second order ODE help

Question

I am trying to solve this ODE:

$$W''(t)+2W'(t)-(\lambda-1)W(t)=0.$$

where $\lambda=-(n\pi)^2$

I used the quadratic formula to give me $r=-1+n\pi i$ and $ r=-1-n\pi i$ hence $W=e^{-t}(C_1\cos(n\pi t)+ C_2\sin(n\pi t))$.

Is this correct?

Answer 1

The caharacteristic equation is $r^2+2r-(\lambda-1)=0$ and the roots are $r=-1+\sqrt{-\lambda}$ and $r=-1-\sqrt{-\lambda}$.

If $\lambda< 0$ then there are real roots.

If $\lambda>0$, there are complex roots: $r=-1-i\sqrt{\lambda}$ and $r=-1-i\sqrt{\lambda}$.

In this case the corresponding solution is $$W=\displaystyle e^{-t}\big( (c_1\sin(\sqrt{\lambda} t)+c_2\cos(\sqrt{\lambda} t)\big).$$