DE-BOOK Chapter 7 Exercises

Section Chapter 7 Exercises

Reading Questions ᯓ★❓ Quick-Answer Questions

1. True-False.

(a) Select all the true statements below.

Select all the true statements below.

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An LHCC equation must have constant coefficients.
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Correct! Constant coefficients are one of the defining features of LHCC equations.
An LHCC equation could contain the independent variable, \(x \text{.}\)
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Incorrect, LHCC equations are linear, meaning they cannot contain non-linear terms like \(y^2 \text{.}\)
\(\ds\quad y' + 3y = 0 \) is an LHCC equation.
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Correct! This is a first-order linear homogeneous differential equation.
A non-homogeneous equation has a non-zero free term.
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Correct! If the free term is not zero, the equation is non-homogeneous.

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(b) Linear or Nonlinear.

The equation \(y'' + y \cdot y' - 3y = 0 \) is linear.

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True.
Incorrect. The term \(y \cdot y' \) makes this equation nonlinear because the function \(y \) and its derivative are multiplied together.
False.
Incorrect. The term \(y \cdot y' \) makes this equation nonlinear because the function \(y \) and its derivative are multiplied together.

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(c) Polynomial Solutions.

Polynomial functions are commonly solutions to LHCC equations.

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True.
Incorrect. Polynomial functions do not satisfy LHCC equations because their derivatives do not remain proportional to the original function.
False.
Incorrect. Polynomial functions do not satisfy LHCC equations because their derivatives do not remain proportional to the original function.

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(d) Understanding Second-Order LHCC Equations.

A second-order LHCC equation has the form \(a y'' + b y' + c y = 0\text{,}\) where \(a, b,\) and \(c\) are constants.

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True.
Correct! Second-order LHCC equations involve a second derivative and have constant coefficients.
False.
Correct! Second-order LHCC equations involve a second derivative and have constant coefficients.

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2. Multiple-Choice.

(a) True-or-False.

The hundredth derivative of \(e^{7x} \) is a like-term with \(e^{7x}\text{.}\)

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True
Correct!
False
Incorrect.

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(b) Building Solutions.

If \(y_1\) and \(y_2\) are solutions to a second-order LHCC equation, then

\begin{equation*} y = 5y_1 - 2y_2 \end{equation*}

is also a solution.

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True.
Correct! By the superposition principle, this combination is also a solution.
False.
Correct! By the superposition principle, this combination is also a solution.

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(c) Match the Label to the DE.

Match each label on the left to an appropriate DE on the right.

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Key: L = linear, H = homogeneous, CC = constant coefficient

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Note: Multiple matches can be correct, but there is only one perfect matching where all are correct.

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CC
\(\ds r'' + 3r' + 2r^2 = 0 \)
LH, order 1
\(\ds 3t r' = 0 \)
LHCC
\(\ds r'' + 3r' + 2r = 0 \)
LH, order 2
\(\ds r'' + 3t r' = 2r \)
LCC
\(\ds y''' - 7y = e^x \)
L
\(\ds t^5\frac{dw}{dt} = \sin(3t) \)

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(d) Select the LHCC Equations.

Select all of the LHCC Differential Equations.

\(\ds y'' + \sin(y) = 17t \)	\(\ds y'' + \frac{y'}{t^2} + y = 7t \)	\(\ds y'' + 3y' + 2y = 0 \)
\(\ds y'' + y^2 = 17t \)	\(\ds y'' + y'y = 0 \)	\(\ds y = y' \)
\(\ds \frac15 y^{(8)} - \sqrt{15}y' = y \)	\(\ds s'''+\pi s = \frac{7}{w} \)	\(\ds t\frac{dP}{dt} + P^2 = \sqrt{t} \)
\(\ds \frac{dg}{dx} + 3x^2 = 0 \)	\(\ds \frac15 y^{(8)} - \sqrt{15}x y' = y \)	\(\ds \frac{d^2s}{dt^2} + \frac{ds}{dt} = 4s \)

Hint.

There are only 4 LHCC equations in this set.

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(e) Identifying Homogeneous Equations.

Which of the following equations is homogeneous?

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\(\quad y'' + 3y' + 2y = x \)
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Incorrect. The right-hand side is \(x \text{,}\) which makes the equation nonhomogeneous.
\(\quad y' + e^x = 0 \)
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Incorrect. The term \(e^x \) on the right-hand side makes the equation nonhomogeneous.
\(\quad y'' + 5y' + 7y = 0 \)
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Correct! The equation is homogeneous because the right-hand side is zero.
\(\quad y'' + y' + \sin x = 0 \)
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Incorrect. The \(\sin x \) term on the right-hand side makes the equation nonhomogeneous.

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(f) Characteristic equation for first-order LHCC.

What is the characteristic equation for \(y' - 5y = 0\text{?}\)

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\(\ds\quad r - 5 = 0\)
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Correct! The characteristic equation is \(r - 5 = 0\text{.}\)
\(\ds\quad r + 5 = 0\)
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Incorrect. Check the sign of the coefficient of \(y\text{.}\)
\(\ds\quad r^2 - 5 = 0\)
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Incorrect. The characteristic equation for a first-order LHCC is linear, not quadratic.
\(\ds\quad 5r - 1 = 0\)
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Incorrect. Make sure to use the correct coefficients from the original equation.

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(g) Identify the first-order LHCC equation.

Which of the following is a first-order LHCC equation?

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\(\ds\quad y'' + y' - y = 0\)
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Incorrect. This is a second-order equation.
\(\ds\quad 3y' + 5y = 0\)
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Correct! This is a first-order linear homogeneous equation with constant coefficients.
\(\ds\quad 2y + y' = 3\)
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Incorrect. This equation is not homogeneous.
\(\ds\quad y' + xy = 0\)
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Incorrect. This is not a constant coefficient equation.

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(h) Match the DE to Its Characteristic Equation.

Match each LHCC differential equation on the left to its corresponding characteristic equation on the right.

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Note: Each DE has a unique characteristic equation.

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\(2y'' - 3y' + y = 0 \)
\(2r^2 - 3r + 1 = 0 \)
\(y'' - 5y' + 6y = 0 \)
\(r^2 - 5r + 6 = 0 \)
\(2y' + 7y = 0 \)
\(2r + 7 = 0 \)
\(y'' + 4y' = 0 \)
\(r^2 + 4r = 0 \)

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(i) The Characteristic Equation.

What is the characteristic equation of the differential equation \(y'' - 5y' + 6y = 0\text{?}\)

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\(\quad\ds r^2 - 6r + 5 = 0\)
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Incorrect. Check the coefficients in the original equation.
\(\quad\ds r^2 - 5r + 6 = 0\)
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Correct! The characteristic equation is formed by replacing \(y''\) with \(r^2\text{,}\) \(y'\) with \(r\text{,}\) and \(y\) with 1.
\(\quad\ds r^2 - 5r - 6 = 0\)
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Incorrect. Be careful with the sign of the free term.

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(j) Give the general form.

Give the general form of a second-order LHCC equation if the characteristic equation has the solution: \(r = -1 \pm i\text{.}\)

\(y = C_1 e^{-x} + C_2 e^{ix}\)
Incorrect. Complex roots require both cosine and sine terms.
\(y = (C_1 + C_2 x) e^{-x}\)
Incorrect. This form is used for repeated real roots.
\(y = C_1 e^{-x} + C_2 e^{x}\)
Incorrect. This form is used for distinct real roots.
\(y = e^{-x} (C_1 \cos(x) + C_2 \sin(x))\)
Correct! This form is used when the roots are complex.

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(k) Give the general form.

Give the general form of a second-order LHCC equation if the characteristic equation has the solutions \(r_1 = 1\) and \(r_2 = -1\text{.}\)

\(y = C_1 e^{x} + C_2 e^{-x}\)
Correct! This form is used when the characteristic equation has distinct real roots.
\(y = C_1 e^{x} + C_2 x e^{x}\)
Incorrect. This form is used for repeated real roots.
\(y = (C_1 + C_2 x) e^{x}\)
Incorrect. This form is also used for repeated real roots.
\(y = e^{x} (C_1 \cos(x) + C_2 \sin(x))\)
Incorrect. This form is used for complex roots.

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(l) Roots of the characteristic equation.

What are the roots of the characteristic equation \(r^2 - 5r + 6 = 0\text{?}\)

\(r = 2\) and \(r = 3\)
Correct! The roots are \(r = 2\) and \(r = 3\text{.}\)
\(r = -2\) and \(r = -3\)
Incorrect. Check the signs of the roots.
\(r = 1\) and \(r = 6\)
Incorrect. Ensure you solve the quadratic equation correctly.
\(r = 5\) and \(r = 1\)
Incorrect. Revisit the quadratic formula to solve for the roots.

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(m) General solution for second-order LHCC.

What is the general solution for \(y'' - 5y' + 6y = 0\text{?}\)

\(y = C_1 e^{2x} + C_2 e^{3x}\)
Correct! The general solution is \(y = C_1 e^{2x} + C_2 e^{3x}\text{.}\)
\(y = C_1 e^{-2x} + C_2 e^{-3x}\)
Incorrect. Check the signs of the exponents.
\(y = C_1 e^{5x} + C_2 e^{x}\)
Incorrect. Make sure to use the correct roots.
\(y = C_1 e^{x} + C_2 e^{-x}\)
Incorrect. Revisit the roots of the characteristic equation.

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(n) Classifying Practice.

Select each classification label that applies to the equation

\begin{equation*} y'' = y' + 6y \end{equation*}

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Linear
Correct, each of the terms are linear.
Homogeneous
Correct, the free term is zero.
Constant Coefficients
Correct, each coefficient is constant.
LHCC
Correct!

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(o) Classifying Practice.

Select each classification label that applies to the equation

\begin{equation*} 3y''' + y'- \sin(y) = 0 \end{equation*}

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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.

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(p) Classifying Practice.

Select each classification label that applies to the equation

\begin{equation*} y''- 6 = 0 \end{equation*}

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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.

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(q) 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*}

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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.

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(r) Natural Solutions.

Why do exponentials naturally arise as solutions to LHCC equations?

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Because their derivatives preserve the same functional form.
Correct! Exponentials are eigenfunctions of the derivative operator.
Because polynomial solutions cannot satisfy these equations.
Incorrect. Polynomial solutions can satisfy LHCC equations.
Because sine and cosine functions do not work.
Incorrect. Sine and cosine functions are also solutions to LHCC equations.
Because they minimize the characteristic equation.
Incorrect. Exponentials are not solutions because they minimize the characteristic equation.

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(s) What exponential term is in the solution.

What is the fundamental exponential solution for the equation

\begin{equation*} -2y' - 7y = 0\text{?} \end{equation*}

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\(\ds\quad e^{-7/2x}\)
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Correct! Solving \(-2r - 7 = 0\) gives \(r = -\frac{7}{2}\text{.}\) So \(\ds e^{-7/2x}\) is the exponential term in the solution.
\(\ds\quad e^{7/2x}\)
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Incorrect. Check the signs when solving the characteristic equation.
\(\ds\quad e^{-2/7x}\)
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Incorrect. Ensure you are solving the characteristic equation correctly.
\(\ds\quad e^{7x}\)
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Incorrect. Write down the characteristic equation and solve for \(r\text{.}\)

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(t) Structure of the General Solution.

What is the general solution of a second-order LHCC equation if the characteristic equation has unequal real solutions \(r_1\) and \(r_2\text{?}\)

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\(\quad y = c_1 e^{r_1 x} + c_1 e^{r_2 x}\)
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Incorrect. The constants must be different for each term.
\(\quad y = c_1 e^{r_1 x} + c_2 e^{r_2 x}\)
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Correct! Each fundamental solution is multiplied by an arbitrary constant.
\(\quad y = e^{r_1 x} + e^{r_2 x}\)
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Incorrect. The general solution includes arbitrary constants.

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(u) Differential & Characteristic Equation Mathching.

Match characteristic solution to the term it would correspond to in the general solution of an LHCC equation.

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\(r = 2 \) (repeats 2 times)
\(c_1 e^{2x} + c_2 x e^{2x} \)
\(c_1 e^{-\pi x} \)
\(r = -4 \pm i \)
\(e^{-4x} (c_1 \cos(x) + c_2 \sin(x))\)
\(r = 0 \) (repeats 4 times)
\(c_1 + c_2 x + c_3 x^2 + c_4 x^3 \)
\(r=\pi \)
\(c_1 e^{\pi x} \)
\(c_1 e^{-4x} (\cos(x) + \sin(x)) \)
\(c_1 e^{2x} + c_2 x e^{2x} + c_3 x^2 e^{2x} \)
\(c_1 x^2 e^{2x} \)

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(v) 📖❓ General Solution from Repeated Roots.

Suppose the characteristic equation has a triple root at \(r = -1\text{.}\) What is the corresponding part of the general solution?

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\begin{equation*} c_1 e^{-x} + c_2 x e^{-x} + c_3 x^2 e^{-x} \end{equation*}
Correct! Each repeated root contributes an extra power of \(x\text{.}\)
\begin{equation*} c_1 e^{-x} + c_2 e^{-2x} + c_3 e^{-3x} \end{equation*}
Incorrect. These are distinct roots, not repeated instances of \(r = -1\text{.}\)
\begin{equation*} c_1 e^{-x} + c_2 x^2 e^{-x} + c_3 x^3 e^{-x} \end{equation*}
Incorrect. The powers should stop at \(x^2\) for a triple root.
\begin{equation*} c_1 + c_2 x + c_3 x^2 \end{equation*}
This form is correct for a root at \(r = 0\text{,}\) not \(r = -1\text{.}\)

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(w) 📖❓ When to Use a Factoring Tool.

Which of the following is the best reason to use a factoring tool or computer algebra system when solving an LHCC equation?

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The characteristic equation is high degree and does not factor easily by hand.
Correct! Technology saves time and avoids algebraic errors.
You already know all the roots from memory.
Then you wouldn’t need to factor at all.
You want to verify that exponentials are like-terms.
That’s a useful idea, but not the purpose of a factoring tool.
The roots are all real and distinct.
That makes constructing the solution easier—but you still need the roots first.

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(x) 📖❓ Interpreting Mixed Roots.

A characteristic equation has the following roots:

\begin{equation*} r = 0 \ (\text{double}),\quad r = 2 \pm i\text{.} \end{equation*}

What is the correct form of the general solution?

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\begin{equation*} y = c_1 + c_2 x + e^{2x}(c_3 \cos x + c_4 \sin x) \end{equation*}
Correct! The double root at 0 gives \(1\) and \(x\text{,}\) and the complex pair gives the oscillatory exponential term.
\begin{equation*} y = c_1 + c_2 x + c_3 \cos(2x) + c_4 \sin(2x) \end{equation*}
Incorrect. The sine and cosine terms must be multiplied by \(e^{2x}\text{.}\)
\begin{equation*} y = c_1 e^{0x} + c_2 x e^{0x} + c_3 e^{ix} + c_4 e^{-ix} \end{equation*}
While technically valid, this form is not simplified. Use real-valued functions.
\begin{equation*} y = c_1 e^{x} + c_2 e^{-x} + c_3 e^{2x} \end{equation*}
These roots do not match the ones given.

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3. Short-Answer.

(a) Definition of Homogeneity.

What makes a differential equation homogeneous?

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4. Other.

(a)

Select the linear differential equations that have constant coefficients.

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Hint.

Identify and ignore the “free term” when determining if the coefficients are constant.

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\begin{equation} \vphantom{.} \end{equation}
\(\dfrac15 y^{(8)} - \sqrt{15}y' = y\)	\(s'''+\pi s = \dfrac{7}{w}\)	\(t\dfrac{dP}{dt} + P^2 = \sqrt{t}\)
\begin{equation} \vphantom{.} \end{equation}
\(\ds \frac{dg}{dx} + 3x^2 = 0\)	\(\ds \frac15 y^{(8)} - \sqrt{15}x y' = y\)	\(\ds \frac{d^2s}{dt^2} + \frac{ds}{dt} = 4s\)
\begin{equation} \vphantom{.} \end{equation}

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(b)

Select the linear homogeneous differential equations with constant coefficients.

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\(\ds y'' + 4y' + 6y = 0\)	\(\ds x^2 y'' + xy' + y = \ln(x)\)	\(\ds \frac{dg}{dx} + 3x^2 = 0\)

\(\ds 3y'' - 2ty' + y = 0\)	\(\ds u'' + 2xu' + u = 5x\)	\(\ds \frac{d^2s}{dt^2} + \frac{ds}{dt} = 4s\)

\(\ds 3y'' - 2\tau y' + y = \tau\)	\(\ds \sqrt{t} - \frac{dP}{dt} - \frac{P^2}{2} = 1\)	\(\ds s''' = \frac{7s}{w}\)

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Reading Questions 🏗️ Warm-ups & Drills

1. Identifying LHCC Equations.

(a) Identify the Linear Equations.

Click on the equations that are linear.

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Remember, a linear equation cannot contain products or compositions of \(y \) and its derivatives.

\(y'' + 3y' - 2y = 0\)	\(y' + y^2 = 5\)	\(x^2 y'' - 4x y' + 6y = 0\)

\(y'' + y y' = \sin x\)	\(y' + 4xy = 0\)	\(y'' + \cos(y) = x\)

\(y''' - 2y' + y = e^x\)	\(y' + x = 0\)	\(3y''' + y' - 3y = 0\)

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(b) Identify the Linear Homogeneous Equations.

Click on the linear and homogeneous equations.

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Remember, a linear equation cannot contain products or compositions of \(y \) and its derivatives.

\(y'' + 3y' - 2y = 0\)	\(y' + y^2 = 5\)	\(x^2 y'' - 4x y' + 6y = 0\)

\(y'' + y y' = \sin x\)	\(y' + 4xy = 0\)	\(y'' + \cos(y) = x\)

\(y''' - 2y' + y = e^x\)	\(y' + x = 0\)	\(3y''' + y' - 3y = 0\)

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(c) Identify the Linear Constant Coefficient Equations.

Click on the linear equations with constant coefficients.

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Remember, a linear equation cannot contain products or compositions of \(y \) and its derivatives.

\(y'' + 3y' - 2y = 0\)	\(y' + y^2 = 5\)	\(x^2 y'' - 4x y' + 6y = 0\)

\(y'' + y y' = \sin x\)	\(y' + 4xy = 0\)	\(y'' + \cos(y) = x\)

\(y''' - 2y' + y = e^x\)	\(y' + x = 0\)	\(3y''' + y' - 3y = 0\)

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(d) Identify the LHCC Equations.

Click on the LHCC equations.

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Remember, a linear equation cannot contain products or compositions of \(y \) and its derivatives.

\(y'' + 3y' - 2y = 0\)	\(y' + y^2 = 5\)	\(x^2 y'' - 4x y' + 6y = 0\)

\(y'' + y y' = \sin x\)	\(y' + 4xy = 0\)	\(y'' + \cos(y) = x\)

\(y''' - 2y' + y = e^x\)	\(y' + x = 0\)	\(3y''' + y' - 3y = 0\)

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Differential Equation \(\to\) Characteristic Equation.

Give the characteristic equation corresponding to each differential equation below.

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Use \(r\) for the variable in your answer. Don’t forget the “\(=\)” sign.

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2. \(y'' + 5y' - y = 0\).

\(\displaystyle y'' + 5y' - y = 0\)

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characteristic equation:

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3. \(w''' - 7.1w'' + 0.1w = 0\).

\(\displaystyle w''' - 7.1w'' + 0.1w = 0\)

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characteristic equation:

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4. \(2 p'' + 5p' = 0\).

\(\displaystyle 2 p'' + 5p' = 0\)

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characteristic equation:

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Factor the Characteristic Equation.

Fully factor each characteristic equation.

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⚠️ Warning: The numbers on these problems are randomly generated when you press “Activate”. So, click “Activate” before you start!

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5. \(2\)nd Degree.

\(\displaystyle {r^{2}+10r+21}\ =\ \)

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6. \(3\)rd Degree.

\(\displaystyle {r^{3}-7r^{2}+6r}\ =\ \)

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7. \(4\)th Degree.

\(\displaystyle {r^{4}-13r^{2}+36}\ =\ \)

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8. \(3\)rd Degree.

\(\displaystyle {r^{3}-5r^{2}-4r+20}\ =\ \)

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Exercises Charateristic Polynomials

From Charateristic Roots \(\to\) General Solution.

Suppose the solutions to a characteristic equation for an LHCC equation are given below. In each case, find the corresponding general solution.

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1.

\(r = 3, -3, 5.3\)

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2.

\(r = 6 \pm i\sqrt{7.7}, 0\)

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3.

\(r = -4\ (3\ \text{repeats}), 5.3\)

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4.

\(r = \pm\dfrac{i}{2}, 2 \pm i\)

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5.

\(r = 0\ (2\ \text{repeats}), 3\ (5\ \text{repeats})\)

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6.

\(r = \pm i, \pi\ (2\ \text{repeats}), 5\)

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Exercises ✍🏻 Solve the Differential Equations

1. Verifying Superposition.

Consider the second-order LHCC equation:

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\begin{equation} y'' - 3y' + 2y = 0.\tag{38} \end{equation}

Suppose we have already determined that \(y_1 = e^x\) and \(y_2 = e^{2x}\) are solutions. Show that their linear combination

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\begin{equation*} y = c_1 e^x + c_2 e^{2x} \end{equation*}

is also a solution for arbitrary constants \(c_1\) and \(c_2\text{.}\)

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Solution.

Differentiating \(y\text{:}\)

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\begin{gather*} y' = c_1 e^x + 2c_2 e^{2x} \\ y'' = c_1 e^x + 4c_2 e^{2x} \end{gather*}

Substituting into (38):

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\begin{gather*} (c_1 e^x + 4c_2 e^{2x}) - 3(c_1 e^x + 2c_2 e^{2x}) + 2(c_1 e^x + c_2 e^{2x}) = 0 \\ c_1 e^x + 4c_2 e^{2x} - 3c_1 e^x - 6c_2 e^{2x} + 2c_1 e^x + 2c_2 e^{2x} = 0 \\ (c_1 e^x - 3c_1 e^x + 2c_1 e^x) + (4c_2 e^{2x} - 6c_2 e^{2x} + 2c_2 e^{2x}) = 0 \\ 0 = 0. \end{gather*}

Since the equation holds for all values of \(c_1\) and \(c_2\text{,}\) this confirms that their linear combination is also a solution.

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Find the general solution \(\ds y(t) \) for a linear, homogeneous DEwith constant coefficients which has the given characteristic equation.

2.

\(\ds (r-1)^2(r+3)(r^2 + 2r + 5)^2 = 0 \)

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3.

\(\ds (r+1)^2 (r-6)^3(r^2+1)(r^2 + 4) = 0 \)

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4. \(\ds y^{(4)} - 5y'' + 4y = 0\).

Solution.

The characteristic equation is:

\begin{gather*} r^4 - 5r^2 + 4 = 0 \text{.} \end{gather*}

Let \(u = r^2\text{,}\) then we can rewrite the equation as:

\begin{gather*} u^2 - 5u + 4 = 0 \text{.} \end{gather*}

Solving for \(u\text{,}\) we get:

\begin{gather*} u = 1, \quad u = 4 \text{.} \end{gather*}

Thus, \(r^2 = 1\) gives \(r = \pm 1\text{,}\) and \(r^2 = 4\) gives \(r = \pm 2\text{.}\) The general solution is:

\begin{equation*} y = c_1 e^{x} + c_2 e^{-x} + c_3 e^{2x} + c_4 e^{-2x} \text{.} \end{equation*}

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5. \(\ds y^{(4)} - 5y'' + 4y = 0\).

Solution.

The characteristic equation is:

\begin{gather*} r^4 - 5r^2 + 4 = 0 \text{.} \end{gather*}

Let \(u = r^2\text{,}\) then we can rewrite the equation as:

\begin{gather*} u^2 - 5u + 4 = 0 \text{.} \end{gather*}

Solving for \(u\text{,}\) we get:

\begin{gather*} u = 1, \quad u = 4 \text{.} \end{gather*}

Thus, \(r^2 = 1\) gives \(r = \pm 1\text{,}\) and \(r^2 = 4\) gives \(r = \pm 2\text{.}\) The general solution is:

\begin{equation*} y = c_1 e^{x} + c_2 e^{-x} + c_3 e^{2x} + c_4 e^{-2x} \text{.} \end{equation*}

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6.

\(\ds (r-1)^2(r+3)(r^2 + 2r + 5)^2 = 0 \)

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7.

\(\ds (r+1)^2 (r-6)^3(r^2+1)(r^2 + 4) = 0 \)

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General Solution.

8.

\(\ds 4y'' -36y' = 0 \)

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9.

\(\ds 4y'' -36y = 0. \)

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10.

\(\ds y''- y' - 11y = 0 \)

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11.

\(\ds 2\frac{d^2 \theta}{dt^2} -6\frac{d\theta}{dt} - 8\theta = 0 \)

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12. LHCC Equation General Solution.

Find the general solution to each of the equations below.

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\(\ds y' + \pi y = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} r + \pi = 0 \end{equation*}

Solving gives \(r = -\pi\text{.}\) Since this is a first-order equation, the general solution is:

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\begin{equation*} y = c_1 e^{-\pi x} \end{equation*}

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\(\ds \omega'' - 7\omega' = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} r^2 - 7r = 0 \end{equation*}

Factoring gives:

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\begin{equation*} r(r - 7) = 0 \end{equation*}

So the roots are \(r = 0\) and \(r = 7\text{.}\) Thus:

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\begin{equation*} \omega = c_1 + c_2 e^{7x} \end{equation*}

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\(\ds 2\frac{d^3 M}{dt^3} - 6\frac{dM}{dt} = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} 2r^3 - 6r = 0 \end{equation*}

Factoring gives:

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\begin{equation*} 2r(r^2 - 3) = 0 \end{equation*}

So the roots are \(r = 0\text{,}\) \(\sqrt{3}\text{,}\) and \(-\sqrt{3}\text{.}\) Therefore:

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\begin{equation*} M(t) = c_1 + c_2 e^{\sqrt{3}t} + c_3 e^{-\sqrt{3}t} \end{equation*}

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13. LHCC Equation General Solution.

Find the general solution to each of the equations below.

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\(\ds y' + \pi y = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} r + \pi = 0 \end{equation*}

Solving gives \(r = -\pi\text{.}\) Since this is a first-order equation, the general solution is:

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\begin{equation*} y = c_1 e^{-\pi x} \end{equation*}

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\(\ds \omega'' - 7\omega' = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} r^2 - 7r = 0 \end{equation*}

Factoring gives:

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\begin{equation*} r(r - 7) = 0 \end{equation*}

So the roots are \(r = 0\) and \(r = 7\text{.}\) Thus:

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\begin{equation*} \omega = c_1 + c_2 e^{7x} \end{equation*}

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\(\ds 2\frac{d^3 M}{dt^3} - 6\frac{dM}{dt} = 0\)

Solution.

The characteristic equation is:

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\begin{equation*} 2r^3 - 6r = 0 \end{equation*}

Factoring gives:

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\begin{equation*} 2r(r^2 - 3) = 0 \end{equation*}

So the roots are \(r = 0\text{,}\) \(\sqrt{3}\text{,}\) and \(-\sqrt{3}\text{.}\) Therefore:

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\begin{equation*} M(t) = c_1 + c_2 e^{\sqrt{3}t} + c_3 e^{-\sqrt{3}t} \end{equation*}

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14.

\(\ds w'' + 6w' + 9w = 0 \)

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15.

\(\ds m'' = 2m' - m \)

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16.

\(\ds y'' + 4y' + 53y = 0 \)

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17.

\(\ds z''=-36z \)

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18.

\(\ds y'' = -24y' - 144y \)

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19.

\(\ds \frac{d^2w}{dx^2} - 49w = 0 \)

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20.

\(\ds \frac{d^2w}{dx^2} + 49w = 0 \)

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21.

\(\ds z''- z' - 42z = 0 \)

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22. \(\quad\ds 9y' + 2y = 0\).

Solution.

The characteristic equation is:

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\begin{equation*} 9r + 2 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = -\frac{2}{9}\text{.}\) Therefore, the general solution is:

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\begin{equation*} y = c e^{-\frac{2}{9}x} \end{equation*}

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Answer.

\begin{equation*} y = c e^{-\frac{2}{9}x} \end{equation*}

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23. \(\quad\ds 3y'' + 4y' = 0\).

Solution.

The characteristic equation is:

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\begin{equation*} 3r^2 + 4r = 0 \end{equation*}

Factoring out an \(r\) from the equation, we get:

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\begin{equation*} r(3r + 4) = 0 \end{equation*}

Therefore, the roots are \(r = 0\) and \(r = -\frac{4}{3}\text{.}\) The general solution is:

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\begin{equation*} y = c_1 + c_2 e^{-\frac{4}{3}x} \end{equation*}

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Answer.

\begin{equation*} y = c_1 + c_2 e^{-\frac{4}{3}x} \end{equation*}

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24. \(\quad\ds 4y'' -36y = 0\).

Solution.

The characteristic equation is:

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\begin{equation*} 4r^2 - 36 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = \pm 3\text{.}\) Therefore, the general solution is:

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\begin{equation*} y = c_1 e^{3x} + c_2 e^{-3x} \end{equation*}

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Answer.

\begin{equation*} y = c_1 e^{3x} + c_2 e^{-3x} \end{equation*}

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25. \(\quad\ds y''- y' - 11y = 0\).

Solution.

The characteristic equation is:

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\begin{equation*} r^2 - r - 11 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = \frac{1 \pm \sqrt{45}}{2}\text{.}\) Therefore, the general solution is:

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\begin{equation*} y = c_1 e^{\frac{1 + \sqrt{45}}{2}x} + c_2 e^{\frac{1 - \sqrt{45}}{2}x} \end{equation*}

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Answer.

\begin{equation*} y = c_1 e^{\frac{1 + \sqrt{45}}{2}x} + c_2 e^{\frac{1 - \sqrt{45}}{2}x} \end{equation*}

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26. \(\quad\ds 2\frac{d^2 \theta}{dt^2} -6\frac{d\theta}{dt} - 8\theta = 0\).

Solution.

The characteristic equation is:

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\begin{equation*} 2r^2 - 6r - 8 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = 2\) and \(r = -2\text{.}\) Therefore, the general solution is:

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\begin{equation*} \theta = c_1 e^{2t} + c_2 e^{-2t} \end{equation*}

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Answer.

\begin{equation*} \theta = c_1 e^{2t} + c_2 e^{-2t} \end{equation*}

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27. \(\quad\ds w'' + 6w' + 9w = 0\).

Solution.

The characteristic equation is:

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\begin{equation*} r^2 + 6r + 9 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = -3\text{.}\) Therefore, the general solution is:

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\begin{equation*} w = c_1 e^{-3t} + c_2 t e^{-3t} \end{equation*}

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Answer.

\begin{equation*} w = c_1 e^{-3t} + c_2 t e^{-3t} \end{equation*}

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28. \(\quad\ds m'' = 2m' - m\).

Solution.

The characteristic equation is:

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\begin{equation*} r^2 - 2r + 1 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = 1\text{.}\) Therefore, the general solution is:

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\begin{equation*} m = c_1 e^t + c_2 t e^t \end{equation*}

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Answer.

\begin{equation*} m = c_1 e^t + c_2 t e^t \end{equation*}

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29. \(\quad\ds y'' + 4y' + 53y = 0\).

Solution.

The characteristic equation is:

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\begin{equation*} r^2 + 4r + 53 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = -2 \pm 7i\text{.}\) Therefore, the general solution is:

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\begin{equation*} y = e^{-2x}(c_1 \cos(7x) + c_2 \sin(7x)) \end{equation*}

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Answer.

\begin{equation*} y = e^{-2x}(c_1 \cos(7x) + c_2 \sin(7x)) \end{equation*}

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30. \(\quad\ds z''=-36z\).

Solution.

The characteristic equation is:

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\begin{equation*} r^2 + 36 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = \pm 6i\text{.}\) Therefore, the general solution is:

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\begin{equation*} z = c_1 \cos(6x) + c_2 \sin(6x) \end{equation*}

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Answer.

\begin{equation*} z = c_1 \cos(6x) + c_2 \sin(6x) \end{equation*}

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31. \(\quad\ds y'' = -24y' - 144y\).

Solution.

The characteristic equation is:

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\begin{equation*} r^2 + 24r + 144 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = -12\text{.}\) Therefore, the general solution is:

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\begin{equation*} y = c_1 e^{-12x} + c_2 x e^{-12x} \end{equation*}

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Answer.

\begin{equation*} y = c_1 e^{-12x} + c_2 x e^{-12x} \end{equation*}

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32. \(\quad\ds \frac{d^2w}{dx^2} - 49w = 0\).

Solution.

The characteristic equation is:

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\begin{equation*} r^2 - 49 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = \pm 7\text{.}\) Therefore, the general solution is:

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\begin{equation*} w = c_1 e^{7x} + c_2 e^{-7x} \end{equation*}

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Answer.

\begin{equation*} w = c_1 e^{7x} + c_2 e^{-7x} \end{equation*}

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33. \(\quad\ds z''- z' - 42z = 0\).

Solution.

The characteristic equation is:

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\begin{equation*} r^2 - r - 42 = 0 \end{equation*}

Solving for \(r\text{,}\) we get \(r = 7\) and \(r = -6\text{.}\) Therefore, the general solution is:

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\begin{equation*} z = c_1 e^{7x} + c_2 e^{-6x} \end{equation*}

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Answer.

\begin{equation*} z = c_1 e^{7x} + c_2 e^{-6x} \end{equation*}

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Solve the following initial value problems.

34.

\(\ds 4y'' -36y = 0 \hspace{1cm} c_1 \) an \(\ds c_2 \) that satisfy thegiven initial conditions.

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35.

\(\ds \ds 2\frac{d^2 \theta}{dt^2} -6\frac{d\theta}{dt} - 8\theta = 0,\hspace{1cm} \theta(0) = 12, \hspace{1cm} \theta'(0) = -2 \)

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36.

Solve the following initial value problem.\(\ds \frac{d^2z}{dx^2} - 4\frac{dz}{dx} + 4z = 0, \hspace{0.5cm} z(1) = 1, \hspace{0.5cm} z'(1) = 1\)

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37. \(\quad\ds 4y'' -36y = 0, \quad y(0) = 4, \quad y'(0) = -6\).

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 \(\ds y(0) = 4. \)

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\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 \(\ds y'(0) = -6. \)

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\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{eq15}) and (\ref{eq16}) 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{eq15}) and then substitute into equation (\ref{eq16}) as follows.

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\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.

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Answer.

\(\ds y = e^{3t} + 3e^{-3t} \) o \(\ds y = e^{3x} + 3e^{-3x} \)

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38. \(\quad\ds 2\frac{d^2 \theta}{dt^2} -6\frac{d\theta}{dt} - 8\theta = 0,\quad \theta(0) = 12, \quad \theta'(0) = -2\).

Solution.

We already have the general solution \(\ds \theta = c_1e^{2t} + c_2e^{-2t} \) . In order to use the initial conditions, we will eventually need the first derivative, so let’s find that now. \(\ds \theta' = 2c_1e^{2t} -2 c_2e^{-2t} \) Now we can see what comes of the first initial condition \(\ds \theta(0) = 12. \)

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\begin{align*} \theta(0) \amp = 12 \\ c_1e^{2\cdot 0} + c_2e^{-2\cdot 0} \amp = 12 \\ c_1 + c_2 \amp = 12 \end{align*}

Now we can use the other initial condition \(\ds \theta'(0) = -2. \)

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\begin{align*} \theta'(0) \amp = -2 \\ 2c_1e^{2\cdot 0} -2 c_2e^{-2\cdot 0} \amp = -2 \\ 2c_1 - 2c_2 \amp = -2 \end{align*}

Notice that the resulting equations, (\ref{eq17}) and (\ref{eq18}) 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{eq17}) and then substitute into equation (\ref{eq18}) as follows.

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\begin{align*} c_2 \amp = 12 - c_1 \\ 2c_1 - 2(12 - c_1) \amp = -2 \\ 2c_1 - 24 + 2c_1 \amp = -2 \\ 4c_1 - 24 \amp = -2 \\ 4c_1 \amp = 22 \\ c_1 \amp = \frac{22}{4} \\ c_1 \amp = \frac{11}{2} \\ c_2 \amp = 12 - \frac{11}{2} \\ \amp = \frac{13}{2} \end{align*}

Hence, we have the solution \(\ds \theta = \frac{11}{2}e^{2t} + \frac{13}{2}e^{-2t}. \) 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.

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Answer.

\begin{equation*} \theta = \frac{11}{2}e^{2t} + \frac{13}{2}e^{-2t}, \quad \theta = \frac{11}{2}e^{2x} + \frac{13}{2}e^{-2x} \end{equation*}

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39. \(\quad\ds \frac{d^2z}{dx^2} - 4\frac{dz}{dx} + 4z = 0,\quad z(1) = 1,\quad z'(1) = 1\).

Solution 1.

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).

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\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

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\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.

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\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 \(\ds z(1) = 1. \)

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\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 \(\ds \theta'(0) = -2. \)

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\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.

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\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} \end{align*}

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Solution 2.