Laplace transform to solve an ODE

[fluidistic]

Homework Statement

I must solve the following diff. eq. $tx''-(4t+1)x'+(4t+2)x=0$ with the initial condition $x(0)=0$ and the relations $\mathcal {L }[tx]=-\frac{d \mathcal{L}[x]}{ds}$, $\mathcal {L} [tx']=-\frac{d [s \mathcal {L}[x]]}{ds}$ and $\mathcal{L}[x']=s \mathcal {L}[x]-x(0)$. Where x is a function of t.

Homework Equations

Already given.

The Attempt at a Solution

Well I've calculated $\mathcal {L}[x''t]$ which gave me $-s^2 \frac{d \mathcal {L}[x]}{ds}-2s \mathcal {L}[x]$.
Then I used the linearity of the Laplace transform and I applied the Laplace transform over the equation. Which eventually lead me to $-\frac{d(\mathcal{L}[x])}{ds} \cdot s^2 + \mathcal{L}[x] (2-3s)=0$.
This is where I was stuck when starting to write this thread. Because I'm used to obtain an algebraic expression for $\mathcal {L}[x]$, not a differential equation. I just solved it and reached $\mathcal{L}[x]=\frac{1}{s^3e^{\frac{2}{s}}}$. Is this the way to go? To finish, I should take the inverse Laplace transform of that.

 

[kurt.physics]

How should I do it? I'm having trouble calculating the inverse of that expression. Thanks in advance.