What was the IVP?

Question

When the Laplace Transform is applied to a certain IVP, the resulting equation is $(s^2Y - 2s - 1) + (sY - 2)- 2Y = \frac{4}{s}$

What was the IVP?

So I think I've solved this, but just want to make sure I got the correct answer.

I got:

$y(0) = 2$ , $y'(0) = 1$, & $C = 2$

$\therefore y'' + y' - 2y = 4$ was the IVP correct?

Answer 1

$$ \underbrace{(s^2Y - 2s - 1)}_{\mathcal{L}(y'') = s^2 \mathcal{L}(y) - s y(0) - y'(0)} + \underbrace{(sY - 2)}_{\mathcal{L}(y') = s \mathcal{L}(y) - y(0)} - 2\underbrace{Y}_{\mathcal{L}(y)} = \underbrace{\frac{4}{s}}_{\mathcal{L}(4 H)} $$ So $y(0) = 2$ and $y'(0) = 1$ fit. And the untransformed ODE was $$ y''+ y' - 2y = 4 H = 4 $$ as we only consider $y$ for non-negative arguments. (Note: $H$ is the Heaviside step function)

If $C$ means the coefficient for $y$ in the ODE, then I would set $C = -2$.