# 5.2.1: Constant Coefficient Homogeneous Equations (Exercises)

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## 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. $$y''+14y'+50y=0, \quad y(0)=2,\quad y'(0)=-17$$

14. $$6y''-y'-y=0, \quad y(0)=10,\quad y'(0)=0$$

15. $$6y''+y'-y=0, \quad y(0)=-1,\quad y'(0)=3$$

16. $$4y''-4y'-3y=0, \quad y(0)={13\over 12},\quad y'(0)={23 \over 24}$$

17. $$4y''-12y'+9y=0, \quad y(0)=3,\quad y'(0)={5\over 2}$$

## Q5.2.3

In Exercises 5.2.18-5.2.21 solve the initial value problem and graph the solution.

18. $$y''+7y'+12y=0, \quad y(0)=-1,\quad y'(0)=0$$

19. $$y''-6y'+9y=0, \quad y(0)=0,\quad y'(0)=2$$

20. $$36y''-12y'+y=0, \quad y(0)=3,\quad y'(0)={5\over2}$$

21. $$y''+4y'+10y=0, \quad y(0)=3,\quad y'(0)=-2$$

## Q5.2.4

22.

1. Suppose $$y$$ is a solution of the constant coefficient homogeneous equation $ay''+by'+cy=0. \tag{A}$ Let $$z(x)=y(x-x_0)$$, where $$x_0$$ is an arbitrary real number. Show that $az''+bz'+cz=0.\nonumber$
2. Let $$z_1(x)=y_1(x-x_0)$$ and $$z_2(x)=y_2(x-x_0)$$, where $$\{y_1,y_2\}$$ is a fundamental set of solutions of (A). Show that $$\{z_1,z_2\}$$ 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 $ay''+by'+cy=0, \quad y(0)=k_0,\quad y'(0)=k_1,\nonumber$ where the initial conditions are imposed at $$x_0=0$$. However, if the initial value problem is $ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1, \tag{B}$ where $$x_0\ne0$$, then determining the constants in $y=c_1e^{r_1x}+c_2e^{r_2x}, \quad y=e^{r_1x}(c_1+c_2x),\mbox{ or } y=e^{\lambda x}(c_1\cos\omega x+c_2\sin\omega x)\nonumber$ (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. $$y''+3y'+2y=0, \quad y(1)=-1,\quad y'(1)=4$$

24. $$y''-6y'-7y=0, \quad y(2)=-{1\over3},\quad y'(2)=-5$$

25. $$y''-14y'+49y=0, \quad y(1)=2,\quad y'(1)=11$$

26. $$9y''+6y'+y=0, \quad y(2)=2,\quad y'(2)=-{14\over3}$$

27. $$9y''+4y=0, \quad y(\pi/4)=2,\quad y'(\pi/4)=-2$$

28. $$y''+3y=0, \quad y(\pi/3)=2,\quad y'(\pi/3)=-1$$

## Q5.2.6

29. Prove: If the characteristic equation of

$ay''+by'+cy=0 \tag{A}$

has a repeated negative root or two roots with negative real parts, then every solution of (A) approaches zero as $$x\to\infty$$.

30. Suppose the characteristic polynomial of $$ay''+by'+cy=0$$ has distinct real roots $$r_1$$ and $$r_2$$. Use a method suggested by Exercise 5.2.22 to find a formula for the solution of

$ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1.\nonumber$

31 Suppose the characteristic polynomial of $$ay''+by'+cy=0$$ has a repeated real root $$r_1$$. Use a method suggested by Exercise 5.2.22 to find a formula for the solution of

$ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1.\nonumber$

32. Suppose the characteristic polynomial of $$ay''+by'+cy=0$$ has complex conjugate roots $$\lambda\pm i\omega$$. Use a method suggested by Exercise 5.2.22 to find a formula for the solution of

$ay''+by'+cy=0, \quad y(x_0)=k_0,\quad y'(x_0)=k_1.\nonumber$

33. Suppose the characteristic equation of

$ay''+by'+cy=0 \tag{A}$ has a repeated real root $$r_1$$. Temporarily, think of $$e^{rx}$$ as a function of two real variables $$x$$ and $$r$$.

1. Show that $a{\partial^2\over\partial^2 x}(e^{rx})+b{\partial \over\partial x}(e^{rx}) +ce^{rx}=a(r-r_1)^2e^{rx}. \tag{B}$
2. Differentiate (B) with respect to $$r$$ to obtain $a{\partial\over\partial r}\left({\partial^2\over\partial^2 x}(e^{rx})\right)+b{\partial\over\partial r}\left({\partial \over\partial x}(e^{rx})\right) +c(xe^{rx})=[2+(r-r_1)x]a(r-r_1)e^{rx}. \tag{C}$
3. Reverse the orders of the partial differentiations in the first two terms on the left side of (C) to obtain $a{\partial^2\over\partial x^2}(xe^{rx})+b{\partial\over\partial x}(xe^{rx})+c(xe^{rx})=[2+(r-r_1)x]a(r-r_1)e^{rx}. \tag{D}$
4. Set $$r=r_1$$ in (B) and (D) to see that $$y_1=e^{r_1x}$$ and $$y_2=xe^{r_1x}$$ are solutions of (A)

34. In calculus you learned that $$e^u$$, $$\cos u$$, and $$\sin u$$ can be represented by the infinite series

$e^u=\sum_{n=0}^\infty {u^n\over n!} =1+{u\over 1!}+{u^2\over 2!}+{u^3\over 3!}+\cdots+{u^n\over n!}+\cdots \tag{A}$

$\cos u=\sum_{n=0}^\infty (-1)^n{u^{2n}\over(2n)!} =1-{u^2\over2!}+{u^4\over4!}+\cdots+(-1)^n{u^{2n}\over(2n)!} +\cdots, \tag{B}$

and

$\sin u=\sum_{n=0}^\infty (-1)^n{u^{2n+1}\over(2n+1)!} =u-{u^3\over3!}+{u^5\over5!}+\cdots+(-1)^n {u^{2n+1}\over(2n+1)!} +\cdots \tag{C}$

for all real values of $$u$$. Even though you have previously considered (A) only for real values of $$u$$, we can set $$u=i\theta$$, where $$\theta$$ is real, to obtain

$e^{i\theta}=\sum_{n=0}^\infty {(i\theta)^n\over n!}. \tag{D}$

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 $$\theta$$.

1. Recalling that $$i^2=-1,$$ write enough terms of the sequence $$\{i^n\}$$ to convince yourself that the sequence is repetitive: $1,i,-1,-i,1,i,-1,-i,1,i,-1,-i,1,i,-1,-i,\cdots.\nonumber$ Use this to group the terms in (D) as \begin{aligned} e^{i\theta}&=\left(1-{\theta^2\over2}+{\theta^4\over4}+\cdots\right) +i\left(\theta-{\theta^3\over3!}+{\theta^5\over5!}+\cdots\right)\\ &=\sum_{n=0}^\infty (-1)^n{\theta^{2n}\over(2n)!} +i\sum_{n=0}^\infty (-1)^n{\theta^{2n+1}\over(2n+1)!}.\end{aligned}\nonumber 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).

This page titled 5.2.1: Constant Coefficient Homogeneous Equations (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench.