I'm trying to solve an IVP with non-constant coefficients $$ y'' + 3ty' - 6y = 1, \quad y(0) = 0, \; y'(0) = 0 $$
Taking the Laplace yields $$ s^2Y + 3(Y + sY') - 6Y = \frac{1}{s}$$ $$ Y' + \left(\frac{s}{3} - \frac{1}{s}\right)Y = \frac{1}{3s^2}$$
The integrating factor is $$ \mu(s) = \exp\left(\frac{s^2}{6} -\ln s\right) = \frac{1}{s}e^{\frac{s^2}{6}}$$
Thus $$ \frac{d}{ds}\left( \frac{1}{s}e^{\frac{s^2}{6}}Y \right) = \frac{1}{3s^3}e^{\frac{s^2}{6}}$$
This where I'm stuck because I'm pretty sure the integral on the RHS is non-elementary.
I do know that the solution supposed to be $y = t^2/2$ and $Y = 1/s^3$
I think you meant $y'' + 3 t y' - 6 y = 1$. Then you should get $$ s^2 Y(s) - 9 Y(s) - 3 s Y'(s) = \dfrac{1}{s}$$