Solving $u''-tu'+(t-1)u=e^{\frac{t^2}{2}}$ 
Given the fact that $e^t$ solves the equation $u''-tu'+(t-1)u=0$, find the general solution of $$u''-tu'+(t-1)u=e^{\frac{t^2}{2}}$$

I get stuck in this problem. I can only use constant coefficients methods (no power series expansion).
I tried finding another independent solution for the homogeneous equation via its charateristic polynomial (hopping $t's$ will disapear when choosing the $\lambda$ for a solution $e^{\lambda t}$), I also try solutions like $e^tv(t)$ without success.

Trying with $u(t)=e^tv(t)$, you will get the differential equation
$$
v''(t)+(2-t)v'(t)=\exp(t^2/2-t).
$$
Next, if we set $w(t)=v'(t)$, we get
$$
w'(t)+(2-t)w'(t)=\exp(t^2/2-t).
$$
This differential equation can be solved using integrating factor, and we find that
$$
w(t)=C\exp(t^2/2-2t)+\exp(t^2/2-t).
$$
Thus, $v$ is given by
$$
v(t)=\int w(t)\,dt=\int\bigl(C\exp(t^2/2-2t)+\exp(t^2/2-t)\bigr)\,dt
$$
and $u$ by
$$
u(t)=e^t v(t)=e^t\int\bigl(C\exp(t^2/2-2t)+\exp(t^2/2-t)\bigr)\,dt.
$$
This integral will give you non-elementary functions (more specifically, some "imaginary error function"). I leave the details to you.
