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Exercises 9.7 Exercises
💡 Conceptual Quiz.
Answer the following questions to test your understanding of LHCC equations.
📝: Abbreviations.
Differential Equation
Initial Value Problem
Linear Homogeneous Constant Coefficients
1.
(a) Select all the true statements below.
Select all the true statements below.
An LHCC equation must have constant coefficients.
Correct! Constant coefficients are one of the defining features of LHCC equations.
An LHCC equation could contain the independent variable, \(x \text{.}\)
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.
Correct! This is a first-order linear homogeneous differential equation.
A non-homogeneous equation has a non-zero free term.
Correct! If the free term is not zero, the equation is non-homogeneous.
(b) Linear or Nonlinear.
The equation
\(y'' + y \cdot y' - 3y = 0 \) is linear.
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.
(c) True-or-False.
The hundredth derivative of
\(e^{7x} \) is a like-term with
\(e^{7x}\text{.}\)
True
Correct!
False
Incorrect.
(d) 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.
True.
Correct! By the superposition principle, this combination is also a solution.
False.
Correct! By the superposition principle, this combination is also a solution.
(e) Identifying Homogeneous Equations.
Which of the following equations is homogeneous?
\(\quad y'' + 3y' + 2y = x \)
Incorrect. The right-hand side is \(x \text{,}\) which makes the equation nonhomogeneous.
\(\quad y' + e^x = 0 \)
Incorrect. The term \(e^x \) on the right-hand side makes the equation nonhomogeneous.
\(\quad y'' + 5y' + 7y = 0 \)
Correct! The equation is homogeneous because the right-hand side is zero.
\(\quad y'' + y' + \sin x = 0 \)
Incorrect. The \(\sin x \) term on the right-hand side makes the equation nonhomogeneous.
(f) Characteristic equation for first-order LHCC.
What is the characteristic equation for
\(y' - 5y = 0\text{?}\)
\(\ds\quad r - 5 = 0\)
Correct! The characteristic equation is \(r - 5 = 0\text{.}\)
\(\ds\quad r + 5 = 0\)
Incorrect. Check the sign of the coefficient of \(y\text{.}\)
\(\ds\quad r^2 - 5 = 0\)
Incorrect. The characteristic equation for a first-order LHCC is linear, not quadratic.
\(\ds\quad 5r - 1 = 0\)
Incorrect. Make sure to use the correct coefficients from the original equation.
(g) Identify the first-order LHCC equation.
Which of the following is a first-order LHCC equation?
\(\ds\quad y'' + y' - y = 0\)
Incorrect. This is a second-order equation.
\(\ds\quad 3y' + 5y = 0\)
Correct! This is a first-order linear homogeneous equation with constant coefficients.
\(\ds\quad 2y + y' = 3\)
Incorrect. This equation is not homogeneous.
\(\ds\quad y' + xy = 0\)
Incorrect. This is not a constant coefficient equation.
(h) The Characteristic Equation.
What is the characteristic equation of the differential equation
\(y'' - 5y' + 6y = 0\text{?}\)
\(\quad\ds r^2 - 6r + 5 = 0\)
Incorrect. Check the coefficients in the original equation.
\(\quad\ds r^2 - 5r + 6 = 0\)
Correct! The characteristic equation is formed by replacing \(y''\) with \(r^2\text{,}\) \(y'\) with \(r\text{,}\) and \(y\) with 1.
\(\quad\ds r^2 - 5r - 6 = 0\)
Incorrect. Be careful with the sign of the free term.
(i) 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.
(j) 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.
(k) Roots of the characteristic equation.
What are the roots of the characteristic equation
\(r^2 - 5r + 6 = 0\text{?}\)
(l) 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 \(y = C_1 e^{2x} + C_2 e^{3x}\) .
\(y = C_1 e^{-2x} + C_2 e^{-3x}\)
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.
(m) 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!
(n) 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.
(o) 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, \(6\) , is non-zero.
Constant Coefficients
Correct, each coefficient is constant.
LHCC
Incorrect.
(p) Classifying Practice.
Select each classification label that applies to the equation
\begin{equation*}
\dfrac{d^3y}{dt^3} + k\dfrac{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, \(t\) , is not constant.
LHCC
Incorrect.
(q) Natural Solutions.
Why do exponentials naturally arise as solutions to LHCC equations?
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.
(r) What exponential term is in the solution.
What is the fundamental exponential solution for the equation
\begin{equation*}
-2y' - 7y = 0\text{?}
\end{equation*}
\(\ds\quad e^{-7/2x}\)
Correct! Solving \(-2r - 7 = 0\) gives \(r = -\dfrac{7}{2}\text{.}\) So \(\ds e^{-7/2x}\) is the exponential term in the solution.
\(\ds\quad e^{7/2x}\)
Incorrect. Check the signs when solving the characteristic equation.
\(\ds\quad e^{-2/7x}\)
Incorrect. Ensure you are solving the characteristic equation correctly.
\(\ds\quad e^{7x}\)
Incorrect. Write down the characteristic equation and solve for \(r\text{.}\)
(s) 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
\(r_2\text{?}\)
\(\quad y = c_1 e^{r_1 x} + c_1 e^{r_2 x}\)
Incorrect. The constants must be different for each term.
\(\quad y = c_1 e^{r_1 x} + c_2 e^{r_2 x}\)
Correct! Each fundamental solution is multiplied by an arbitrary constant.
\(\quad y = e^{r_1 x} + e^{r_2 x}\)
Incorrect. The general solution includes arbitrary constants.
(t) 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?
\(\quad c_1 e^{-x} + c_2 x e^{-x} + c_3 x^2 e^{-x}\)
Correct! Each repeated root contributes an extra power of \(x\) .
\(\quad c_1 e^{-x} + c_2 e^{-2x} + c_3 e^{-3x}\)
Incorrect. These are distinct roots, not repeated instances of \(r = -1\) .
\(\quad c_1 e^{-x} + c_2 x^2 e^{-x} + c_3 x^3 e^{-x}\)
Incorrect. The powers should stop at \(x^2\) for a triple root.
\(\quad c_1 + c_2 x + c_3 x^2\)
This form is correct for a root at \(r = 0\) , not \(r = -1\) .
(u) 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?
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.
(v) 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?
\(\quad y = c_1 + c_2 x + e^{2x}(c_3 \cos x + c_4 \sin x)\)
Correct! The double root at 0 gives \(1\) and \(x\) , and the complex pair gives the oscillatory exponential term.
\(\quad y = c_1 + c_2 x + c_3 \cos(2x) + c_4 \sin(2x)\)
Incorrect. The sine and cosine terms must be multiplied by \(e^{2x}\) .
\(\quad y = c_1 e^{0x} + c_2 x e^{0x} + c_3 e^{ix} + c_4 e^{-ix}\)
While technically valid, this form is not simplified. Use real-valued functions.
\(\quad y = c_1 e^{x} + c_2 e^{-x} + c_3 e^{2x}\)
These roots do not match the ones given.
(w) Definition of Homogeneity.
What makes a differential equation homogeneous?
(x)
Suppose an LHCC equation has
\(x^2\) as a solution. Can you conclude anything specific about the structure of such an LHCC equation?
🏋️♂️ Practice Drills.
More practice to strengthen your understanding.
2. Identifying LHCC Equations.
(a) Identify the Linear Equations .
Click on the equations that are linear.
\(y'' + 3y' - 2y = 0\) \(y' + y^2 = 5\) \(x^2 y'' - 4x y' + 6y = 0\)
\(\) \(\) \(\) \(\) \(\)
\(y'' + y y' = \sin x\) \(y' + 4xy = 0\) \(y'' + \cos(y) = x\)
\(\) \(\) \(\) \(\) \(\)
\(y''' - 2y' + y = e^x\) \(y' + x = 0\) \(3y''' + y' - 3y = 0\)
(b) Identify the Linear Homogeneous Equations .
Click on the linear and homogeneous equations.
\(y'' + 3y' - 2y = 0\) \(y' + y^2 = 5\) \(x^2 y'' - 4x y' + 6y = 0\)
\(\) \(\) \(\) \(\) \(\)
\(y'' + y y' = \sin x\) \(y' + 4xy = 0\) \(y'' + \cos(y) = x\)
\(\) \(\) \(\) \(\) \(\)
\(y''' - 2y' + y = e^x\) \(y' + x = 0\) \(3y''' + y' - 3y = 0\)
(c) Identify the Linear Constant Coefficient Equations .
Click on the linear equations with constant coefficients.
\(y'' + 3y' - 2y = 0\) \(y' + y^2 = 5\) \(x^2 y'' - 4x y' + 6y = 0\)
\(\) \(\) \(\) \(\) \(\)
\(y'' + y y' = \sin x\) \(y' + 4xy = 0\) \(y'' + \cos(y) = x\)
\(\) \(\) \(\) \(\) \(\)
\(y''' - 2y' + y = e^x\) \(y' + x = 0\) \(3y''' + y' - 3y = 0\)
(d) Identify the LHCC Equations .
Click on the LHCC equations.
\(y'' + 3y' - 2y = 0\) \(y' + y^2 = 5\) \(x^2 y'' - 4x y' + 6y = 0\)
\(\) \(\) \(\) \(\) \(\)
\(y'' + y y' = \sin x\) \(y' + 4xy = 0\) \(y'' + \cos(y) = x\)
\(\) \(\) \(\) \(\) \(\)
\(y''' - 2y' + y = e^x\) \(y' + x = 0\) \(3y''' + y' - 3y = 0\)
(e) Match the Label to the DE.
Match each label on the left to an appropriate DE on the right.
Key:
L = linear,
H = homogeneous,
CC = constant coefficient
Note: Multiple matches can be correct, but there is only one perfect matching where all are correct.
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
\(\ds t^5\dfrac{dw}{dt} = \sin(3t) \)
(f) Select the LHCC Equations.
Select all of the LHCC Differential Equations.
\(\ds y'' + \sin(y) = 17t \) \(\ds y'' + \dfrac{y'}{t^2} + y = 7t \) \(\ds y'' + 3y' + 2y = 0 \)
\(\ds y'' + y^2 = 17t \) \(\ds y'' + y'y = 0 \) \(\ds y = y' \)
\(\ds \dfrac15 y^{(8)} - \sqrt{15}y' = y \) \(\ds s'''+\pi s = \dfrac{7}{w} \) \(\ds t\dfrac{dP}{dt} + P^2 = \sqrt{t} \)
\(\ds \dfrac{dg}{dx} + 3x^2 = 0 \) \(\ds \dfrac15 y^{(8)} - \sqrt{15}x y' = y \) \(\ds \dfrac{d^2s}{dt^2} + \dfrac{ds}{dt} = 4s \)
Hint .
There are only 4 LHCC equations in this set.
(g) Match the DE to Its Characteristic Equation.
Match each LHCC differential equation on the left to its corresponding characteristic equation on the right.
Note: Each DE has a unique characteristic equation.
\(2y'' - 3y' + y = 0 \) \(2r^2 - 3r + 1 = 0 \) \(y'' - 5y' + 6y = 0 \) \(r^2 - 5r + 6 = 0 \) \(2y' + 7y = 0 \) \(2r + 7 = 0 \) \(y'' + 4y' = 0 \) \(r^2 + 4r = 0 \)
(h) Differential & Characteristic Equation Matching.
Match characteristic solution to the term it would correspond to in the general solution of an LHCC equation.
\(r = 2 \) ( 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 )
\(c_1 + c_2 x + c_3 x^2 + c_4 x^3 \) \(r=\pi \) \(c_1 e^{\pi x} \) \(c_1 e^{-4x} (\cos(x) + \sin(x)) \) \(c_1 e^{2x} + c_2 x e^{2x} + c_3 x^2 e^{2x} \) \(c_1 x^2 e^{2x} \)
(i)
\(\dfrac15 y^{(8)} - \sqrt{15}y' = y\) \(s'''+\pi s = \dfrac{7}{w}\) \(t\dfrac{dP}{dt} + P^2 = \sqrt{t}\)
\(\) \(\) \(\) \(\) \(\)
\(\ds \dfrac{dg}{dx} + 3x^2 = 0\) \(\ds \dfrac15 y^{(8)} - \sqrt{15}x y' = y\) \(\ds \dfrac{d^2s}{dt^2} + \dfrac{ds}{dt} = 4s\)
Hint .
Identify and ignore the “
free term ” when determining if the coefficients are constant.
(j)
\(\ds y'' + 4y' + 6y = 0\) \(\ds x^2 y'' + xy' + y = \ln(x)\) \(\ds \dfrac{dg}{dx} + 3x^2 = 0\)
\(\) \(\) \(\) \(\) \(\)
\(\ds 3y'' - 2ty' + y = 0\) \(\ds u'' + 2xu' + u = 5x\) \(\ds \dfrac{d^2s}{dt^2} + \dfrac{ds}{dt} = 4s\)
\(\) \(\) \(\) \(\) \(\)
\(\ds 3y'' - 2\tau y' + y = \tau\) \(\ds \sqrt{t} - \dfrac{dP}{dt} - \dfrac{P^2}{2} = 1\) \(\ds s''' = \dfrac{7s}{w}\)
🏋️♂️ Differential Equation \(\to\) Characteristic Equation.
Give the characteristic equation corresponding to each differential equation below.
Use \(r\) for the variable in your answer. Don’t forget the “\(=\) ” sign.
3. \(y'' + 5y' - y = 0\) .
4. \(w''' - 7.1w'' + 0.1w = 0\) .
5. \(2 p'' + 5p' = 0\) .
🏋️♂️ Factor the Characteristic Equation.
Fully factor each characteristic equation.
⚠️
Warning: The numbers on these problems are randomly generated when you press “Activate”. So, click “Activate” before you start!
6. \(2\) nd Degree.
7. \(3\) rd Degree.
8. \(4\) th Degree.
9. \(3\) rd Degree.
✍🏻 From Characteristic 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.
10.
Answer .
Since we have three different real numbers, we use the standard case to get the general solution as
\begin{equation*}
y = c_1 e^{3x} + c_2 e^{-3x} + c_3 e^{5.3x}
\end{equation*}
11.
\(r = 6 \pm i\sqrt{7.7}, 0\)
Answer .
A complex conjugate pair and a single \(0\) solutions, means we use a combination of the complex and standard cases .
\begin{equation*}
y = e^{6x} (c_1 \sin(\sqrt{7.7}x) + c_2 \cos(\sqrt{7.7}x)) + c_3
\end{equation*}
12.
Answer .
The solution \(-4\) repeats three times and the single \(5.3\) gives the general solution
\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*}
13.
\(r = \pm\dfrac{i}{2}, 2 \pm i\)
Answer .
The two separate complex conjugate pairs implies we use the complex case for both to get
\begin{equation*}
y = c_1 \sin\left(\dfrac{x}{2}\right) + c_2 \cos\left(\dfrac{x}{2}\right) + e^{2x} (c_3 \sin x + c_4 \cos x)
\end{equation*}
14.
Answer .
Two \(0\) ’s and four \(3\) ’s means the general solution is
\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}
\end{equation*}
15.
\(r = \pm i, \pi, \pi, 5\)
Answer .
One complex pair, two \(\pi\) ’s, and a single \(5\text{,}\) gives
\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*}
16. Verifying Superposition.
Consider the second-order LHCC equation:
\begin{equation}
y'' - 3y' + 2y = 0.\tag{9.8}
\end{equation}
Suppose we have already determined that \(y_1 = e^x\) and \(y_2 = e^{2x}\) are solutions. Show that their linear combination
\begin{equation*}
y = c_1 e^x + c_2 e^{2x}
\end{equation*}
is also a solution for arbitrary constants \(c_1\) and \(c_2\text{.}\)
✍🏻 Solve the Differential Equations.
Find the general solution for the LNCC equation with the given characteristic equation.
17.
\(\ds (r-1)^2(r+3)(r^2 + 2r + 5)^2 = 0 \)
Answer .
\begin{equation*}
y = c_1 e^{x} + c_2 x e^{x} + c_3 e^{-3x} + e^{-x}[(c_4 + c_5 x)\cos(2x) + (c_6 + c_7 x)\sin(2x)]
\end{equation*}
18.
\(\ds (r+1)^2 (r-6)^3(r^2+1)(r^2 + 4) = 0\)
Answer .
\begin{equation*}
y = c_1 e^{-x} + c_2 x e^{-x} + c_3 e^{6x} + c_4 x e^{6x} + c_5 x^2 e^{6x} + c_6 \cos(x) + c_7 \sin(x) + c_8 \cos(2x) + c_9 \sin(2x)
\end{equation*}
19.
\(\ds (r-1)^2(r+3)(r^2 + 2r + 5)^2 = 0 \)
Answer .
\begin{equation*}
y = c_1 e^{x} + c_2 x e^{x} + c_3 e^{-3x} + e^{-x}[(c_4 + c_5 x)\cos(2x) + (c_6 + c_7 x)\sin(2x)]
\end{equation*}
20.
\(\ds (r+1)^2 (r-6)^3(r^2+1)(r^2 + 4) = 0\)
Answer .
\begin{equation*}
y = c_1 e^{-x} + c_2 x e^{-x} + c_3 e^{6x} + c_4 x e^{6x} + c_5 x^2 e^{6x} + c_6 \cos(x) + c_7 \sin(x) + c_8 \cos(2x) + c_9 \sin(2x)
\end{equation*}
✍🏻 LHCC Equation General Solution.
Find the general solution to each of the equations below.
21. 22.
\(\ds \omega'' - 7\omega' = 0\)
23.
\(\ds 2\dfrac{d^3 M}{dt^3} - 6\dfrac{dM}{dt} = 0\)
✍🏻 General Solution.
Find the general solution for each LNCC equation.
24.
Answer .
\begin{equation*}
y = c_1 + c_2 e^{9x}
\end{equation*}
25.
Answer .
\begin{equation*}
y = c_1 e^{3x} + c_2 e^{-3x}
\end{equation*}
26.
Answer .
\begin{equation*}
y = c_1 e^{\dfrac{1 + \sqrt{45}}{2}x} + c_2 e^{\dfrac{1 - \sqrt{45}}{2}x}
\end{equation*}
27.
\(2\dfrac{d^2 \theta}{dt^2} -6\dfrac{d\theta}{dt} - 8\theta = 0\)
Answer .
\begin{equation*}
\theta = c_1 e^{2t} + c_2 e^{-2t}
\end{equation*}
28.
Answer .
\begin{equation*}
m = c_1 e^{t} + c_2 t e^{t}
\end{equation*}
29.
\(\quad y'' + 4y' + 53y = 0\)
Answer .
\begin{equation*}
y = e^{-2x}(c_1 \cos(7x) + c_2 \sin(7x))
\end{equation*}
30.
Answer .
\begin{equation*}
z = c_1 \cos(6x) + c_2 \sin(6x)
\end{equation*}
31.
\(\quad y'' = -24y' - 144y\)
Answer .
\begin{equation*}
y = c_1 e^{-12x} + c_2 x e^{-12x}
\end{equation*}
32.
\(\quad\dfrac{d^2w}{dx^2} - 49w = 0\)
Answer .
\begin{equation*}
w = c_1 e^{7x} + c_2 e^{-7x}
\end{equation*}
33.
\(\quad\dfrac{d^2w}{dx^2} + 49w = 0 \)
Answer .
\begin{equation*}
w = c_1 \cos(7x) + c_2 \sin(7x)
\end{equation*}
34.
\(\quad z''- z' - 42z = 0\)
Answer .
\begin{equation*}
z = c_1 e^{7x} + c_2 e^{-6x}
\end{equation*}
35.
Answer .
\begin{equation*}
y = c e^{-\dfrac{2}{9}x}
\end{equation*}
36.
Answer .
\begin{equation*}
y = c_1 + c_2 e^{-\dfrac{4}{3}x}
\end{equation*}
37.
\(\quad w'' + 6w' + 9w = 0\)
Answer .
\begin{equation*}
w = c_1 e^{-3t} + c_2 t e^{-3t}
\end{equation*}
38.
Answer .
\(m = c_1 e^t + c_2 t e^t\)
39.
\(\quad y^{(4)} - 5y'' + 4y = 0\)
Answer .
\(y = c_1 e^{x} + c_2 e^{-x} + c_3 e^{2x} + c_4 e^{-2x}\)
✍🏻 Initial-Value Problems.
Solve the following initial value problems.
40.
\(\quad 4y'' -36y = 0, \ \ y(0) = 4,\ \ y'(0) = -6 \)
Answer .
\(\ds y = e^{3t} + 3e^{-3t} \) o
\(\ds y = e^{3x} + 3e^{-3x} \)
41.
\(\quad\dfrac{d^2z}{dx^2} - 4\dfrac{dz}{dx} + 4z = 0, \ \ z(1) = 1, \ \ z'(1) = 1\)
Answer .
\(\ds z = (2e^{-2} - e^{-2}x)e^{2x} \) or
\(\ds z = (2 - x)e^{2x - 2} \)
42.
\(\quad 2\dfrac{d^2 \theta}{dt^2} -6\dfrac{d\theta}{dt} - 8\theta = 0,\ \ \theta(0) = 12, \ \ \theta'(0) = -2\)
Answer .
\begin{equation*}
\theta = \dfrac{11}{2}e^{2t} + \dfrac{13}{2}e^{-2t}, \quad \theta = \dfrac{11}{2}e^{2x} + \dfrac{13}{2}e^{-2x}
\end{equation*}
43. The General Solution to First-Order LHCC Equations.
Derive the general solution to a first order linear homogeneous constant coefficient (LHCC) equation of the form:
\begin{equation*}
a_1\ y' + a_0\ y = 0\text{,}
\end{equation*}
where \(a_1,\ a_0\) are constants.
✍🏻 Boundary Value Problems.
Solve the following boundary value problems. Explain whether you found a unique solution, an infinite number of solutions, or no solution.
44.
Consider the following differential equation:
\begin{equation*}
y'' + y = 0, \quad y(0)=0, \quad y\left(b\right)=y_b, \quad b > 0\text{.}
\end{equation*}
Notice that instead of initial conditions at
\(t=0\) only, there is one at
\(t=b\text{.}\) This is called a
boundary-value problem (BVP) because it has conditions over two different
\(t\) -values. Now, for this equation find the following, if possible:
(a)
Answer .
\(\ds y = c_1\sin(t) + c_2\cos(t) \)
(b) The solution with boundary conditions:
\(y(0)=0\) and
\(y\left(\sfrac{\pi}{2}\right)=0.4\text{.}\)
Answer .
the unique solution is
\(\ds y = 0.4\cos(t) \)
(c) The solution with boundary conditions:
\(y(0)=0\) and
\(y(pi)=5\text{.}\)
(d) The solution with boundary conditions:
\(y(0)=0\) and
\(y(pi)=0\text{.}\)
Answer .
there are an infinite number of solutions of the form
\(\ds y = c_1\sin(t) \)
45.
If possible, solve the following boundary value problem:
\begin{equation*}
y'' - y = 0, \quad y(0) = 1, \quad y(1) = 2e- \dfrac{1}{e}
\end{equation*}
Answer .
\(\ds y = 2e^{t} - e^{-t} \)
46. Show why
\(x\) is needed in the general solution for repeated roots of the CE
Answer .
For a repeated root
\(r\text{,}\) the solutions
\(e^{rx}\) and
\(e^{rx}\) are not independent (they’re the same function). Multiplying by
\(x\) gives
\(xe^{rx}\text{,}\) which is linearly independent from
\(e^{rx}\) and also satisfies the differential equation, providing the second solution needed for the general solution.
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