Question

Solve y' + 2xy = 2x^3, y(0) = 1.

Answer 1

This is an ODE of the form y' + P(x)y = Q(x)
IF = $e^{ \int\limits {P(x)} \, dx }=e^{ \int\limits {2x} \, dx }=e^{x^2}$
Multiply through by IF
$e^{x^2}y'+2xe^{x^2}y=2x^3e^{x^2} \\ (e^{x^2}y)'=2x^3e^{x^2} \\ \int {(e^{x^2}y)'} \, dx = \int {2x^3e^{x^2}} \, dx \\ e^{x^2}y=x^2e^{x^2}-e^{x^2} + c \\ y=c_1e^{-x^2}+x^2-1$
But y(0) = 1
$1=c_1e^{-(0)^2}+(0)^2-1 = c_1-1 \\ c_1=1+1=2$
Therefore, solution is
 $y=2e^{-x^2}+x^2-1$