Skip to main content
Contents Index
Calc
Dark Mode Prev Up Next Scratch ActiveCode Profile
\(\newcommand\DLGray{\color{Gray}}
\newcommand\DLO{\color{BurntOrange}}
\newcommand\DLRa{\color{WildStrawberry}}
\newcommand\DLGa{\color{Green}}
\newcommand\DLGb{\color{PineGreen}}
\newcommand\DLBa{\color{RoyalBlue}}
\newcommand\DLBb{\color{Cerulean}}
\newcommand\ds{\displaystyle}
\newcommand\ddx{\frac{d}{dx}}
\newcommand\os{\overset}
\newcommand\us{\underset}
\newcommand\ob{\overbrace}
\newcommand\obt{\overbracket}
\newcommand\ub{\underbrace}
\newcommand\ubt{\underbracket}
\newcommand\ul{\underline}
\newcommand\laplacesym{\mathscr{L}}
\newcommand\lap[1]{\laplacesym\left\{#1\right\}}
\newcommand\ilap[1]{\laplacesym^{-1}\left\{#1\right\}}
\newcommand\tikznode[3][]
{\tikz[remember picture,baseline=(#2.base)]
\node[minimum size=0pt,inner sep=0pt,#1](#2){#3};
}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\newcommand{\sfrac}[2]{{#1}/{#2}}
\)
Section Chapter 8 Exercises
Reading Questions α―β
β Quick-Answer Questions
1. Multiple-Choice.
(a) Modifying the Initial Guess.
When solving \(y'' - 2y' + y = e^x\text{,}\) the initial guess for \(y_p\) is modified to \(Ax^2 e^x\) because \(e^x\) overlaps with .
\(c_1 e^x\)
\(c_2 x^2 e^x\)
\(y_h\)
\(c_1 \cos(x)\)
(b) Purpose of the Method.
What is the main purpose of the Method of Undetermined Coefficients?
To solve the homogeneous solution \(y_h\) directly.
To determine the coefficients in the particular solution \(y_p\text{.}\)
To find the characteristic equation roots.
To verify initial conditions are satisfied.
(c) When It Works.
The Method of Undetermined Coefficients applies when
\(f(x)\) is:
A polynomial, exponential, sine, or cosine function.
A logarithmic function like \(\ln x\text{.}\)
A tangent function like \(\tan x\text{.}\)
A sum or product of polynomials, exponentials, and trig functions.
(d) Choose \(y_p\) .
Suppose
\begin{equation*}
y'' - 4y' + 3y = 6e^x\text{.}
\end{equation*}
Which of the following is the correct selection for \(y_p\text{?}\)
\(\ds\quad y_p = Ae^x\)
Incorrect. Remember to consider whether \(e^x\) overlaps with any terms in \(y_h\text{.}\)
\(\ds\quad y_p = Axe^x\)
Correct! Since \(e^x\) is a term in \(y_h\text{,}\) we need to multiply the particular solution by \(x\) to ensure independence.
\(\ds\quad y_p = Ax + B\)
Incorrect. This form is more suited for a polynomial right-hand side like \(f(x) = 6x\text{.}\)
\(\ds\quad y_p = A\)
Incorrect. A constant form of \(y_p\) would only be appropriate if \(f(x)\) were a constant.
(e) Selecting the particular form.
Select the initial form of the particular solution for an LNCC equation with forcing function:
\begin{equation*}
f(x) = 5x^2 + 3x\text{.}
\end{equation*}
\(\ds\quad y_p = Ax^2 + Bx + C\)
Correct! Since the forcing function is a 2nd degree polynomial, the particular solution should be the most general 2nd degree polynomial.
\(\ds\quad y_p = A e^{x} + B\)
Incorrect. The forcing function does not contain an exponential function, so neither should the particular solution.
\(\ds\quad y_p = A x^2 + B x\)
Incorrect. This form is missing a free term.
\(\ds\quad y_p = Ax^3 + Bx^2 + Cx + D\)
Incorrect. While this form technically works, the \(x^3\) term is unnecessary and makes the problem more difficult.
(f) Select the correct form.
What is the most appropriate form for the particular solution to
\begin{equation*}
y'' - 4y' + 3y = 9x + 6\text{?}
\end{equation*}
Correct! The forcing function is a first degree polynomial, so the particular solution should generalize this form as \(Ax + B\text{.}\)
Incorrect. The forcing function is a polynomial, not an exponential function.
Incorrect. The forcing function is a polynomial, not a trigonometric function.
Incorrect. The forcing function is not a constant.
(g) Classifying Practice.
Select each classification label that applies to the equation
\begin{equation*}
y'' = y' + 6y
\end{equation*}
Linear
Correct, each of the terms are linear.
Homogeneous
Correct, the free term is zero.
Constant Coefficients
Correct, each coefficient is constant.
LHCC
Correct!
(h) Classifying Practice.
Select each classification label that applies to the equation
\begin{equation*}
3y''' + y'- \sin(y) = 0
\end{equation*}
Linear
Incorrect, \(\sin(y)\) is a nonlinear term.
Homogeneous
Technically, only linear equations can be labeled as homogeneous or not. Since the equation is nonlinear, we do not select it.
Constant Coefficients
Technically, only linear equations can be labeled as having constant coefficients or not. Since the equation is nonlinear, we do not select it.
LHCC
Incorrect.
(i) Classifying Practice.
Select each classification label that applies to the equation
\begin{equation*}
y''- 6 = 0
\end{equation*}
Linear
Correct, both terms are linear.
Homogeneous
Incorrect, the free term, \(6\text{,}\) is non-zero.
Constant Coefficients
Correct, each coefficient is constant.
LHCC
Incorrect.
(j) Classifying Practice.
Select each classification label that applies to the equation
\begin{equation*}
\frac{d^3y}{dt^3} + k\frac{dy}{dt} = ty, \qquad k \text{ is constant}
\end{equation*}
Linear
Correct, all terms are linear.
Homogeneous
Correct, the free term is zero.
Constant Coefficients
Incorrect, the \(y\) term coefficient, \(t\text{,}\) is not constant.
LHCC
Incorrect.
2. Short-Answer.
(a)
Suppose you knew that the final form of the particular solution of an LHCC equation was
\(y_p = Axe^x
\text{.}\) Using only this information, list out everything you can deduce about the differential equation.
(b) Why do we modify the particular solution \(y_p\) by multiplying it by \(x\) when solving LNCC equations?
Why do we modify the particular solution \(y_p\) by multiplying it by \(x\) when solving LNCC equations?
(c) In the general solution of an LNCC equation, what roles do \(y_h\) and \(y_p\) play, and why is it important that they are independent?
In the general solution of an LNCC equation, what roles do \(y_h\) and \(y_p\) play, and why is it important that they are independent?
(d)
Describe the difference between a homogeneous and a non-homogeneous differential equation.
Exercises ποΈ Warm-ups & Drills
Write Down the Form of \(y_p\) .
For each forcing function
\(g(t)\text{,}\) write down the correct
form of the particular solution
\(y_p\text{.}\) Donβt solve for coefficients, just show the form of
\(y_p\text{.}\)
1.
\(g(t) = e^{7t} + 6\) 2.
\(g(t) = 6t^2 - 7\sin(3t) + 9\) 3.
\(g(t) = 10e^t - 5t e^{-8t} + 2 e^{-8t}\) 4.
\(g(t) = t^2\cos t - 5t\sin t\) 5.
\(g(t) = 16e^{7t}\sin(10t)\) 6.
\(g(t) = (9t^2 - 103t)\cos t\) 7.
\(g(t) = - e^{-2t}(3 - 5t)\cos(9t)\) 8.
\(g(t) = 4\cos(6t) - 9\sin(6t)\) 9.
\(g(t) = - 2\sin t + \sin(14t) - 5\cos(14t)\) 10.
\(g(t) = 5e^{-3t} + e^{-3t}\cos(6t) - \sin(6t)\)
Find the Initial Form of \(y_p\) .
For each differential equation below, determine the correct
form of the particular solution
\(y_p\text{.}\) Do NOT solve for coefficients like \(A, B, C\text{.}\)
11.
\(3z'' - 4z' - 12z = 17 + 2\cos(2\theta)\) 12.
\(y'' - 3y' - 17y = x^2 + \cos x \) 13.
\(y'' + y' - 12y = 4e^{3t} + \cos(2t) \) 14.
\(y'' - 4y = 2t - e^{2t}\sin(3t) \) 15.
\(w'' - 4w' + 13w = e^{3x} + x^2e^{-x} \) 16.
\(y'' - 2y' + y = x^2 + e^x \) 17.
\(z'' - 4z' + 5z = te^{2t} + 2e^{2t}\sin t \) 18.
\(x^3 \sin(4x) + x^2 \sin(4x) + 7y'' - y' + 2y = 0\)
Particular Solution.
In these exercises, go one step beyond identifying the
form of
\(y_p\) and actually solve for the coefficients and write down a particular solution.
Remember:
Find \(y_c\) first, to identify duplicates.
Guess \(y_p\) based on the forcing term.
Modify \(y_p\) if thereβs overlap with \(y_c\text{.}\)
Plug into the DE, solve for the coefficients, and write the final \(y_p\text{.}\)
19.
\(y'' - 4y' - 12y = 3e^{5t}\) 20.
\(y'' - 4y' - 12y = \sin(2t)\) 21.
\(y'' - 4y' - 12y = 2t^3 - t + 3\) 22.
\(y'' - 4y' - 12y = e^{6t}\)
Exercises βπ» Solve the Differential Equations
General Solutions β Method of Undetermined Coefficients.
Find the general solution to each differential equation using the five-step Method of Undetermined Coefficients:
Find the homogeneous solution \(y_h\text{.}\)
Guess a form for \(y_p\) based on the forcing function.
Modify \(y_p\) if it overlaps with \(y_h\text{.}\)
Substitute \(y_p\) into the DE and solve for coefficients.
Write the general solution \(y = y_h + y_p\text{.}\)
1.
\(2x' + x = 3t^2\) 2.
\(y'' - 3y' + 2y = 5x + 1\) 3.
\(y'' - 4y' - 5y = t + 2e^{-t}\) 4.
\(z'' - 6z' + 34z = 650 \sin(6x)\) 5.
\(y'' - 4y' + 4y = 8\cos x + 12 e^{2x}\) 6.
\(y'' - y' - 6y = 2t + 3e^{3t} - e^{-2t}\) 7.
\(y'' + 3y' + 2y = 6e^{2x}\) 8.
\(y'' - 3y' + y = 2x^2 + 3x\) 9.
\(y'' - 4y' - 12y = t e^{4t}\) 10.
\(y'' + y = 3 \cos x\) 11.
\(y'' + 3y' - 28y = 7t + e^{4t} - 1\) 12.
\(y'' - 100y = 9 t^2 e^{10t} + \cos t - t \sin t\) 13.
\(4y'' + y = e^{-2t} \sin \left( \frac{t}{2} \right) + 6 t \cos \left( \frac{t}{2} \right)\) 14.
\(4y'' + 16y' + 17y = e^{-2t} \sin \left( \frac{t}{2} \right) + 6 t \cos \left( \frac{t}{2} \right)\) 15.
\(y'' + 8y' + 16y = e^{-4t} + (t^2 + 5)e^{-4t}\)
Initial Value Problems β Easy to Moderate.
These IVPs involve either first-order linear DEs or simpler second-order equations with single forcing terms. Follow the UC workflow:
Find \(y_h\) (the homogeneous solution).
Guess \(y_p\) (the particular solution).
Adjust \(y_p\) if it overlaps with \(y_h\text{.}\)
Substitute \(y_p\) and solve for coefficients.
Write \(y = y_h + y_p\text{.}\)
Apply initial condition(s) to find the specific solution.
16.
\(y' - 3y = 6,\quad y(0) = 5\) 17.
\(2x' + x = 3t^2,\quad x(0) = 15\) 18.
\(e^t z' = 1 - 4e^t z,\quad z(0) = \frac{4}{3}\) 19.
\(2z'' + z = 9e^{2t}, \ z(0)=3, \ z'(0)=-1\)
Initial Value Problems β Advanced.
These IVPs involve second-order DEs with multi-term forcing functions (polynomials, exponentials, trig). They require careful handling of overlaps and sometimes multiple adjustments to
\(y_p\text{.}\)
20.
\(y'' - 4y' - 12y = 3 e^{5t}, \ y(0)=\frac{18}{7}, \ y'(0)=-\frac{1}{7}\) 21.
\(y'' - 4y' - 6y = 2t + 3e^{3t} - e^{-2t},\ y(0)=0, \ y'(0)=0\) 22.
\(4y'' + 16y' + 17y = e^{-2t} \sin \frac{t}{2} + 6 t \cos \frac{t}{2},\ y(0)=-1, \ y'(0)=0\)
You have attempted
of
activities on this page.