17.R: Chapter 17 Review Exercises
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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:
6z1+3z2=84z1+2z2=4.
3. y=excos(3x)+exsin(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 asy=\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}