Laplace transform on a non-standard sort of problem

Question

I don't know where a laplace comes into play here:

$\ddot{a}+2a=0,a(0)=b_1,\dot{a}(0)=b_2$

I am meant to solve the above using a Laplace transform, but I don't see how I would use it here?

I see that $\dddot{a}+2=0$ and hence $\ddot{a}=-2x+c$, so then $-2x+c+2a=0$

$$a=\frac{2x-c}{2}$$ with $a(0)=b_1$ and hence $\frac{-c}{2}=b_1$, $c=-2b_1$, $a=\frac{2x-2b_1}{2}=x-2b_1$

This doesn't use my other IC? Where is my Laplace?

Answer 1

Take the transform of both sides:

$L(a'' - 2a) =0$

$\implies s^2 L(a) - sa(0) - a'(0) -2L(a)=0 $

$\implies s^2 L(a) - sb_1 - b_2 - 2L(a) = 0$

$\implies L(a) = \frac{sb_1+b_2}{s^2-2}$

Can you take it from here?