Question

dy/dx = xy + 2

Answer 1

Equation $y'+p(x)y=q(x)$ has a solution:

$y=e^{-\int{p(x)dx}}(C+\int{q(x)}e^{\int{p(x)dx}}dx)$

for equation $y'-xy=2 :$

$p(x)=-x; q(x)=2$

$\int{p(x)dx}=\int{(-x)dx}=-\frac{x^2}{2}$

$y=e^{\frac{x^2}{2}}(C+\int{2e^{-\frac{x^2}{2}}dx})$ is a solution of the equation.

The integral $\int{e^{-\frac{x^2}{2}}dx}$ is not expressed explicitly.