how do you determine the solution of the this differential eqation:
xy'-2y=x^3e^x
Are you familiar with the method of integrating factor?
$\displaystyle
xy' - 2y = x^3 e^x $
divide the ODE by x:
$\displaystyle y' - \frac{2}
{x}y = x^2 e^x $
thus the integrating factor is:
$\displaystyle
\mu = e^{ - \int {\frac{2}
{x}dx} } = e^{ - 2\ln x} = \frac{1}
{{x^2 }}$
read this and try to continue:
Integrating factor - Wikipedia, the free encyclopedia