} - 3\lap{y'} + 2\lap{y} \amp = \dfrac{1}{s - 2} \end{align*}
where \(\lap{y(t)} = Y(s)\) and
  • \(\lap{y'} = sY(s) - y(0) = sY(s) - 1\text{,}\)
    πŸ”—
  • \(\ds \lap{y''} = s^2Y(s) - sy(0) - y'(0) = s^2Y(s) - s\text{.}\)
    πŸ”—
So the complete forward transform is
\begin{equation*} s^2Y(s) - s - 3[sY(s) - 1] + 2Y(s) = \dfrac{1}{s - 2} \end{equation*}
πŸ”—
Step 2β€” Solve & Prepare \(Y(s)\). Next, we isolate \(Y(s)\) on one side of the equation:
\begin{gather*} s^2Y(s) - s - 3sY(s) + 3 + 2Y(s) = \dfrac{1}{s - 2}\\ [s^2 - 3s + 2]Y(s) - s + 3 = \dfrac{1}{s - 2}\\ Y(s)[(s - 1)(s - 2)] = \dfrac{1}{s - 2} + s - 3\\ Y(s) = \dfrac{1}{(s - 1)(s - 2)}\ub{\left(\dfrac{1}{s - 2} + s - 3\right)}_{(*)} \end{gather*}
πŸ”—
Step 2β€” Solve & Prepare \(Y(s)\). Now, we need to express \(Y(s)\) as a sum of functions that match known forms in the Laplace transform table. This can be simplified slightly by first combining the terms in \((*)\) as a single fraction, like so
\begin{equation*} (*) = \dfrac{1}{s - 2} + \dfrac{(s - 3)(s - 2)}{s - 2} = \dfrac{1 + s^2 - 5s + 6}{s - 2} = \dfrac{s^2 - 5s + 7}{s - 2}\text{.} \end{equation*}
Plugging this back into the equation for \((*)\) gives the new \(Y(s)\text{,}\)
\begin{equation*} Y(s) = \dfrac{1}{(s - 1)(s - 2)}\cdot \dfrac{s^2 - 5s + 7}{s - 2} = \dfrac{s^2 - 5s + 7}{(s - 1)(s - 2)^2} \end{equation*}
and we are ready to apply partial fraction decomposition. The form of the decomposition is
\begin{align*} \dfrac{s^2 - 5s + 7}{(s - 1)(s - 2)^2} \amp = \dfrac{A}{s - 1} + \dfrac{B}{s - 2} + \dfrac{C}{(s - 2)^2} \quad \text{or}\\ s^2 - 5s + 7 \amp = A(s - 2)^2 + B(s - 1)(s - 2) + C(s - 1). \end{align*}
Now, we find \(A, B, \) and \(C\) by plugging in values of \(s\text{,}\)
πŸ”—
\(s=1\) : \(1 - 5 + 7\) \(=\) \(A(1 - 2)^2 + B(1 - 1)(1 - 2) + C(1 - 1)\)
\(3\) \(=\) \(A\)
\(s=2\) : \(4 - 10 + 7\) \(=\) \(3(2 - 2)^2 + B(2 - 1)(2 - 2) + C(2 - 1)\)
\(1\) \(=\) \(C\)
\(s=0\) : \(7\) \(=\) \(3(0 - 2)^2 + B(0 - 1)(0 - 2) + 1(0 - 1)\)
\(7\) \(=\) \(12 + 2B - 1\)
\(-2\) \(=\) \(B\)
Thus, the prepared \(Y(s)\) is
\begin{equation*} Y(s) = \dfrac{3}{s - 1} + \dfrac{-2}{s - 2} + \dfrac{1}{(s - 2)^2} \end{equation*}
πŸ”—
Step 3β€” Inverse Transform. Finally, we perform the backward step to get the solution:
\begin{align*} \ilap{Y(s)}\\ \amp = \ilap{\dfrac{3}{s - 1}} + \ilap{\dfrac{-2}{s - 2}} + \ilap{\dfrac{1}{(s - 2)^2}}\\ y(t) \amp = 3\, e^{t} - 2\, e^{2t} + t\, e^{2t} \end{align*}
πŸ”—
πŸ”—
14.
Solution.
Step 1β€” Forward Transform. Applying \(\laplacesym\) to both sides:
\begin{gather*} \lap{x'' - 4x' + 13x} = \lap{54e^{-t}}\\ \lap{x''} - 4\lap{x'} + 13\lap{x} = \dfrac{54}{s + 1} \end{gather*}
where \(\lap{x(t)} = X(s)\) and
  • \(\lap{x'} = sX(s) - x(0) = sX(s)\text{,}\)
    πŸ”—
  • \(\ds \lap{x''} = s^2X(s) - sx(0) - x'(0) = s^2X(s)\text{.}\)
    πŸ”—
So the complete forward transform is:
\begin{equation*} s^2X(s) - 4sX(s) + 13X(s) = \dfrac{54}{s + 1} \end{equation*}
πŸ”—
Step 2β€” Solve & Prepare \(Y(s)\). Next, we isolate \(X(s)\) on one side of the equation:
\begin{gather*} s^2X(s) - 4sX(s) + 13X(s) = \dfrac{54}{s + 1}\\ [s^2 - 4s + 13]X(s) = \dfrac{54}{s + 1}\\ X(s) = \dfrac{54}{(s + 1)(s^2 - 4s + 13)} \end{gather*}
πŸ”—
Step 2β€” Solve & Prepare \(Y(s)\). Now, we need to express \(X(s)\) as a sum of functions that match known forms in the Laplace transform table, which requires partial fraction decomposition. We start by writing down the form of the decomposition,
\begin{gather*} \dfrac{54}{(s + 1)(s^2 - 4s + 13)} = \dfrac{A}{s + 1} + \dfrac{Bs + C}{s^2 - 4s + 13} \quad \text{or}\\ 54 = A(s^2 - 4s + 13) + (Bs + C)(s + 1) \end{gather*}
Now, we find \(A, B, \) and \(C\) by plugging in values of \(s\text{,}\)
πŸ”—
\(s=-1\text{:}\)
πŸ”—
\(54 = A(-1)^2 - 4(-1) + 13 \)
\(54 = A + 17 \)
πŸ”—
\(\vphantom{A} \)
\(A = 37 \)
πŸ”—
\(s=0\text{:}\)
πŸ”—
\(54 = 37(13) + (B(0) + C)(0 + 1) \)
\(54 = 481 + C \)
πŸ”—
\(\vphantom{C} \)
\(C = -427 \)
πŸ”—
\(s=1\text{:}\)
πŸ”—
\(54 = 37(-10) + (B(1) - 427)(1 + 1) \)
\(54 = 2B - 1224 \)
πŸ”—
\(\vphantom{B} \)
\(B = 585 \)
πŸ”—
πŸ”—
So the updated \(X(s)\) is
\begin{equation*} X(s) = \dfrac{37}{s + 1} + \dfrac{585s - 427}{s^2 - 4s + 13}\text{.} \end{equation*}
Note, the second term is not yet ready and we need to complete the square of its denominator before we can do the backward step.
\begin{align*} X(s) \amp = \dfrac{37}{s + 1} + \dfrac{585s - 427}{(s - 2)^2 + 9}\\ \amp = \dfrac{37}{s + 1} + \dfrac{585s - 2(585) + 2(585)- 427}{(s - 2)^2 + 9}\\ \amp = \dfrac{37}{s + 1} + \dfrac{585s - 2(585)}{(s - 2)^2 + 9} + \dfrac{2(585)- 427}{(s - 2)^2 + 9}\\ \amp = \dfrac{37}{s + 1} + 585\dfrac{s - 2}{(s - 2)^2 + 9} + 743\dfrac{1}{(s - 2)^2 + 9}\\ \amp = \dfrac{37}{s + 1} + 585\dfrac{s - 2}{(s - 2)^2 + 9} + \dfrac{743}{3}\dfrac{3}{(s - 2)^2 + 9} \end{align*}
πŸ”—
Step 3β€” Inverse Transform. Now for the backward step to get the solution:
\begin{align*} x(t) \amp = 37e^{-t} + 585\cos(3t) + \dfrac{743}{3}\sin(3t) \end{align*}
πŸ”—
πŸ”—

11 Laplace Method for Piecewise Functions
Chapter 11 Exercises

πŸ—οΈ Unit Step Functions

1.
Solution.
Since \(t = 2.7 \ge 0\) (positive),
\begin{equation*} u_0(2.7) = 1\text{.} \end{equation*}
πŸ”—
πŸ”—
2.
Solution.
Since \(-5.32 < 0\) (negative),
\begin{equation*} u_0(-5.32) = 0\text{.} \end{equation*}
πŸ”—
πŸ”—
3.
Solution.
\begin{equation*} u_0(2)\cdot f(2) = 1\cdot f(2) = (2)^2 - 2 = (4 - 2) = 2\text{.} \end{equation*}
πŸ”—
πŸ”—
4.
Solution.
\begin{equation*} u_0(-3)\cdot f(-3) = 0\cdot ((-3)^2 - 2) = 0\cdot (9 - 2) = 0\text{.} \end{equation*}
πŸ”—
πŸ”—
5.
Solution.
Since \(k \ge 0,\) we know that \(u_0(k) = 1,\) so
\begin{equation*} u_0(k)\cdot f(k) = 1\cdot f(k) = k^2 - 2\text{.} \end{equation*}
πŸ”—
πŸ”—
6.
Solution.
Since \(k \lt 0,\) we know that \(u_0(k) = 0,\) so
\begin{equation*} u_0(k)\cdot f(k) = 0\cdot f(k) = 0\text{.} \end{equation*}
πŸ”—
πŸ”—
7.
Solution.
Since \(U(t-2)\) is \(U(t)\) shifted to the right by \(2\text{,}\) the jump in the graph occurs at \(t=2\) (see below). Additionally, multiplying by \(-2\) reflects the graph about the \(x\)-axis and scales it by \(2\text{,}\) as shown below.
πŸ”—
πŸ”—
8.
Solution.
Use the shifting property of the unit step function:
\begin{equation*} \lap{u(t - 3)} = \dfrac{e^{-3s}}{s}\text{.} \end{equation*}
πŸ”—
πŸ”—
12.
Solution.
We start by distributing:
\begin{equation*} e^{3t}(1 - u_1(t)) = e^{3t} - e^{3t} \cdot u_1(t)\text{.} \end{equation*}
πŸ”—
Applying the Laplace transform to this gives us
\begin{equation*} \lap{e^{3t}(1 - u_1(t))} = \lap{e^{3t}} - \lap{e^{3t} \cdot u_1(t)}\text{.} \end{equation*}
πŸ”—
The first term uses a standard Laplace transform:
\begin{equation*} \lap{e^{3t}} = \dfrac{1}{s - 3}\text{.} \end{equation*}
πŸ”—
The second term requires the new rule with \(c=1\text{,}\) which gives us
πŸ”—
\begin{align*} \lap{{\DLO e^{3t}} \cdot u_1(t)} \amp = e^{-s} \lap{{\DLO f(t + 1)}}\\ \amp = e^{-s} \lap{{\DLO e^{3t} \cdot e^{3}}}\\ \amp = e^{-s} \cdot e^{3} \lap{e^{3t}}\\ \amp = e^{-s + 3} \left(\dfrac{1}{s - 3}\right) \end{align*}
πŸ”—
\begin{align*} \DLO f(t)\ \amp\DLO = e^{3t}\\ \DLO f(t + 1)\ \amp\DLO = e^{3(t + 1)} \\ \amp\DLO = e^{3t + 3} = e^{3t} \cdot e^{3}\\ \text{Note:}\ e^3\ \amp \text{is constant} \end{align*}
πŸ”—
πŸ”—
Putting it all together, we arrive at the desired transformation:
\begin{equation*} \lap{e^{3t}(1 - u_1(t))} = \dfrac{1}{s - 3} - \dfrac{e^{3 - s}}{s - 3} \end{equation*}
πŸ”—
πŸ”—

πŸ—οΈ Piecewise Functions

1.
Solution.
Since \(1 - u_2(t)\) switches OFF at \(t = 2\text{,}\) this function is ON before \(2\) and OFF afterward:
\begin{equation*} g(t) = \left\{ \begin{array}{ll} 3 - t^2, \amp t \lt 2 \\ 0, \amp t \ge 2 \end{array} \right. \end{equation*}
πŸ”—
πŸ”—
2.
Solution.
\(u_3(t)\) turns ON at \(t = 3\text{,}\) so:
\begin{equation*} h(t) = \left\{ \begin{array}{ll} 0, \amp t \lt 3 \\ t^2 - 4, \amp t \ge 3 \end{array} \right. \end{equation*}
πŸ”—
πŸ”—
3.
Solution.
Since this piecewise function is of the form
\begin{equation*} h(t) = \left(t - 2\right)\cdot \left\{ \begin{array}{ll} 0, & t \lt 2\\ 1, & t \ge 2 \end{array} \right. \quad = \left(t - 2\right)\cdot u_2(t)\text{.} \end{equation*}
πŸ”—
To compute \(\lap{h(t)}\) using the transform rule with \(c=2\text{,}\) we first label \(f(t)\text{:}\)
πŸ”—
\begin{equation*} \lap{h(t)} = \laplacesym \big\{{\DLO \us{f(t)}{\us{\uparrow}{(t - 2)}}}\, u_2(t)\big\} = e^{-2s} \lap{{\DLO f(t+2)}} \end{equation*}
Since \(g(t) = t - 2\text{,}\) then \(f(t + 2) = (t+2) - 2 = t\) and we have
\begin{equation*} \lap{h(t)} = e^{-2s} \lap{t} = e^{-2s} \cdot \dfrac{1}{s^2} = \dfrac{e^{-2s}}{s^2}\text{.} \end{equation*}
πŸ”—
πŸ”—
4.
Solution.
We can rewrite this function as
\begin{equation*} m(t) = t^2 \cdot \left\{ \begin{array}{ll} 0, \amp t \lt 1\\ 1, \amp 1 \le t \lt 3\\ 0, \amp t \ge 3 \end{array} \right. \quad = t^2 \cdot [u_1(t) - u_3(t)]\text{.} \end{equation*}
πŸ”—
Now we compute
πŸ”—
\begin{align*} \laplacesym\amp \big\{m(t)\big\} = \laplacesym \big\{{\DLO \us{f(t)}{\us{\uparrow}{\ul{t^2}}}}\, u_1(t)\big\} - \laplacesym \big\{{\DLO \us{f(t)}{\us{\uparrow}{\ul{t^2}}}}\, u_3(t)\big\}\\ \amp = e^{-s} \lap{{\DLO f(t+1)}} - e^{-3s} \lap{{\DLO f(t+3)}}\\ \amp = e^{-s} \lap{{\DLO t^2 + 2t + 1}} - e^{-3s} \lap{{\DLO t^2 + 6t + 9}} \end{align*}
πŸ”—
\begin{align*} \DLO f(t+1)\ \amp\DLO = (t+1)^2\\ \amp\DLO = t^2 + 2t + 1\\ \DLO f(t+3)\ \amp\DLO = (t+3)^2\\ \amp\DLO = t^2 + 6t + 9 \end{align*}
πŸ”—
\begin{equation*} \lap{m(t)} = e^{-s}\left(\dfrac{6}{s^4} + \dfrac{4}{s^3} + \dfrac{1}{s^2}\right) - e^{-3s}\left(\dfrac{6}{s^4} + \dfrac{12}{s^3} + \dfrac{9}{s^2}\right) \end{equation*}
πŸ”—
5.
Solution.
The function is initially \(2e^{-\sfrac{t^2}{2}}\) and then becomes \(0\) at \(t=1\text{.}\) So the piecewise function, \(f(t)\text{,}\) is equivalent to
\begin{align*} f(t) \amp = \left\{ \begin{array}{ll} 2e^{-\sfrac{t^2}{2}}, & t \lt 1\\ 0, & t \ge 1 \end{array} \right.\\ \\ \amp = 2e^{-\sfrac{t^2}{2}} \cdot {\color{BurntOrange}\ub{ \left\{ \begin{array}{ll} 1, & t \lt 1\\ 0, & t \ge 1 \end{array} \right.}_{\large 1 - u_{1}(t)}} \end{align*}
and therefore, the piecewise function, \(f(t)\text{,}\) can be written as
\begin{equation*} f(t) = 2e^{-\sfrac{t^2}{2}} \cdot (1 - u_{1}(t))\text{,} \end{equation*}
and has the following graph
πŸ”—
Figure 277. Graph of \(2e^{-\sfrac{t^2}{2}} \cdot u_{1}(t)\)
πŸ”—
πŸ”—
πŸ”—

Piecewise Inverse Transforms

1.
Solution.
This is already in the form \(e^{-cs}F(s)\) with \(c = 3\) and \(F(s) = \dfrac{2}{s^2 + 9}\text{.}\)
πŸ”—
The inverse transform of \(\dfrac{2}{s^2 + 9}\) is \(\sin(3t)\text{,}\) since this matches the standard form:
\begin{equation*} \ilap{ \dfrac{b}{s^2 + b^2} } = \sin(bt)\text{.} \end{equation*}
πŸ”—
Applying the shift rule:
\begin{equation*} y(t) = u_3(t) \cdot \sin(3(t - 3)) = u_3(t)\cdot \sin(3t - 9) \end{equation*}
πŸ”—
πŸ”—
2.
Solution.
We identify this as \(e^{-2s} \cdot F(s)\text{,}\) where
\begin{equation*} F(s) = \dfrac{s}{s^2 + 4}. \end{equation*}
This matches the form for \(\cos(2t)\text{:}\)
\begin{equation*} \ilap{ \dfrac{s}{s^2 + b^2} } = \cos(bt). \end{equation*}
πŸ”—
So the inverse is:
\begin{equation*} f(t) = u_2(t) \cdot \cos(2(t - 2)) = u_2(t)\cdot \cos(2t - 4) \end{equation*}
πŸ”—
πŸ”—
6.
Solution.
\(\ds X(s) = \dfrac{e^{-s}(s-5)}{(s+1)(s+2)} \) vspace{1cm}
πŸ”—
πŸ”—

Solving Differential Equations

1.
Solution.
πŸ”—
Step 1: Apply the Laplace Transform.
πŸ”—
\begin{align*} \lap{y'' + 4y} = \lap{u_3(t)}\\ s^2Y + 4Y \amp = \dfrac{e^{-3s}}{s} \end{align*}
πŸ”—
\begin{align*} \amp\text{ Laplace} \\ \leftarrow \amp\text{ Equation} \end{align*}
πŸ”—
πŸ”—
Step 2: Solve for \(Y(s)\).
πŸ”—
πŸ”—
\begin{align*} (s^2 + 4)Y \amp = \dfrac{e^{-3s}}{s} \\ Y(s) \amp = \dfrac{e^{-3s}}{s(s^2 + 4)} \end{align*}
πŸ”—
\begin{gather*} \leftarrow\text{ Laplace Solution}\vphantom{\dfrac12} \end{gather*}
πŸ”—
πŸ”—
Step 3: Prepare for the Inverse Transform.
πŸ”—
The exponential factor \(e^{-3s}\) indicates a delay. Focus on the base transform:
\begin{equation*} \dfrac{1}{s(s^2 + 4)} \end{equation*}
From the table, this matches the transform of \(1 - \cos(2t)\text{.}\) Therefore:
πŸ”—
\begin{equation*} \ilap{ \dfrac{1}{s(s^2 + 4)} } = 1 - \cos(2t) \end{equation*}
The delay is applied using the shift theorem.
πŸ”—
πŸ”—
Step 4: Take the Inverse Transform.
πŸ”—
\begin{gather*} y(t) = \ilap{ \dfrac{e^{-3s}}{s(s^2 + 4)} } = u_3(t)\left[1 - \cos(2(t - 3))\right] \end{gather*}
πŸ”—
Therefore, the solution is:
\begin{equation*} y(t) = u_3(t)\left(1 - \cos(2(t - 3))\right) \end{equation*}
πŸ”—
πŸ”—
4.
Solution.
Prepare the Forcing Function for Step 1.
We begin by rewriting \(h(t)\) using unit step functions. Since \(h(t)\) equals 1 on \([2, 3)\) and 0 elsewhere, we express it as:
\begin{equation*} h(t) = u_2(t) - u_3(t) \end{equation*}
This transforms the differential equation into:
\begin{equation*} y'' + 9y = u_2(t) - u_3(t), \qquad y(0) = 0,\quad y'(0) = 0 \end{equation*}
πŸ”—
πŸ”—
Step 1: Apply the Laplace Transform to Both Sides.
Apply the Laplace transform to both sides:
\begin{equation*} \lap{y''} + 9\lap{y} = \lap{u_2(t)} - \lap{u_3(t)} \end{equation*}
Using standard rules:
\begin{equation*} s^2Y(s) + 9Y(s) = \dfrac{e^{-2s}}{s} - \dfrac{e^{-3s}}{s} \end{equation*}
πŸ”—
πŸ”—
Step 2: Solve for \(Y(s)\).
Factor out \(Y(s)\text{:}\)
\begin{gather*} (s^2 + 9)Y(s) = \dfrac{e^{-2s}}{s} - \dfrac{e^{-3s}}{s} \end{gather*}
Then divide:
\begin{equation*} Y(s) = \dfrac{1}{s^2 + 9} \left( \dfrac{e^{-2s}}{s} - \dfrac{e^{-3s}}{s} \right) \end{equation*}
πŸ”—
πŸ”—
Step 3: Prepare \(Y(s)\) for an Inverse Transform.
Step 3 focuses on rewriting \(Y(s)\) to match inverse Laplace rules. We separate the exponentials:
\begin{equation*} Y(s) = \dfrac{1}{s(s^2 + 9)}\left(e^{-2s} - e^{-3s}\right) \end{equation*}
πŸ”—
Let the rational part be:
\begin{equation*} F(s) = \dfrac{1}{s(s^2 + 9)} \end{equation*}
Then:
\begin{equation*} Y(s) = F(s)\left(e^{-2s} - e^{-3s}\right) \end{equation*}
πŸ”—
This means the inverse will take the form:
\begin{equation*} y(t) = \ilap{F(s)e^{-2s}} - \ilap{F(s)e^{-3s}} \end{equation*}
but we must first identify \(f(t) = \ilap{F(s)}\) using partial fractions. The details are shown below:
πŸ”—
Solution. Partial Fraction Decomposition of \(F(s)\)
Write:
\begin{equation*} F(s) = \dfrac{1}{s(s^2 + 9)} = \dfrac{A}{s} + \dfrac{Bs + C}{s^2 + 9} \end{equation*}
Multiply by \(s(s^2 + 9)\) and group:
\begin{align*} 1 \amp = A(s^2 + 9) + (Bs + C)s\\ \amp = (A + B)s^2 + Cs + 9A \end{align*}
Match coefficients:
\begin{align*} A + B \amp = 0\\ C \amp = 0\\ 9A \amp = 1 \end{align*}
πŸ”—
\begin{equation*} \Rightarrow \end{equation*}
πŸ”—
\begin{align*} A \amp = \dfrac{1}{9}\\ B \amp = -\dfrac{1}{9}\\ C \amp = 0 \end{align*}
πŸ”—
So:
\begin{equation*} F(s) = \dfrac{1}{9}\left(\dfrac{1}{s} - \dfrac{s}{s^2 + 9}\right) \end{equation*}
πŸ”—
πŸ”—
πŸ”—
This completes Step 3. We’ll now apply the inverse transform.
πŸ”—
πŸ”—
Step 4: Invert \(Y(s)\) to Recover \(y(t)\).
We now compute:
\begin{equation*} y(t) = \ilap{F(s)e^{-2s}} - \ilap{F(s)e^{-3s}} \end{equation*}
By rule \(L15\), this gives:
\begin{equation*} y(t) = f(t - 2)\cdot u_2(t) - f(t - 3)\cdot u_3(t) \end{equation*}
where:
\begin{gather*} f(t) = \ilap{F(s)} = \dfrac{1}{9}\ilap{\dfrac{1}{s}} - \dfrac{1}{9}\ilap{\dfrac{s}{s^2 + 9}}\\ = \dfrac{1}{9} - \dfrac{1}{27}\cos(3t) \end{gather*}
πŸ”—
Substitute:
\begin{equation*} y(t) = \left(\dfrac{1}{9} - \dfrac{1}{27}\cos(3(t-2))\right)u_2(t) - \left(\dfrac{1}{9} - \dfrac{1}{27}\cos(3(t-3))\right)u_3(t) \end{equation*}
πŸ”—
This is the complete solution for all \(t \ge 0\text{.}\)
πŸ”—
πŸ”—
πŸ”—
5.
Solution.
Step 1 β€” Into the Laplace Domain.
Recall that our first step is to apply the Laplace transform of both sides:
\begin{equation*} \lap{y'' + y} = \lap{g(t)} \end{equation*}
πŸ”—
It’s clear how to handle the left-side, but the right-side requires the transform of a piecewise function. Before we can use the rules we derived in this chapter, we first need to write \(g(t)\) in unit step form.
πŸ”—
Apply the Laplace transform to both sides of the differential equation:
\begin{equation*} y'' + y = \left\{ \begin{array}{ll} 0, \amp t \lt 1 \\ 3t, \amp 1 \le t \lt 2 \\ 0, \amp t \ge 2 \end{array} \right., \qquad y(0) = 0,\quad y'(0) = 0 \end{equation*}
πŸ”—
The forcing function is active only from \(t = 1\) to \(t = 2\text{,}\) so we rewrite it using a unit step window:
\begin{equation*} g(t) = 3t \cdot \left(u_1(t) - u_2(t)\right). \end{equation*}
πŸ”—
Apply the Laplace transform to both sides of the DE:
\begin{equation*} \lap{y'' + y} = \lap{3t \cdot (u_1(t) - u_2(t))}. \end{equation*}
On the left-hand side, using \(y(0) = y'(0) = 0\text{,}\) we have:
\begin{equation*} s^2 Y(s) + Y(s). \end{equation*}
πŸ”—
On the right-hand side, distribute the Laplace transform:
\begin{equation*} \lap{3t \cdot (u_1(t) - u_2(t))} = 3\lap{t \cdot u_1(t)} - 3\lap{t \cdot u_2(t)}. \end{equation*}
Use the shift rule for each term:
\begin{gather*} \lap{t \cdot u_1(t)} = e^{-s} \lap{t + 1} = e^{-s} \cdot \lap{t + 1} \\ \lap{t \cdot u_2(t)} = e^{-2s} \lap{t + 2} = e^{-2s} \cdot \lap{t + 2} \end{gather*}
Now apply linearity and known transforms:
\begin{gather*} \lap{t + c} = \lap{t} + c\lap{1} = \dfrac{1}{s^2} + \dfrac{c}{s} \end{gather*}
So:
\begin{align*} \lap{3t \cdot (u_1 - u_2)} \amp = 3e^{-s}\left(\dfrac{1}{s^2} + \dfrac{1}{s}\right) - 3e^{-2s}\left(\dfrac{1}{s^2} + \dfrac{2}{s}\right) \end{align*}
πŸ”—
Combining both sides, we arrive at the Laplace equation:
\begin{equation*} (s^2 + 1)Y(s) = 3e^{-s}\left(\dfrac{1}{s^2} + \dfrac{1}{s}\right) - 3e^{-2s}\left(\dfrac{1}{s^2} + \dfrac{2}{s}\right). \end{equation*}
This completes Step 1.
πŸ”—
Now that we’ve built the Laplace equation, the remaining steps are familiar. We’ll solve for \(Y(s)\text{,}\) simplify, and invert. But there’s one twist: the presence of \(e^{-cs}\) terms means we’ll need special care when taking the inverse Laplace transform. Let’s explore that next.
πŸ”—
πŸ”—
Step 2 β€” Solving in the Laplace Domain.
πŸ”—
Step 3 β€” Leaving the Laplace Domain.
Once we’ve solved for \(Y(s)\) in the Laplace domain, the final challenge is to take the inverse Laplace transform. When the forcing function was continuous, this step relied on partial fractions and matching terms from the Laplace table. That still works here, but now we’ll also see terms involving exponentials like \(e^{-cs}F(s)\text{.}\)
πŸ”—
These exponentials come directly from transforming unit step expressions. To move back to the \(t\)-domain, we need a new tool: a rule that undoes the exponential shift.
πŸ”—
πŸ”—
πŸ”—
6.
Solution.
Roadmap-Summary.
\begin{gather*} y'' + 2y' = g(t) \\ \small y(0) = 0,\ y'(0) = 0 \end{gather*}
πŸ”—
\begin{equation*} \os{\Large\text{1️⃣}}{\us{\large\text{forward}}{\os{\large\laplacesym}{\longrightarrow}}} \end{equation*}
πŸ”—
\begin{equation*} s^2Y + 2sY = \dfrac{e^{-1s}}{s} - \dfrac{e^{-2s}}{s} \end{equation*}
πŸ”—
πŸ”—
\begin{equation*} \downarrow \end{equation*}
πŸ”—
2️⃣ Solve for \(Y\)
πŸ”—
πŸ”—
πŸ”—
πŸ”—
\begin{equation*} Y(s) = \dfrac{1}{s(s + 2)}(e^{-s} - e^{-2s}) \end{equation*}
πŸ”—
πŸ”—
πŸ”—
\begin{equation*} \downarrow \end{equation*}
πŸ”—
3️⃣ Prepare for \(\laplacesym^{-1}\)
πŸ”—
\begin{equation*} y(t) = \left(1 - e^{-2(t - 1)}\right)u_1(t) - \left(1 - e^{-2(t - 2)}\right)u_2(t) \end{equation*}
πŸ”—
\begin{equation*} \os{\Large\text{4️⃣}}{\us{\large\text{backward}}{\os{\large\laplacesym^{-1}}{\longleftarrow}}} \end{equation*}
πŸ”—
\begin{equation*} Y(s) = \dfrac{1}{s(s + 2)}(e^{-s} - e^{-2s}) \end{equation*}
πŸ”—
πŸ”—
πŸ”—
Preparation β€” Forcing Function in Step Form.
We begin by rewriting \(g(t)\) with unit step functions. Since it equals \(3\) on \([0,1)\text{,}\) \(0\) on \([1,4)\text{,}\) and \(1\) after \(t=4\text{:}\)
πŸ”—
\begin{equation*} g(t) = 3(u_0(t) - u_1(t)) + (1 - 3)(u_4(t)) \end{equation*}
which simplifies to:
πŸ”—
\begin{equation*} g(t) = 3u_0(t) - 3u_1(t) + u_4(t). \end{equation*}
πŸ”—
Step 1 β€” Into the Laplace Domain.
Take the Laplace transform of both sides:
\begin{equation*} \lap{y''} + 2\lap{y'} = \lap{3u_0(t) - 3u_1(t) + u_4(t)}. \end{equation*}
Using the initial conditions \(y(0)=0\) and \(y'(0)=0\text{,}\) the left side becomes:
\begin{equation*} s^2 Y(s) + 2s Y(s). \end{equation*}
Apply L\(_9\) to each step term:
\begin{equation*} \lap{3u_0(t)} = \dfrac{3}{s}, \quad \lap{-3u_1(t)} = -\dfrac{3e^{-s}}{s}, \quad \lap{u_4(t)} = \dfrac{e^{-4s}}{s}. \end{equation*}
This gives the Laplace-domain equation:
\begin{equation*} (s^2 + 2s)Y(s) = \dfrac{3}{s} - \dfrac{3e^{-s}}{s} + \dfrac{e^{-4s}}{s}. \end{equation*}
πŸ”—
πŸ”—
Step 2 β€” Solve for \(Y(s)\).
Factor out \(Y(s)\) and divide through by \((s^2 + 2s)\text{:}\)
πŸ”—
\begin{gather*} Y(s) = \dfrac{1}{s(s + 2)}\left( 3 - 3e^{-s} + e^{-4s} \right). \end{gather*}
To prepare for inversion, write:
πŸ”—
\begin{equation*} Y(s) = F(s)\big( 3 - 3e^{-s} + e^{-4s} \big), \end{equation*}
\begin{equation*} F(s) = \dfrac{1}{s(s + 2)}. \end{equation*}
We’ll find \(f(t) = \ilap{F(s)}\) so we can invert \(Y(s)\text{.}\)
πŸ”—
Solution. Partial Fraction Decomposition of \(F(s)\)
Decompose:
\begin{equation*} \dfrac{1}{s(s + 2)} = \dfrac{A}{s} + \dfrac{B}{s + 2}. \end{equation*}
Multiply through:
\begin{equation*} 1 = A(s + 2) + B(s). \end{equation*}
Sub in convenient \(s\)-values:
  • At \(s = 0\text{:}\) \(1 = 2A\) β†’ \(A = \dfrac{1}{2}\)
    πŸ”—
  • At \(s = -2\text{:}\) \(1 = -2B\) β†’ \(B = -\dfrac{1}{2}\)
    πŸ”—
Thus:
\begin{equation*} F(s) = \dfrac{1}{2}\dfrac{1}{s} - \dfrac{1}{2}\dfrac{1}{s+2}. \end{equation*}
πŸ”—
Take the inverse Laplace:
\begin{equation*} f(t) = \ilap{F(s)} = \dfrac{1}{2} - \dfrac{1}{2}e^{-2t}. \end{equation*}
πŸ”—
πŸ”—
πŸ”—
Step 3 β€” Back to the Time Domain.
Each exponential factor in \(Y(s)\) corresponds to a shifted version of \(f(t)\) multiplied by the appropriate step function using L\(_11\).
πŸ”—
\begin{equation*} \ilap{Y(s)} = 3f(t) - 3f(t-1)u_1(t) + f(t-4)u_4(t). \end{equation*}
Substitute \(f(t) = \dfrac{1}{2} - \dfrac{1}{2}e^{-2t}\) into each term:
πŸ”—
\begin{equation*} y(t) = 3\left(\dfrac{1}{2} - \dfrac{1}{2}e^{-2t}\right) - 3\left(\dfrac{1}{2} - \dfrac{1}{2}e^{-2(t-1)}\right)u_1(t) + \left(\dfrac{1}{2} - \dfrac{1}{2}e^{-2(t-4)}\right)u_4(t). \end{equation*}
This simplifies to the complete solution for all \(t \ge 0\text{.}\) It clearly shows how each piece of the forcing term triggers its own shifted response in the solution.
πŸ”—
πŸ”—
πŸ”—

Using the Definition of the Laplace Transform

1.
Solution.
\begin{align*} D(s) \amp = \lap{ d(t) } \\ \amp = \int_0^{\infty}e^{-st}d(t) dt \\ \amp = \int_0^3 e^{-st}\cdot 7 dt + \int_3^{\infty} e^{-st}\cdot 0 dt \\ \amp = 7\int_0^3 e^{-st}dt + 0 \\ \amp = 7 \cdot \left[\dfrac{1}{-s}e^{-st}\right]_0^3 \\ \amp = 7 \cdot \left[\left(\dfrac{1}{-s}e^{-s\cdot 3}\right) - \left(\dfrac{1}{-s}e^{-s\cdot 0}\right)\right] \\ \amp = 7 \cdot \left[\left(\dfrac{1}{-s}e^{-s\cdot 3}\right) - \left(\dfrac{1}{-s}\cdot 1\right)\right] \\ \amp = 7 \cdot \left[-\dfrac{1}{s}e^{-3s} +\dfrac{1}{s}\cdot 1\right] \\ \amp = 7 \cdot \left[-\dfrac{1}{s}e^{-3s} +\dfrac{1}{s}\right] \\ \amp = \dfrac{7}{s} - \dfrac{7}{s}e^{-3s} \end{align*}
πŸ”—
πŸ”—

V Systems of Differential Equations
12 First-Order Linear Systems
Chapter 12 Exercises

Exercises

1.
Solution.
We can take the LT of each DE. The result is two algebraic equations in two unknowns \(\ds X(s) \) an \(\ds Y(s). \) We will solve the equations simultaneously as shown below.
\begin{align*} \frac{dx}{dt} \amp = -x+y \amp \frac{dy}{dt} \amp = 2x \\ \lap{\frac{dx}{dt}} \amp = \lap{-x+y} \amp \lap{\frac{dy}{dt}} \amp = \lap{2x} \\ sX(s) - x(0) \amp = -\lap{x} + \lap{y} \amp sY(s) - y(0) \amp = 2 \lap{x} \\ sX(s) - 0 \amp = -X(s) + Y(s) \amp sY(s) - 1 \amp = 2X(s) \\ Y(s) \amp = sX(s) + X(s) \amp \amp \\ \amp \amp s\Big[ sX(s) + X(s) \Big] - 1 \amp = 2X(s) \\ \amp \amp s^2 X(s) + sX(s) - 1 \amp = 2X(s) \\ \amp \amp s^2X(s) + sX(s) - 2X(s) \amp = 1 \\ \amp \amp X(s) [s^2 + s - 2] \amp = 1 \\ \amp \amp X(s) \amp = \frac{1}{s^2 + s - 2} \\ \amp \amp \amp = \frac{1}{(s+2)(s-1)} \\ Y(s) \amp = sX(s) + X(s) \amp \amp \\ \amp = X(s)[s+1] \amp \amp \\ \amp = \frac{1}{(s+2)(s-1)}\cdot (s+1) \amp \amp \\ \amp = \frac{s+1}{(s+2)(s-1)} \amp \amp \end{align*}
We need only take the inverse LT of each function in order to solve for the desired function \(\ds x(t) \) an \(\ds y(t). \) This means we will need to find a partial fraction decomposition for each.
\begin{align*} \frac{1}{(s+2)(s-1)} \amp = \frac{A}{s+2} + \frac{B}{s-1} \\ 1 \amp = A(s-1) + B(s+2) \\ 0s + 1 \amp = As - A + Bs + 2B \\ \amp = (A+B)s + (-A + 2B) \end{align*}
\begin{align*} A+B \amp = 0 \amp -A + 2B \amp = 1 \\ B \amp = -A \amp \amp \\ \amp \amp -A + 2(-A) \amp = 1 \\ \amp \amp -3A \amp = 1 \\ \amp \amp A \amp = -\frac{1}{3} \\ B \amp = -\left( -\frac{1}{3} \right)\amp \amp \\ \amp = \frac{1}{3}\amp \amp \end{align*}
Hence,
\begin{align*} X(s) \amp = \frac{-\frac{1}{3}}{s+2} + \frac{\frac{1}{3}}{s-1}\\ \amp = \frac{1}{3}\left[ \frac{1}{s-1} - \frac{1}{s+2} \right] \end{align*}
Similarly, we will find a partial fraction decomposition for\(\ds Y(s). \)
\begin{align*} \frac{s+1}{(s+2)(s-1)} \amp = \frac{A}{s+2} + \frac{B}{s-1} \\ s + 1 \amp = A(s-1) + B(s+2) \\ \amp = As - A + Bs + 2B \\ \amp = (A+B)s + (-A + 2B) \end{align*}
\begin{align*} A+B \amp = 1 \amp -A + 2B \amp = 1 \\ B \amp = 1-A \amp \amp \\ \amp \amp -A + 2(1-A) \amp = 1 \\ \amp \amp -3A + 2 \amp = 1 \\ \amp \amp -3A \amp = -1 \\ \amp \amp A \amp = \frac{1}{3} \\ B \amp = 1-\left( \frac{1}{3} \right)\amp \amp \\ \amp = \frac{2}{3}\amp \amp \end{align*}
Hence,
\begin{align*} Y(s) \amp = \frac{\frac{1}{3}}{s+2} + \frac{\frac{2}{3}}{s-1}\\ \amp = \frac{1}{3}\left[ \frac{1}{s+2} + 2\cdot \frac{1}{s-1} \right] \end{align*}
Now we need only find the inverse LT of equations.
\begin{align*} x(t) \amp = \lap^{-1}\left\{ X(s) \right\} \\ \amp = \lap^{-1}\left\{ \frac{1}{3}\left[ \frac{1}{s-1} - \frac{1}{s+2} \right] \right\} \\ \amp = \frac{1}{3}\left[ \lap^{-1}\left\{\frac{1}{s-1}\right\} - \lap^{-1}\left\{\frac{1}{s+2}\right\} \right] \\ \amp = \frac{1}{3}\left[ e^t - e^{-2t} \right] \\ y(t) \amp = \lap^{-1}\left\{ Y(s) \right\} \\ \amp = \lap^{-1}\left\{ \frac{1}{3}\left[ \frac{1}{s+2} +2\cdot\frac{1}{s-1} \right] \right\} \\ \amp = \frac{1}{3}\left[ \lap^{-1}\left\{\frac{1}{s+2}\right\} +2 \lap^{-1}\left\{\frac{1}{s-1}\right\} \right] \\ \amp = \frac{1}{3}\left[ e^{-2t} + 2e^{t} \right] \end{align*}
This, the solution to this system is
\begin{align*} x(t) \amp = \frac{1}{3}e^t - \frac{1}{3}e^{-2t} \\ y(t) \amp = \frac{1}{3}e^{-2t} + \frac{2}{3}e^t \end{align*}
We can verify our solution.
\begin{align*} \mbox{LHS of first DE} \amp = \frac{dx}{dt} \\ \amp = \frac{d}{dt}\left( \frac{1}{3}e^t - \frac{1}{3}e^{-2t} \right) \\ \amp = \frac{1}{3}e^{t} + \frac{2}{3}e^{-2t} \\ \mbox{RHS of first DE} \amp = -x+y \\ \amp = -\left( \frac{1}{3}e^t - \frac{1}{3}e^{-2t} \right) + \left( \frac{1}{3}e^{-2t} + \frac{2}{3}e^t \right) \\ \amp = \frac{1}{3}e^t + \frac{2}{3}e^{-2t} \\ \amp = \mbox{LHS of first DE} \\ \mbox{LHS of second DE} \amp = \frac{dy}{dt} \\ \amp = \frac{d}{dt}\left( \frac{1}{3}e^{-2t} + \frac{2}{3}e^t \right) \\ \amp = -\frac{2}{3}e^{-2t} + \frac{2}{3}e^t \\ \mbox{RHS of second DE} \amp = 2x \\ \amp = 2\left( \frac{1}{3}e^t - \frac{1}{3}e^{-2t} \right) \\ \amp = \frac{2}{3}e^t - \frac{2}{3}e^{-2t} \\ \amp = \mbox{LHS of second DE} \\ x(0) \amp = \frac{1}{3}e^0 - \frac{1}{3}e^{-2\cdot 0} \\ \amp = \frac{1}{3}\cdot 1 - \frac{1}{3}\cdot 1 \\ \amp = 0 \\ y(0) \amp = \frac{1}{3}e^{-2\cdot 0} + \frac{2}{3}e^0 \\ \amp = \frac{1}{3}\cdot 1 + \frac{2}{3}\cdot 1 \\ \amp = 1 \end{align*}
Hence, the solution satisfies both DEs and both initial conditions.
πŸ”—
πŸ”—
2.
Solution.
We can take the LT of each DE, starting with the first DE. We will then solve for\(\ds X(s). \)
\begin{align*} \frac{dx}{dt} \amp = x-2y\\ \lap{\frac{dx}{dt}} \amp = \lap{x-2y}\\ sX(s) - x(0) \amp = \lap{x} - 2 \lap{y}\\ sX(s) + 1 \amp = X(s) - 2Y(s)\\ sX(s) - X(s) \amp = -2Y(s) - 1\\ X(s)[s-1] \amp = -2Y(s) - 1\\ X(s) \amp = \frac{-2}{s-1}Y(s) - \frac{1}{s-1} \end{align*}
Now we take the LT of the second DE. We will substitute in for\(\ds X(s) \) using the result we found above.
\begin{align*} \frac{dy}{dt} \amp = 5x - y \\ \lap{\frac{dy}{dt}} \amp = \lap{5x - y} \\ sY(s) - y(0) \amp = 5\lap{x} - \lap{y} \\ sY(s) - 2 \amp = 5X(s) - Y(s) \\ sY(s) - 2 \amp = 5\left[ \frac{-2}{s-1}Y(s) - \frac{1}{s-1} \right] - Y(s) \\ sY(s) - 2 \amp = \frac{-10}{s-1}Y(s) - \frac{5}{s-1}- Y(s) \\ (s-1) \cdot \left( sY(s) - 2 \right) \amp = \left( \frac{-10}{s-1}Y(s) - \frac{5}{s-1}- Y(s) \right) \cdot (s-1) \\ (s^2 - s)Y(s) - 2s + 2 \amp = -10Y(s) - 5 - (s-1)Y(s) \\ (s^2 - s)Y(s) + 10Y(s) + (s-1)Y(s) \amp = 2s - 2 - 5 \\ Y(s)[s^2 - s + 10 + s - 1] \amp = 2s - 7 \\ Y(s)[s^2 + 9] \amp = 2s - 7 \\ Y(s) \amp = \frac{2s - 7}{s^2 + 9} \end{align*}
We now substitute this back into equation.
\begin{align*} X(s) \amp = \frac{-2}{s-1}Y(s) - \frac{1}{s-1} \\ \amp = \frac{-2}{s-1}\cdot\frac{2s - 7}{s^2 + 9} - \frac{1}{s-1} \\ \amp = \frac{-4s+14}{(s-1)(s^2 + 9)} - \frac{1}{s-1}\cdot \frac{s^2 + 9}{s^2 + 9} \\ \amp = \frac{(-4s + 14) - (s^2 + 9) }{(s-1)(s^2 + 9)} \\ \amp = \frac{-4s + 14 - s^2 - 9) }{(s-1)(s^2 + 9)} \\ \amp = \frac{-s^2 - 4s + 5}{(s-1)(s^2 + 9)} \\ \amp = \frac{-(s^2 + 4s - 5)}{(s-1)(s^2 + 9)} \\ \amp = \frac{-(s+5)(s-1)}{(s-1)(s^2 + 9)} \\ \amp = \frac{-(s+5)}{s^2 + 9} \\ \amp = \frac{-s-5}{s^2 + 9} \end{align*}
We need only take the inverse LT of each function in order to solve for the desired functions \(\ds x(t) \) an \(\ds y(t). \)
\begin{align*} x(t) \amp = \lap^{-1}\left\{ X(s) \right\} \\ \amp = \lap^{-1}\left\{ \frac{-s-5}{s^2 + 9} \right\} \\ \amp = - \lap^{-1}\left\{ \frac{s}{s^2 + 9} \right\} - 5 \lap^{-1}\left\{ \frac{1}{s^2 + 9} \right\} \\ \amp = - \cos(3t) - 5 \lap^{-1}\left\{ \frac{1}{s^2 + 9} \right\}\cdot \frac{3}{3} \\ \amp = - \cos(3t) - 5 \lap^{-1}\left\{ \frac{3}{s^2 + 3^2} \right\}\cdot \frac{1}{3} \\ \amp = - \cos(3t) - \frac{5}{3} \sin(3t) \end{align*}
\begin{align*} y(t) \amp = \lap^{-1}\left\{ Y(s) \right\} \\ \amp = \lap^{-1}\left\{ \frac{2s - 7}{s^2 + 9} \right\} \\ \amp = 2\lap^{-1}\left\{ \frac{s}{s^2 + 9} \right\} - 7\lap^{-1}\left\{ \frac{1}{s^2 + 9} \right\} \\ \amp = 2\cos(3t) - 7\lap^{-1}\left\{ \frac{1}{s^2 + 9} \right\}\cdot \frac{3}{3} \\ \amp = 2\cos(3t) - 7\lap^{-1}\left\{ \frac{3}{s^2 + 3^2} \right\}\cdot \frac{1}{3} \\ \amp = 2\cos(3t) - \frac{7}{3}\sin(3t) \end{align*}
Thus, the solution to this system is
\begin{align*} x(t) \amp = -\cos(3t) - \frac{5}{3} \sin(3t) \\ y(t) \amp = 2\cos(3t) - \frac{7}{3}\sin(3t) \end{align*}
We can verify our solution.
\begin{align*} \mbox{LHS of first DE} \amp = \frac{dx}{dt} \\ \amp = \frac{d}{dt}\left( -\cos(3t) - \frac{5}{3} \sin(3t) \right) \\ \amp = 3\sin(3t) - 5\cos(3t) \\ \mbox{RHS of first DE} \amp = x-2y \\ \amp = \left( -\cos(3t) - \frac{5}{3} \sin(3t) \right) -2 \left( 2\cos(3t) - \frac{7}{3}\sin(3t) \right) \\ \amp = -\cos(3t) - \frac{5}{3} \sin(3t) - 4\cos(3t) + \frac{14}{3}\sin(3t) \\ \amp = -5\cos(3t) + \frac{9}{3}\sin(3t) \\ \amp = -5\cos(3t) + 3\sin(3t) \\ \amp = \mbox{LHS of first DE} \\ \mbox{LHS of second DE} \amp = \frac{dy}{dt} \\ \amp = \frac{d}{dt}\left( 2\cos(3t) - \frac{7}{3}\sin(3t) \right) \\ \amp = -6\sin(3t) - 7\cos(3t) \\ \mbox{RHS of second DE} \amp = 5x - y \\ \amp = 5\left( -\cos(3t) - \frac{5}{3} \sin(3t) \right) - \left( 2\cos(3t) - \frac{7}{3}\sin(3t) \right) \\ \amp = -5\cos(3t) - \frac{25}{3} \sin(3t) - 2\cos(3t) + \frac{7}{3}\sin(3t) \\ \amp = -7\cos(3t)- \frac{18}{3} \sin(3t) \\ \amp = -7\cos(3t) - 6\sin(3t) \\ \amp = \mbox{LHS of second DE} \\ x(0) \amp = -\cos(3\cdot 0) - \frac{5}{3} \sin(3\cdot 0) \\ \amp = -1 - 0 \\ \amp = -1 \\ y(0) \amp = 2\cos(3\cdot 0) - \frac{7}{3}\sin(3\cdot 0) \\ \amp = 2\cdot 1 - 0 \\ \amp = 2 \end{align*}
Hence, the solution satisfies both DEs and both initial conditions.
πŸ”—
πŸ”—
3.
Solution.
We can take the LT of each DE, starting with the first DE. We will then solve for\(\ds X(s). \)
\begin{align*} y' - 2x \amp = 1\\ \lap{ y' - 2x } \amp = \lap{ 1 }\\ sY(s) -y(0) - 2X(s) \amp = \frac{1}{s}\\ sY(s) - 0 - 2X(s) \amp = \frac{1}{s}\\ sY(s) \amp = \frac{1}{s}+ 2X(s)\\ Y(s) \amp = \frac{1}{s^2} + \frac{2}{s}X(s) \end{align*}
Now we take the LT of the second DE. We will substitute in for\(\ds Y(s) \) using the result we found above.
\begin{align*} x' + y' - 3x - 3y \amp = 2 \\ \lap{ x' + y' - 3x - 3y } \amp = \lap{2} \\ sX(s) - x(0) + sY(s) - y(0) - 3X(s) - 3Y(s) \amp = \frac{2}{s} \\ sX(s) - 0 + sY(s) - 0 - 3X(s) - 3Y(s) \amp = \frac{2}{s} \\ sX(s) - 3X(s) + (s-3) Y(s) \amp = \frac{2}{s} \\ sX(s) - 3X(s) + (s-3)\left[ \frac{1}{s^2} + \frac{2}{s}X(s) \right] \amp = \frac{2}{s} \\ sX(s) - 3X(s) + \frac{s-3}{s^2} + \frac{2(s-3)}{s}X(s) \amp = \frac{2}{s} \\ sX(s) - 3X(s) + \frac{2(s-3)}{s}X(s) \amp = \frac{2}{s} - \frac{s-3}{s^2} \\ s^2 \cdot \left[sX(s) - 3X(s) + \frac{2(s-3)}{s}X(s)\right] \amp = s^2\cdot \left[\frac{2}{s} - \frac{s-3}{s^2}\right] \\ s^3X(s) - 3s^2X(s) + 2s(s-3)X(s) \amp = 2s - (s-3) \\ X(s) [s^3 - 3s^2 + (2s^2 - 6s)] \amp = 2s - s + 3 \\ X(s) [s^3 - s^2 - 6s] \amp = s+3 \\ X(s) \amp = \frac{s+3}{s^3 - s^2 - 6s} \\ \amp = \frac{s+3}{s(s^2 - s - 6)} \\ \amp = \frac{s+3}{s(s-3)(s+2)} \end{align*}
We now substitute this back into equation.
\begin{align*} Y(s) \amp = \frac{1}{s^2} + \frac{2}{s}X(s) \\ \amp = \frac{1}{s^2} + \frac{2}{s}\cdot \frac{s+3}{s(s-3)(s+2)} \\ \amp = \frac{1}{s^2} + \frac{2(s+3)}{s^2(s-3)(s+2)} \\ \amp = \frac{1}{s^2}\cdot \frac{(s-3)(s+2)}{(s-3)(s+2)} + \frac{2(s+3)}{s^2(s-3)(s+2)} \\ \amp = \frac{(s-3)(s+2) + 2(s+3)}{s^2(s-3)(s+2)} \\ \amp = \frac{s^2 - s - 6 + 2s + 6}{s^2(s-3)(s+2)} \\ \amp = \frac{s^2 + s}{s^2(s-3)(s+2)} \\ \amp = \frac{s(s+1)}{s^2(s-3)(s+2)} \\ \amp = \frac{s+1}{s(s-3)(s+2)} \end{align*}
We need only take the inverse LT of each function in order to solve for the desired function \(\ds x(t) \) an \(\ds y(t). \) This means we will need to find a partial fraction decomposition for each.
\begin{align*} \frac{s+3}{s(s-3)(s+2)} \amp = \frac{A}{s} + \frac{B}{s-3} + \frac{C}{s+2} \\ s+3 \amp = A(s-3)(s+2) + B(s)(s+2) + C(s)(s-3) \\ 0s^2 + s + 3 \amp = A(s^2 - s - 6) + B(s^2 + 2s) + C(s^2 - 3s) \\ \amp = As^2 - As - 6A + Bs^2 + 2Bs + Cs^2 - 3Cs \\ \amp = (A+B+C)s^2 + (-A + 2B- 3C)s + (-6A) \end{align*}
\begin{align*} A+B+C \amp = 0 \amp -A + 2B - 3C \amp = 1 \amp -6A \amp = 3 \\ \amp \amp \amp \amp A \amp = -\frac{1}{2} \\ B + C \amp = -A \amp 2B - 3C \amp = 1 + A \amp \amp \\ B + C \amp = -\left( -\frac{1}{2} \right) \amp 2B - 3C \amp = 1 + \left( -\frac{1}{2} \right) \amp \amp \\ B + C \amp = \frac{1}{2} \amp 2B - 3C \amp = \frac{1}{2} \amp \amp \\ C \amp = \frac{1}{2} - B \amp \amp \amp \amp \\ \amp \amp 2B - 3\left( \frac{1}{2} - B \right) \amp = \frac{1}{2} \amp \amp \\ \amp \amp 2B - \frac{3}{2} + 3B \amp = \frac{1}{2} \amp \amp \\ \amp \amp 5B \amp = \frac{1}{2} + \frac{3}{2} \amp \amp \\ \amp \amp 5B \amp = 2 \amp \amp \\ \amp \amp B \amp = \frac{2}{5} \amp \amp \\ C \amp = \frac{1}{2} - B \amp \amp \amp \amp \\ \amp = \frac{1}{2} - \frac{2}{5} \amp \amp \amp \amp \\ \amp = \frac{1}{10} \amp \amp \amp \amp \end{align*}
Hence,
\begin{align*} X(s) \amp = \frac{-\frac{1}{2}}{s} + \frac{\frac{2}{5}}{s-3} + \frac{\frac{1}{10}}{s+2}\\ \amp = -\frac{1}{2}\cdot \frac{1}{s} + \frac{2}{5}\cdot \frac{1}{s-3} + \frac{1}{10}\cdot \frac{1}{s+2} \end{align*}
Similarly, we will find a partial fraction decomposition for\(\ds Y(s). \)
\begin{align*} \frac{s+1}{s(s-3)(s+2)} \amp = \frac{A}{s} + \frac{B}{s-3} + \frac{C}{s+2} \\ s+1 \amp = A(s-3)(s+2) + B(s)(s+2) + C(s)(s-3) \\ 0s^2 + s + 1 \amp = A(s^2 - s - 6) + B(s^2 + 2s) + C(s^2 - 3s) \\ \amp = As^2 - As - 6A + Bs^2 + 2Bs + Cs^2 - 3Cs \\ \amp = (A+B+C)s^2 + (-A + 2B- 3C)s + (-6A) \end{align*}
\begin{align*} A+B+C \amp = 0 \amp -A + 2B - 3C \amp = 1 \amp -6A \amp = 1 \\ \amp \amp \amp \amp A \amp = -\frac{1}{6} \\ B + C \amp = -A \amp 2B - 3C \amp = 1 + A \amp \amp \\ B + C \amp = -\left( -\frac{1}{6} \right) \amp 2B - 3C \amp = 1 + \left( -\frac{1}{6} \right) \amp \amp \\ B + C \amp = \frac{1}{6} \amp 2B - 3C \amp = \frac{5}{6} \amp \amp \\ C \amp = \frac{1}{6} - B \amp \amp \amp \amp \\ \amp \amp 2B - 3\left( \frac{1}{6} - B \right) \amp = \frac{5}{6} \amp \amp \\ \amp \amp 2B - \frac{1}{2} + 3B \amp = \frac{5}{6} \amp \amp \\ \amp \amp 5B \amp = \frac{5}{6} + \frac{1}{2} \amp \amp \\ \amp \amp 5B \amp = \frac{4}{3} \amp \amp \\ \amp \amp B \amp = \frac{4}{15} \amp \amp \\ C \amp = \frac{1}{6} - B \amp \amp \amp \amp \\ \amp = \frac{1}{6} - \frac{4}{15}\amp \amp \amp \amp \\ \amp = \frac{5}{30} - \frac{8}{30}\amp \amp \amp \amp \\ \amp = -\frac{1}{10} \amp \amp \amp \amp \end{align*}
Hence,
\begin{align*} Y(s) \amp = \frac{-\frac{1}{6}}{s} + \frac{\frac{4}{15}}{s-3} + \frac{-\frac{1}{10}}{s+2}\\ \amp = -\frac{1}{6}\cdot \frac{1}{s} + \frac{4}{15}\cdot \frac{1}{s-3} - \frac{1}{10}\cdot \frac{1}{s+2} \end{align*}
Now we need only find the inverse LT of equations.
\begin{align*} x(t) \amp = \lap^{-1}\left\{ X(s) \right\} \\ \amp = \lap^{-1}\left\{ -\frac{1}{2}\cdot \frac{1}{s} + \frac{2}{5}\cdot \frac{1}{s-3} + \frac{1}{10}\cdot \frac{1}{s+2} \right\} \\ \amp = -\frac{1}{2}\lap^{-1}\left\{ \frac{1}{s} \right\} + \frac{2}{5} \lap^{-1}\left\{ \frac{1}{s-3} \right\} + \frac{1}{10}\lap^{-1}\left\{ \frac{1}{s+2} \right\} \\ \amp = -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t} \\ y(t) \amp = \lap^{-1}\left\{ Y(s) \right\} \\ \amp = \lap^{-1}\left\{ -\frac{1}{6}\cdot \frac{1}{s} + \frac{4}{15}\cdot \frac{1}{s-3} - \frac{1}{10}\cdot \frac{1}{s+2} \right\} \\ \amp = -\frac{1}{6}\lap^{-1}\left\{ \frac{1}{s} \right\} + \frac{4}{15}\lap^{-1}\left\{ \frac{1}{s-3} \right\} - \frac{1}{10} \lap^{-1}\left\{ \frac{1}{s+2} \right\} \\ \amp = -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t} \end{align*}
Thus, the solution to this system is
\begin{align*} x(t) \amp = -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t} \\ y(t) \amp = -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t} \end{align*}
We can verify that this is the solution.
\begin{align*} \mbox{LHS of first DE} \amp = y' - 2x \\ \amp = \frac{d}{dt}\left( -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t} \right) - 2\Big( -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t} \Big) \\ \amp = \left( 0 + \frac{4}{5}e^{3t} + \frac{1}{5} e^{-2t} \right) + 1 - \frac{4}{5}e^{3t} - \frac{1}{5}e^{-2t} \\ \amp = 1 \\ \amp = \mbox{RHS of first DE} \\ \mbox{RHS of second DE} \amp = x' + y' - 3x - 3y \\ \amp = \frac{d}{dt}\left( -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t} \right) + \frac{d}{dt}\left( -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t} \right) \\ \amp \mbox{}\hspace{1cm} - 3 \left( -\frac{1}{2} + \frac{2}{5}e^{3t} + \frac{1}{10}e^{-2t} \right) - 3 \left( -\frac{1}{6}+ \frac{4}{15}e^{3t} - \frac{1}{10} e^{-2t} \right) \\ \amp = \left(0 + \frac{6}{5}e^{3t} - \frac{1}{5}e^{-2t} \right) + \left( 0 + \frac{4}{5}e^{3t} + \frac{1}{5} e^{-2t} \right) \\ \amp \mbox{}\hspace{1cm} + \frac{3}{2} - \frac{6}{5}e^{3t} - \frac{3}{10}e^{-2t} + \frac{1}{2} - \frac{4}{5}e^{3t} + \frac{3}{10} e^{-2t} \\ \amp = \left( \frac{6}{5} + \frac{4}{5} - \frac{6}{5} - \frac{4}{5} \right)e^{3t} + \left( - \frac{1}{5} + \frac{1}{5} - \frac{3}{10} + \frac{3}{10} \right)e^{2t} + \left( \frac{3}{2} + \frac{1}{2} \right) \\ \amp = 0 + 0 + 2 \\ \amp = 2 \\ \amp = \mbox{RHS of second DE} \\ x(0) \amp = -\frac{1}{2} + \frac{2}{5}e^{3\cdot 0} + \frac{1}{10}e^{-2\cdot 0} \\ \amp = -\frac{1}{2} + \frac{2}{5} + \frac{1}{10} \\ \amp = -\frac{5}{10} + \frac{4}{10} + \frac{1}{10} \\ \amp = 0 \\ y(0) \amp = -\frac{1}{6} + \frac{4}{15}e^{3\cdot 0} - \frac{1}{10} e^{-2\cdot 0} \\ \amp = -\frac{1}{6} + \frac{4}{15} - \frac{1}{10} \\ \amp = -\frac{5}{30} + \frac{8}{30} - \frac{3}{30} \\ \amp = 0 \end{align*}
Hence, the solution satisfies both DEs and both initial conditions.
πŸ”—
πŸ”—
4.
Solution.
Find the mass of sugar in the tank afte \(\ds t \) minutes.
πŸ”—
πŸ”—
5.
Solution.
Find the concentration of sugar in the tank after 10 minutes.
πŸ”—
πŸ”—
8. From a Second-Order Equation to a System.
Solution.
Let \(u = y\) and \(v = y'\text{.}\) Then:
πŸ”—
\begin{equation*} u' = v \end{equation*}
and
πŸ”—
\begin{equation*} v' = y'' = y = u \end{equation*}
So the system becomes:
πŸ”—
\begin{align*} u' \amp = v \\ v' \amp = u \end{align*}
πŸ”—

13 Numerical & Qualitative Methods for Linear Systems
Chapter 13 Exercises

Exercises

1.
Solution.
Compute two iterations of Euler’s method using step size \(\ds h = 0.1. \)
πŸ”—
πŸ”—
3. Euler’s Method for Systems.
Solution.
\begin{align*} \mbox{Preliminaries: } \amp \amp f_1(t,x,y) \amp = 2x - y + t \\ \amp \amp f_2(t,x,y) \amp = x \\ \amp \amp \Delta t \amp = 0.1 \\ \amp \amp t_0 \amp = 0 \\ \amp \amp x_0 \amp = 6 \\ \amp \amp y_0 \amp = 2 \\ \amp \amp \amp \mbox{Iteration 1:} \amp \amp t_1 \amp = t_0 + \Delta t \\ \amp \amp \amp = 0 + 0.1 \\ \amp \amp \amp = 0.1 \\ \amp \amp x_1 \amp = x_0 + \Delta t \cdot f_1(t_0,x_0,y_0) \\ \amp \amp \amp = 6 + 0.1 \cdot f_1(0,6,2) \\ \amp \amp \amp = 6 + 0.1 \cdot [2(6) - 2 + 0] \\ \amp \amp \amp = 7 \\ \amp \amp y_1 \amp = y_0 + \Delta t\cdot f_2(t_0,x_0,y_0) \\ \amp \amp \amp = 2 + 0.1 \cdot f_2(0,6,2) \\ \amp \amp \amp = 2 + 0.1 \cdot [6] \\ \amp \amp \amp = 2.6 \\ \amp \amp \amp \mbox{Iteration 2:} \amp \amp t_2 \amp = t_1 + \Delta t \\ \amp \amp \amp = 0.1 + 0.1 \\ \amp \amp \amp = 0.2 \\ \amp \amp x_2 \amp = x_1 + \Delta t \cdot f_1(t_1,x_1,y_1) \\ \amp \amp \amp = 7 + 0.1 \cdot f_1(0.1,7,2.6) \\ \amp \amp \amp = 7 + 0.1 \cdot [2(7) - 2.6 + 0.1] \\ \amp \amp \amp = 8.15 \\ \amp \amp y_2 \amp = y_1 + \Delta t\cdot f_2(t_1,x_1,y_1) \\ \amp \amp \amp = 2.6 + 0.1 \cdot f_2(0.1,7,2.6) \\ \amp \amp \amp = 2.6 + 0.1 \cdot [7] \\ \amp \amp \amp = 3.3 \end{align*}
Therefore \(\ds x(0.2) \approx 8.15 \) an \(\ds y(0.2) \approx 3.3. \)
πŸ”—
πŸ”—
6.
Solution.
We have one second-order and one first-order DE, which means we will need three variables to generate a systemof three first-order DEs.\\ Let \(\ds u = y \) \(\ds v = y', \) an \(\ds w = x \) . Then we have the following.
\begin{align*} u' \amp = y' = v \label{eq14-9} \end{align*}
Substituting into the first DE yields the following.
\begin{align*} y'' + x \amp = 12 \nonumber \\ v' + w \amp = 12 \nonumber \\ v' \amp = 12 - w \label{eq14-10} \end{align*}
Substituting into the second DE yields the following.
\begin{align*} w' + 2ut \amp = \cos t \nonumber \\ w' \amp = \cos t - 2ut \label{eq14-11} \end{align*}
Then equations (\ref{eq14-9}), (\ref{eq14-10}) and (\ref{eq14-11}) constitute a system of first-order DEs in the variable \(\ds u \) \(\ds v \) , and \(\ds w \) , as below:
\begin{align*} u' \amp = v \\ v' \amp = 12 - w \\ w' \amp = \cos t - 2ut \end{align*}
πŸ”—
πŸ”—
7.
Solution.
First we convert the second-order DE to a system of two first-order DEs. Let \(\ds u = y \) and \(\ds v = y' \) . Then \begin{equation} u’ = y’ = v\label{eq14-12} \end{equation} and \begin{equation} v’ = (y’)’ = y’’.\label{eq14-13} \end{equation} Substituting equations (\ref{eq14-12}) and (\ref{eq14-13}) into the DE yields the following.
\begin{align*} y'' + 2y' + y^2 \amp = 0 \nonumber \\ v' + 2v + u^2 \amp = 0 \nonumber \\ v' \amp = -2v - u^2 \label{eq14-14} \end{align*}
Then equations (\ref{eq14-12}) and (\ref{eq14-14}) constitute a system of first-order DEs in the variable \(\ds u \) and \(\ds v \) .
\begin{align*} u' \amp = v \\ v' \amp = -2v - u^2 \end{align*}
Note that we can also convert the initial conditions to the new variables:
\begin{align*} u(0) \amp = 1 \\ v(0) \amp = 2 \end{align*}
Now we can use Euler’s method as follows.
\begin{align*} \mbox{Preliminaries: } \amp \amp f_1(t,u,v) \amp = v \\ \amp \amp f_2(t,u,v) \amp = -2v-u^2 \\ \amp \amp h \amp = 0.1 \\ \amp \amp t_0 \amp = 0 \\ \amp \amp u_0 \amp = 1 \\ \amp \amp v_0 \amp = 2 \\ \amp \amp \amp \mbox{Iteration 1:} \amp \amp t_1 \amp = t_0 + h \\ \amp \amp \amp = 0 + 0.1 \\ \amp \amp \amp = 0.1 \\ \amp \amp u_1 \amp = u_0 + h \cdot f_1(t_0,u_0,v_0) \\ \amp \amp \amp = 1 + 0.1 \cdot f_1(0,1,2) \\ \amp \amp \amp = 1 + 0.1 \cdot [2] \\ \amp \amp \amp = 1.2 \\ \amp \amp v_1 \amp = v_0 + h\cdot f_2(t_0,u_0,v_0) \\ \amp \amp \amp = 2 + 0.1 \cdot f_2(0,1,2) \\ \amp \amp \amp = 2 + 0.1 \cdot [-2(2) - (1)^2] \\ \amp \amp \amp = 1.5 \\ \amp \amp \amp \mbox{Iteration 2:} \amp \amp t_2 \amp = t_1 + h \\ \amp \amp \amp = 0.1 + 0.1 \\ \amp \amp \amp = 0.2 \\ \amp \amp u_2 \amp = u_1 + h \cdot f_1(t_1,u_1,v_1) \\ \amp \amp \amp = 1.2 + 0.1 \cdot f_1(0.1, 1.2, 1.5) \\ \amp \amp \amp = 1.2 + 0.1 \cdot [1.5] \\ \amp \amp \amp = 1.35 \\ \amp \amp v_2 \amp = v_1 + h\cdot f_2(t_1,u_1,v_1) \\ \amp \amp \amp = 1.5 + 0.1 \cdot f_2(0.1, 1.2, 1.5) \\ \amp \amp \amp = 1.5 + 0.1 \cdot [-2(1.5) - (1.2)^2] \\ \amp \amp \amp = 1.0560 \end{align*}
Therefore \(\ds u(0.2) \approx 1.35 \) an \(\ds v(0.2) \approx 1.0560. \) This mean \(\ds y(0.2) \approx 1.35 \) (and, even though we were not asked for it specifically \(\ds y'(0.2) \approx 1.0560 \) ).
πŸ”—
πŸ”—
8.
Solution.
First we convert the second-order DE to a system of two first-order DEs. Let \(\ds u = y \) and \(\ds v = y' \) . Then \begin{equation} u’ = y’ = v\label{eq14-15} \end{equation} and \begin{equation} v’ = (y’)’ = y’’.\label{eq14-16} \end{equation} Substituting equations (\ref{eq14-14}) and (\ref{eq14-15}) into the DE yields the following.
\begin{align*} y'' - 2y \amp = e^{t-3} \cos t \nonumber \\ v' - 2u \amp = e^{t-3}\cos t \nonumber \\ v' \amp = e^{t-3}\cos t +2u \label{eq14-17} \end{align*}
Then equations (\ref{eq14-15}) and (\ref{eq14-17}) constitute a system of first-order DEs in the variable \(\ds u \) and \(\ds v \) .
\begin{align*} u' \amp = v \\ v' \amp = e^{t-3}\cos t +2u \end{align*}
Note that we can also convert the initial conditions to the new variables:
\begin{align*} u(3) \amp = -1 \\ v(3) \amp = 0 \end{align*}
Now we can use Euler’s method as follows.
\begin{align*} \mbox{Preliminaries: } \amp \amp f_1(t,u,v) \amp = v \\ \amp \amp f_2(t,u,v) \amp = e^{t-3}\cos t +2u \\ \amp \amp \Delta t \amp = 0.2 \\ \amp \amp t_0 \amp = 3 \\ \amp \amp u_0 \amp = -1 \\ \amp \amp v_0 \amp = 0 \\ \amp \amp \amp \mbox{Iteration 1:} \amp \amp t_1 \amp = t_0 + \Delta t \\ \amp \amp \amp = 3 + 0.2 \\ \amp \amp \amp = 3.2 \\ \amp \amp u_1 \amp = u_0 + \Delta t \cdot f_1(t_0,u_0,v_0) \\ \amp \amp \amp = -1 + 0.2 \cdot f_1(3, -1, 0) \\ \amp \amp \amp = -1 + 0.2 \cdot [0] \\ \amp \amp \amp = -1 \\ \amp \amp v_1 \amp = v_0 + \Delta t \cdot f_2(t_0,u_0,v_0) \\ \amp \amp \amp = 0 + 0.2 \cdot f_2(3, -1, 0) \\ \amp \amp \amp = 0 + 0.2 \cdot [e^{3-3}\cos(3) + 2(-1)] \\ \amp \amp \amp = -0.5980 \end{align*}
Therefore \(\ds u(3.2) \approx -1 \) , which mean \(\ds y(3.2) \approx -1 \) (and, even though we were not asked for it specifically, we also know \(\ds y'(3.2) \approx -0.5980 \) ).
πŸ”—
πŸ”—
9.
Solution.
We will use the system of DEs developed in part \ref{partA_euler} Now we can use Euler’s method as follows.
\begin{align*} \mbox{Preliminaries: } \amp \amp f_1(t,u,v) \amp = v \\ \amp \amp f_2(t,u,v) \amp = e^{t-3}\cos t +2u \\ \amp \amp h \amp = 0.1 \\ \amp \amp t_0 \amp = 3 \\ \amp \amp u_0 \amp = -1 \\ \amp \amp v_0 \amp = 0 \\ \amp \amp \amp \mbox{Iteration 1:} \amp \amp t_1 \amp = t_0 + h \\ \amp \amp \amp = 3 + 0.1 \\ \amp \amp \amp = 3.1 \\ \amp \amp u_1 \amp = u_0 + h \cdot f_1(t_0,u_0,v_0) \\ \amp \amp \amp = -1 + 0.1 \cdot f_1(3, -1, 0) \\ \amp \amp \amp = -1 + 0.1 \cdot [0] \\ \amp \amp \amp = -1 \\ \amp \amp v_1 \amp = v_0 + h\cdot f_2(t_0,u_0,v_0) \\ \amp \amp \amp = 0 + 0.1 \cdot f_2(3, -1, 0) \\ \amp \amp \amp = 0 + 0.1 \cdot [e^{3-3}\cos(3) + 2(-1)] \\ \amp \amp \amp = -0.2990 \\ \amp \amp \amp \mbox{Iteration 2:} \amp \amp t_2 \amp = t_1 + h \\ \amp \amp \amp = 3.1 + 0.1 \\ \amp \amp \amp = 3.2 \\ \amp \amp u_2 \amp = u_1 + h \cdot f_1(t_1,u_1,v_1) \\ \amp \amp \amp = -1 + 0.1 \cdot f_1(3.1, -1, -0.2990) \\ \amp \amp \amp = -1 + 0.1 \cdot [-0.2990] \\ \amp \amp \amp = -1.0299 \\ \amp \amp v_2 \amp = v_1 + h\cdot f_2(t_1,u_1,v_1) \\ \amp \amp \amp = -0.2990 + 0.1 \cdot f_2(3.1, -1, -0.2990) \\ \amp \amp \amp = -0.2990 + 0.1 \cdot [e^{3.1 - 3}\cos(3.1) + 2(-1)] \\ \amp \amp \amp = -0.6094 \end{align*}
Therefore \(\ds u(3.2) \approx -1.0299 \) , which mean \(\ds y(3.2) \approx -1.0299 \) (and, even though we were not asked for it specifically, we also know \(\ds y'(3.2) \approx -0.6094 \) ).\\ %
πŸ”—
πŸ”—
πŸ”—
PrevTopNext
Feedback