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`