Question

Let $y=y\left(t\right)$ be a solution of the differential equation $y^{\prime }+2ty=t^2$, then $16\underset{t\to \infty }{lim}\frac{y}{t}$ is

Answer 1

$\frac{dy}{dt}+2ty=t^2$
I.F. $=e^{t^2}$
Thus, solution is $y.e^{t^2}=\int t^2{e}^{t^2}dt=\frac{1}{2}\int t.(2t.e^{t^2})dt$
$\therefore y.e^{t^2}=t.\frac{e^{t^2}}{2}-\frac{1}{2}\int e^{t^2}dt+C$
$\therefore y=\frac{t}{2}-e^{-t^2}\int \frac{e^{t^2}}{2}dt+C{e}^{-t^2}$
$\therefore \underset{t\to \infty }{lim}\frac{y}{t}=\frac{1}{2}-\underset{t\to \infty }{lim}\frac{\int \frac{e^{t^2}}{2}}{t{e}^{t^2}}+\frac{c}{t.e^{t^2}}=\frac{1}{2}$