The inverse Laplace transformation of $e^s$

Question

I am solving the differential equation:

$$y'' + 3xy' -6y = 1, \ y(0) = y'(0) = 0$$

Using Laplace transformations.

I arrived at:

$$L(y)(s) = \frac{c}{s^3} e^{s^2 / 6} + \frac1{s^3}$$

Where $c$ is an arbitrary constant. I doubt that this is what it really is, though I ran through my calculations again and nothing seemed to be wrong.

So I did:

$$L(y - \frac{x^2}2 )(s) = \frac{c}{s^3} e^{s^2/6} = L(g \star h)$$

Where:

$$L(g)(s) = \frac{c}{s^3}$$

$$L(h)(s) = e^{s^2/6}$$

Now, all I have to do is find $h$, which seems not easy at all (if possible). Not to mention that I'll have to find the convolution afterwards.

Is there an inverse Laplace transformation for the function $u \rightarrow e^u$? How could it be found?

Thank you.

Answer 1

It is necessary to introduce the condition:

$$L \left( y \right) \left( \infty \right) =0$$

and then this implies that $ c = 0$.

According with this the solution is

$$y \left( x \right) ={\frac {{x}^{2}}{2}}$$