Answer
$y = (t^2 + C)e^{-t^2}$
Work Step by Step
$$y' + 2ty = 2te^{-t^2}$$
We solve this by the integrating factor method:
$$\mu(t) = e^{\int P(t),dt}$$
Here, (P(t) = 2t).
Therefore,
$$\mu(t) = e^{\int 2t,dt} = e^{t^2}$$
Now, multiplying both sides of the equation by (\mu(t)):
$$e^{t^2}(y' + 2ty) = e^{t^2} \cdot 2te^{-t^2}$$
$$e^{t^2}y' + 2te^{t^2}y = 2t$$
Notice the left-hand side of the equation:
$$(e^{t^2}y)' = 2t$$
Then, integrating both sides:
$$\int (e^{t^2}y)',dt = \int 2t,dt$$
$$e^{t^2}y = t^2 + C$$
Dividing both sides by (e^{t^2}):
$$y = (t^2 + C)e^{-t^2}$$
