How do you solve a homogeneous ODE using Laplace Transform?

Question

Suppose I want to solve $$y'' - 4y = 0$$ all I.C. zero

Taking the laplace transform I get $$(s^2 -4)Y(s) = 0$$

So y(t) can be anything.

Why am I running into this problem and how can I get around this issue?

Answer 1

With no forcing and zero initial conditions, this is to be expected. Let's solve the problem in another manner. $$m^2-4=0\Rightarrow m=\pm 2$$ So $$ y(t)=Ae^{2t}+Be^{-2t} $$ Now you say zero IC so \begin{alignat}{2} y(0)&=A+B&&={}0\\ y'(0)&=A-B&&={}0 \end{alignat} Thus, $A = \pm B$ so $A=B=0$ and $y(t) = 0$.

From your Laplace transform, you have $$ Y(s) = \frac{0}{s^2-4} = 0\Rightarrow y(t) = 0 $$

Answer 2

The laplace transform of $y''-4y$ is $s^2 Y-4Y-sy(0)-y'(0)$. Thus you should have gotten $$ Y(s) = \frac{y'(0) + sy(0)}{s^2-4} $$ which you can then take the inverse laplace transform of.