Find the general solution of the differential equation: $\frac{d^2y}{dt^2}-4t \frac{dy}{dt}+(4t^2-2)y=0$ Q. Using method of reduction, find the general solution of the differential equation:
$$\frac{d^2y}{dt^2}-4t \frac{dy}{dt}+(4t^2-2)y=0$$
and
$$ y_1(t)=e^{t^2}$$
I proceeded by the way my professor taught us, he said find $u(t)$
$$u(t)=\frac{e^{-\int -4t dt}}{(e^{t^2})^2}=\frac{e^{2t}}{e^{2t^2}}=e^{2t-2t^2} $$
Then we need to find $y_2(t)$ where
$$ y_2(t)=e^{t^2} \int e^{2t-2t^2}  $$
but i'm having trouble integrating the right hand side, so either I did something wrong or I'm just simply stuck.
If anyone wants the answer, it is:
$$ y(t)=(c_1+c_2t)e^{t^2} $$
 A: If you use the method of reduction by assuming 

$$ y_2=u y_1 = u(t)e^{t^2} $$

and plugging back in the ode you should get

$$ u''(t) = 0  $$

which suggests a possible solution $u(t)=t$ which implies $y_2=uy_1=te^{t^2}$. Then the general solution od the ode is 

$$ y(t) = c_1 e^{t^2}+c_2te^{t^2} = (c_1+c_2 t)e^{t^2}. $$

A: let $$y = e^{t^2}, \frac{dy}{dt} = 2ty, \frac{d^2 y}{dt^2} = 2y + 2t\frac{dy}{dt} = 2y + 4t^2y$$ then we have $$ Ly = \frac{d^2 y}{dt^2} - 4t\frac{dy}{dt}+(4t^2 - 2)y=2y + 4t^2 y -4t(2ty)+(4t^2 - 2)y = 0.$$
we have verified that  $$L\left(e^{t^2} \right)= 0.$$  we will look for a second solution in the form 
$$y = e^{t^2}u,\, y' = 2te^{t^2}u+e^{t^2}u',\, y'' = 2e^{t^2}u + 4t^2e^{t^2}u+4te^{t^2}u' + e^{t^2}u''$$
setting 
$$\begin{align} 0 = Ly &=  2e^{t^2}u+4t^2e^{t^2}u+4te^{t^2}u' + e^{t^2}u''
-4t\left( 2te^{t^2}u+e^{t^2}u' \right) + (4t^2 -2)e^{t^2}u \\
&=u''+u'\left(4t-4t\right) +u\left(2+4t^2-8t^2+4t^2 - 2\right)\\=u''
\end{align}$$
a particular solution is $$u = t,\, y = te^{t^2} \text{ second independent solution.} $$
