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