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4.2.1: Constant Coefficient Homogeneous Equations (Exercises)

( \newcommand{\kernel}{\mathrm{null}\,}\)

Q5.2.1

In Exercises 5.2.1-5.2.12 find the general solution.

1. y+5y6y=0

2. y4y+5y=0

3. y+8y+7y=0

4. y4y+4y=0

5. y+2y+10y=0

6. y+6y+10y=0

7. y8y+16y=0

8. y+y=0

9. y2y+3y=0

10. y+6y+13y=0

11. 4y+4y+10y=0

12. 10y3yy=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. 6yyy=0,y(0)=10,y(0)=0

15. 6y+yy=0,y(0)=1,y(0)=3

16. 4y4y3y=0,y(0)=1312,y(0)=2324

17. 4y12y+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. y6y+9y=0,y(0)=0,y(0)=2

20. 36y12y+y=0,y(0)=3,y(0)=52

21. y+4y+10y=0,y(0)=3,y(0)=2

Q5.2.4

22.

  1. Suppose y is a solution of the constant coefficient homogeneous equation ay+by+cy=0. Let z(x)=y(xx0), where x0 is an arbitrary real number. Show that az+bz+cz=0.
  2. Let z1(x)=y1(xx0) and z2(x)=y2(xx0), where {y1,y2} is a fundamental set of solutions of (A). Show that {z1,z2} is also a fundamental set of solutions of (A).
  3. 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 x00, 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. y6y7y=0,y(2)=13,y(2)=5

25. y14y+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.

  1. Show that a22x(erx)+bx(erx)+cerx=a(rr1)2erx.
  2. Differentiate (B) with respect to r to obtain ar(22x(erx))+br(x(erx))+c(xerx)=[2+(rr1)x]a(rr1)erx.
  3. Reverse the orders of the partial differentiations in the first two terms on the left side of (C) to obtain a2x2(xerx)+bx(xerx)+c(xerx)=[2+(rr1)x]a(rr1)erx.
  4. 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)!=1u22!+u44!++(1)nu2n(2n)!+,

and

sinu=n=0(1)nu2n+1(2n+1)!=uu33!+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.

  1. 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=(1θ22+θ44+)+i(θθ33!+θ55!+)=n=0(1)nθ2n(2n)!+in=0(1)nθ2n+1(2n+1)!. By comparing this result with (B) and (C), conclude that e^{i\theta}=\cos\theta+i\sin\theta. \tag{E} This is Euler’s identity.
  2. Starting from e^{i\theta_1}e^{i\theta_2}=(\cos\theta_1+i\sin\theta_1) (\cos\theta_2+i\sin\theta_2),\nonumber 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 \begin{aligned} \cos(\theta_1+\theta_2)&=\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\\ \sin(\theta_1+\theta_2)&=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2\end{aligned}\nonumber to verify that e^{i(\theta_1+\theta_2)}=e^{i\theta_1}e^{i\theta_2},\nonumber as you would expect from the use of the exponential notation e^{i\theta}.
  3. If \alpha and \beta are real numbers, define e^{\alpha+i\beta}=e^\alpha e^{i\beta}=e^\alpha(\cos\beta+i\sin\beta). \tag{F} Show that if z_1=\alpha_1+i\beta_1 and z_2=\alpha_2+i\beta_2 then e^{z_1+z_2}=e^{z_1}e^{z_2}.\nonumber
  4. Let a, b, and c be real numbers, with a\ne0. 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 \tag{G} 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 \lambda\pm i\omega then z_1=e^{(\lambda+i\omega)x} and z_2=e^{(\lambda-i\omega)x} are both solutions of (G).

4.2.1: Constant Coefficient Homogeneous Equations (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by LibreTexts.

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