General solution of a differential equation nonhomogeneous I need to find the general solution of a differential equation
The equation is 
$$
\frac{dy}{dt} + 2ty = \sin(t)\exp(-t^2)
$$
Any tips on how to begin or full answers? 
 A: if $y'+py=q$ then $u=e^{\int p}$ is a integrating factor. 
In that case,$u=e^{t^2}$ since $p=2t$. so we have,
$$(yu)'=sin(t).exp(-t^2).exp(t^2)$$
$$(yu)'=sin(t)\implies yu=-cos(t)+c$$
$$y=(-cos(t)+c)exp(-t^2)$$
$$=-cos(t)e^{-t^2}+ce^{-t^2} $$
Note: If you are not familiar with this method please check; http://en.wikipedia.org/wiki/Integrating_factor
A: The first order linear ODE can be solved by using integrating factor. The integration factor is
$$
e^{\ \large\int2t\ dt}=e^{\ \large t^2+C}=e^{\ \large t^2}e^{C}=Ae^{\ \large t^2}.
$$
Multiply the ODE by the integration factor, yield
$$
\begin{align}
\frac{dy}{dt}\ Ae^{\ \large t^2}+ 2ty\ Ae^{\ \large t^2}&= \sin t\ e^{\ -\large t^2}\ Ae^{\ \large t^2}\\
\frac{dy}{dt}\ e^{\ \large t^2}+ 2ty\ e^{\ \large t^2}&= \sin t\\
\frac{d}{dt}\left(y\ e^{\ \large t^2}\right)&= \sin t\\
d\left(y\ e^{\ \large t^2}\right)&= \sin t\ dt\\
\int\ d\left(y\ e^{\ \large t^2}\right)&=\int\sin t\ dt\\
y\ e^{\ \large t^2}&=-\cos t+C\\
y(t)&=e^{-\ \large t^2}(-\cos t+C)\\
&=Ce^{\ -\large t^2}-e^{\ -\large t^2}\cos t.
\end{align}
$$
