5.2E: Constant Coefficient Homogeneous Equations (Exercises)

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 $$\begin{aligned} e^{i\theta}&=\left(1-{\theta^2\over2}+{\theta^4\over4}+\cdots\right) +i\left(\theta-{\theta^3\over3!}+{\theta^5\over5!}+\cdots\right)\\[4pt] &=\sum_{n=0}^\infty (-1)^n{\theta^{2n}\over(2n)!} +i\sum_{n=0}^\infty (-1)^n{\theta^{2n+1}\over(2n+1)!}.\end{aligned}\nonumber$$ 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 $$\begin{aligned} \cos(\theta_1+\theta_2)&=\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\\[4pt] \sin(\theta_1+\theta_2)&=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2\end{aligned}\nonumber$$ 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).