17.R: Chapter 17 Review Exercises
- Page ID
- 72453
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)True or False? Justify your answer with a proof or a counterexample.
1. If \(y\) and \(z\) are both solutions to \(y''+2y′+y=0,\) then \(y+z\) is also a solution.
- Answer
- True
2. The following system of algebraic equations has a unique solution:
\(\begin{align*} 6z_1+3z_2 &=8 \\ 4z_1+2z_2 &=4. \end{align*}\)
3. \(y=e^x \cos (3x)+e^x \sin (2x)\) is a solution to the second-order differential equation \(y″+2y′+10=0.\)
- Answer
- False
4. To find the particular solution to a second-order differential equation, you need one initial condition.
In problems 5 - 8, classify the differential equations. Determine the order, whether it is linear and, if linear, whether the differential equation is homogeneous or nonhomogeneous. If the equation is second-order homogeneous and linear, find the characteristic equation.
5. \(y″−2y=0\)
- Answer
- second order, linear, homogeneous, \(λ^2−2=0\)
6. \(y''−3y+2y= \cos (t)\)
7. \(\left(\dfrac{dy}{dt}\right)^2+yy′=1\)
- Answer
- first order, nonlinear, nonhomogeneous
8. \(\dfrac{d^2y}{dt^2}+t \dfrac{dy}{dt}+\sin^2 (t)y=e^t\)
In problems 9 - 16, find the general solution.
9. \(y''+9y=0\)
- Answer
- \(y=c_1 \sin (3x)+c_2 \cos (3x)\)
10. \(y''+2y′+y=0\)
11. \(y''−2y′+10y=4x\)
- Answer
- \(y=c_1e^x \sin (3x)+c_2e^x \cos (3x)+\frac{2}{5}x+\frac{2}{25}\)
12. \(y''= \cos (x)+2y′+y\)
13. \(y''+5y+y=x+e^{2x}\)
- Answer
- \(y=c_1e^{−x}+c_2e^{−4x}+\frac{x}{4}+\frac{e^{2x}}{18}−\frac{5}{16}\)
14. \(y''=3y′+xe^{−x}\)
15. \(y''−x^2=−3y′−\frac{9}{4}y+3x\)
- Answer
- \(y=c_1e^{(−3/2)x}+c_2xe^{(−3/2)x}+\frac{4}{9}x^2+\frac{4}{27}x−\frac{16}{27}\)
16. \(y''=2 \cos x+y′−y\)
In problems 17 - 18, find the solution to the initial-value problem, if possible.
17. \(y''+4y′+6y=0, \; y(0)=0, \; y′(0)=\sqrt{2}\)
- Answer
- \(y=e^{−2x} \sin (\sqrt{2}x)\)
18. \(y''=3y− \cos (x), \; y(0)=\frac{9}{4}, \; y′(0)=0\)
In problems 19 - 20, find the solution to the boundary-value problem.
19. \(4y′=−6y+2y″, \; y(0)=0, \; y(1)=1\)
- Answer
- \(y=\dfrac{e^{1−x}}{e^4−1}(e^{4x}−1)\)
20. \(y''=3x−y−y′, \; y(0)=−3, \; y(1)=0\)
For the following problem, set up and solve the differential equation.
21. The motion of a swinging pendulum for small angles \(θ\) can be approximated by \(\dfrac{d^2θ}{dt^2}+\dfrac{g}{L}θ=0,\) where \(θ\) is the angle the pendulum makes with respect to a vertical line, \(g\) is the acceleration resulting from gravity, and \(L\) is the length of the pendulum. Find the equation describing the angle of the pendulum at time \(t,\) assuming an initial displacement of \(θ_0\) and an initial velocity of zero.
- Answer
- \(θ(t)=θ_0 \cos\left(\sqrt{\frac{g}{l}}t\right)\)
In problems 22 - 23, consider the “beats” that occur when the forcing term of a differential equation causes “slow” and “fast” amplitudes. Consider the general differential equation \(ay″+by= \cos (ωt)\) that governs undamped motion. Assume that \(\sqrt{\frac{b}{a}}≠ω.\)
22. Find the general solution to this equation (Hint: call \(ω_0=\sqrt{b/a}\)).
23. Assuming the system starts from rest, show that the particular solution can be written as\(y=\dfrac{2}{a(ω_0^2−ω^2)} \sin \left(\dfrac{ω_0−ωt}{2}\right) \sin\left(\dfrac{ω_0+ωt}{2}\right).\)
24. [T] Using your solutions derived earlier, plot the solution to the system \(2y″+9y= \cos (2t)\) over the interval \(t=[−50,50].\) Find, analytically, the period of the fast and slow amplitudes.
For the following problem, set up and solve the differential equations.
25. An opera singer is attempting to shatter a glass by singing a particular note. The vibrations of the glass can be modeled by \(y″+ay= \cos (bt)\), where \(y''+ay=0\) represents the natural frequency of the glass and the singer is forcing the vibrations at \( \cos (bt)\). For what value \(b\) would the singer be able to break that glass? (Note: in order for the glass to break, the oscillations would need to get higher and higher.)
- Answer
- \(b=\sqrt{a}\)