5.3.1: Constant Coefficient Homogeneous Equations (Exercises)

In Exercises 1-12 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. \(10y''-3y'-y=0\)

In Exercises 13-21 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. \(4y''-12y'+9y=0, \quad y(0)=3,\quad y'(0)={5\over 2}\)

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

22. 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 \(x\to\infty\).

23. 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 \(\theta\).

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.