If $\frac{dy}{dt}-ty=te^{-t}$, and $y(1)=1$, then what is $\lim_{t \to \infty} \frac{y(t)}{t}$? 
If $\dfrac{dy}{dt}-ty=te^{-t}$, and $y(1)=1$, then what is $\lim_{t \to \infty} \dfrac{y(t)}{t}$?

So I tried the usual technique for a differential equation of the form $\dfrac{dy}{dt}+P(t)y=Q(t)$ with an integrating factor $e^{\int{-t\,dt}}=e^{-t^2/2} $, giving me  $$y(t)=e^{t^2/2} \int te^{-t-t^2/2} \, dt$$
However, I cannot solve the integral, so I am guessing it may have something to do with the limit, or I am just messing up somewhere. I also tried the substitution $x(t)=\dfrac{y(t)}{t}$ with the same outcome.
 A: For every $t\geqslant1$,
$$y(t)=\mathrm e^{t^2/2}\left(\mathrm e^{-1/2}+\int_1^ts\mathrm e^{-s-s^2/2}\mathrm ds\right).$$
When $t\to+\infty$, the integral converges to a positive limit $C$ hence 
$$
\frac{y(t)}t\sim\frac{\mathrm e^{t^2/2}}t\left(\mathrm e^{-1/2}+C\right).$$
The RHS goes to infinity when $t\to+\infty$ hence $\lim\limits_{t\to+\infty}y(t)/t=+\infty$.
A shortcut is to note that the conditions that $y'(t)-ty(t)\geqslant0$ for every $t\geqslant1$ and $y(1)=1$ imply that the function $t\mapsto\mathrm e^{t^2/2}y(t)$ is nondecreasing on $t\geqslant1$ hence $y(t)\geqslant\mathrm e^{t^2/2}\mathrm e^{-1/2}$ for every $t\geqslant1$. This is enough to conclude that $y(t)/t\to+\infty$ when $t\to+\infty$ since $t\ll\mathrm e^{t^2/2}$ when $t\to+\infty$.
A: When $t \gg 1$, $\dot{y} - ty \sim 0$. Then, $y \sim A{\rm e}^{t^{2}/2}$ when $t \gg 1$. Since, $\dot{y} > 0\,, \forall\ t > 1$ and $y\left(1\right) = 1 > 0$ we have $A > 0$. So,
$$
\color{#ff0000}{\large\lim_{t \to \infty}{y\left(t\right) \over t} = \infty}
$$
