Answer
$y = e^{t^2}\int_{0}^{t}e^{-{s^2}}ds + e^{t^2}$
Work Step by Step
Given the differential equation $y'-2ty = 1$, confirm that $y = e^{t^{2}}\int_{0}^{t}e^{-s^{2}}ds + e^{t^{2}}$ is a solution.
$\mathit{Confirmation}.$
First, we differentiate the proposed solution to get an explicit expression for $y'$:
For $y = e^{t^{2}}\int_{0}^{t}e^{-s^{2}}ds+e^{t^{2}}$,
$y' = (2te^{t^{2}}\int_{0}^{t}e^{-s^{2}}ds+e^{t^{2}}\cdot e^{-t^{2}})+ 2te^{t{2}}$,
$=2te^{t^{2}}\int_{0}^{t}e^{-s^{2}}ds+1+2te^{t^{2}}.$
Then we substitute our expressions for $y$ and $y'$ in the given differential equation. Thus we get
$y'-2ty = (2te^{t^{2}}\int_{0}^{t}e^{-s^{2}}ds+1+2te^{t^{2}})-2t(e^{t^{2}}\int_{0}^{t}e^{-s^{2}}ds+e^{t^{2}})$,
which, after cancellation of terms,
$=1$.
Our final equation, then, is the given differential equation, which was to be shown.
