Question

 Find the general solution of the differential equation: dy = 2t y +(cost)et2

Answer 1

$\frac{dy}{dt} = 2t y +(cost)e^{t^2}\\ \frac{dy}{dt} -2t y =(cost)e^{t^2}\\ \text{This is a linear differential equation.}\\ \text{Integrating factor, IF}=e^{\int-2tdt}=e^{-t^2}\\ \text{Therefore, solution is given by}\\ ye^{-t^2}=\int e^{-t^2}(cost)e^{t^2}dt\\ ye^{-t^2}=\int cost \space dt\\ ye^{-t^2}=sint+c\\ y=e^{t^2}(sint+c)$