5.2.1: Constant Coefficient Homogeneous Equations (Exercises)
- Last updated
- Jan 7, 2020
- Save as PDF
- Page ID
- 30725
( \newcommand{\kernel}{\mathrm{null}\,}\)
Q5.2.1
In Exercises 5.2.1-5.2.12 find the general solution.
1. y″+5y′−6y=0
2. y″−4y′+5y=0
3. y″+8y′+7y=0
4. y″−4y′+4y=0
5. y″+2y′+10y=0
6. y″+6y′+10y=0
7. y″−8y′+16y=0
8. y″+y′=0
9. y″−2y′+3y=0
10. y″+6y′+13y=0
11. 4y″+4y′+10y=0
12. 10y″−3y′−y=0
Q5.2.2
In Exercises 5.2.13-5.2.17 solve the initial value problem.
13. y″+14y′+50y=0,y(0)=2,y′(0)=−17
14. 6y″−y′−y=0,y(0)=10,y′(0)=0
15. 6y″+y′−y=0,y(0)=−1,y′(0)=3
16. 4y″−4y′−3y=0,y(0)=1312,y′(0)=2324
17. 4y″−12y′+9y=0,y(0)=3,y′(0)=52
Q5.2.3
In Exercises 5.2.18-5.2.21 solve the initial value problem and graph the solution.
18. y″+7y′+12y=0,y(0)=−1,y′(0)=0
19. y″−6y′+9y=0,y(0)=0,y′(0)=2
20. 36y″−12y′+y=0,y(0)=3,y′(0)=52
21. y″+4y′+10y=0,y(0)=3,y′(0)=−2
Q5.2.4
22.
- Suppose y is a solution of the constant coefficient homogeneous equation ay″+by′+cy=0. Let z(x)=y(x−x0), where x0 is an arbitrary real number. Show that az″+bz′+cz=0.
- Let z1(x)=y1(x−x0) and z2(x)=y2(x−x0), where {y1,y2} is a fundamental set of solutions of (A). Show that {z1,z2} is also a fundamental set of solutions of (A).
- The statement of Theorem 5.2.1 is convenient for solving an initial value problem ay″+by′+cy=0,y(0)=k0,y′(0)=k1, where the initial conditions are imposed at x0=0. However, if the initial value problem is ay″+by′+cy=0,y(x0)=k0,y′(x0)=k1, where x0≠0, then determining the constants in y=c1er1x+c2er2x,y=er1x(c1+c2x), or y=eλx(c1cosωx+c2sinωx) (whichever is applicable) is more complicated. Use (b) to restate Theorem 5.2.1 in a form more convenient for solving (B).
Q5.2.5
In Exercises 5.2.23-5.2.28 use a method suggested by Exercise 5.2.22 to solve the initial value problem.
23. y″+3y′+2y=0,y(1)=−1,y′(1)=4
24. y″−6y′−7y=0,y(2)=−13,y′(2)=−5
25. y″−14y′+49y=0,y(1)=2,y′(1)=11
26. 9y″+6y′+y=0,y(2)=2,y′(2)=−143
27. 9y″+4y=0,y(π/4)=2,y′(π/4)=−2
28. y″+3y=0,y(π/3)=2,y′(π/3)=−1
Q5.2.6
29. Prove: If the characteristic equation of
ay″+by′+cy=0
has a repeated negative root or two roots with negative real parts, then every solution of (A) approaches zero as x→∞.
30. Suppose the characteristic polynomial of ay″+by′+cy=0 has distinct real roots r1 and r2. Use a method suggested by Exercise 5.2.22 to find a formula for the solution of
ay″+by′+cy=0,y(x0)=k0,y′(x0)=k1.
31 Suppose the characteristic polynomial of ay″+by′+cy=0 has a repeated real root r1. Use a method suggested by Exercise 5.2.22 to find a formula for the solution of
ay″+by′+cy=0,y(x0)=k0,y′(x0)=k1.
32. Suppose the characteristic polynomial of ay″+by′+cy=0 has complex conjugate roots λ±iω. Use a method suggested by Exercise 5.2.22 to find a formula for the solution of
ay″+by′+cy=0,y(x0)=k0,y′(x0)=k1.
33. Suppose the characteristic equation of
ay″+by′+cy=0 has a repeated real root r1. Temporarily, think of erx as a function of two real variables x and r.
- Show that a∂2∂2x(erx)+b∂∂x(erx)+cerx=a(r−r1)2erx.
- Differentiate (B) with respect to r to obtain a∂∂r(∂2∂2x(erx))+b∂∂r(∂∂x(erx))+c(xerx)=[2+(r−r1)x]a(r−r1)erx.
- Reverse the orders of the partial differentiations in the first two terms on the left side of (C) to obtain a∂2∂x2(xerx)+b∂∂x(xerx)+c(xerx)=[2+(r−r1)x]a(r−r1)erx.
- Set r=r1 in (B) and (D) to see that y1=er1x and y2=xer1x are solutions of (A)
34. In calculus you learned that eu, cosu, and sinu can be represented by the infinite series
eu=∞∑n=0unn!=1+u1!+u22!+u33!+⋯+unn!+⋯
cosu=∞∑n=0(−1)nu2n(2n)!=1−u22!+u44!+⋯+(−1)nu2n(2n)!+⋯,
and
sinu=∞∑n=0(−1)nu2n+1(2n+1)!=u−u33!+u55!+⋯+(−1)nu2n+1(2n+1)!+⋯
for all real values of u. Even though you have previously considered (A) only for real values of u, we can set u=iθ, where θ is real, to obtain
eiθ=∞∑n=0(iθ)nn!.
Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in (D) converges for all real θ.
- Recalling that i2=−1, write enough terms of the sequence {in} to convince yourself that the sequence is repetitive: 1,i,−1,−i,1,i,−1,−i,1,i,−1,−i,1,i,−1,−i,⋯. Use this to group the terms in (D) as eiθ=(1−θ22+θ44+⋯)+i(θ−θ33!+θ55!+⋯)=∞∑n=0(−1)nθ2n(2n)!+i∞∑n=0(−1)nθ2n+1(2n+1)!. By comparing this result with (B) and (C), conclude that eiθ=cosθ+isinθ. This is Euler’s identity.
- Starting from eiθ1eiθ2=(cosθ1+isinθ1)(cosθ2+isinθ2), collect the real part (the terms not multiplied by i) and the imaginary part (the terms multiplied by i) on the right, and use the trigonometric identities cos(θ1+θ2)=cosθ1cosθ2−sinθ1sinθ2sin(θ1+θ2)=sinθ1cosθ2+cosθ1sinθ2 to verify that ei(θ1+θ2)=eiθ1eiθ2, as you would expect from the use of the exponential notation eiθ.
- If α and β are real numbers, define eα+iβ=eαeiβ=eα(cosβ+isinβ). Show that if z1=α1+iβ1 and z2=α2+iβ2 then ez1+z2=ez1ez2.
- Let a, b, and c be real numbers, with a≠0. Let z=u+iv where u and v are real-valued functions of x. Then we say that z is a solution of ay″+by′+cy=0 if u and v are both solutions of (G). Use Theorem 5.2.1 (c) to verify that if the characteristic equation of (G) has complex conjugate roots λ±iω then z1=e(λ+iω)x and z2=e(λ−iω)x are both solutions of (G).