Answer
$y=\frac{x^6}{3}+C_1x^3+C_2$
Work Step by Step
We are given:
$y''-2x^{-1}y'=6x^4$
Suppose that
$u=\frac{dy}{dx}$
Now, $\frac{du}{dx}=\frac{d^2y}{dx^2}$
Subsitute to the equation:
$\frac{du}{dx}-2x^{-1}u=6x^4$
Integrating factor:
$I(x)=e^{\int =2x^{-1}dx}=e^{-2\ln x}=x^{-2}$
The equation becomes:
$\frac{d}{dx}(ux^{-2})=6x^2$
Integrating both sides:
$ux^{-2}=2x^3+C$
$ u=2x^5+Cx^2$
$\frac{dy}{dx}=2x^5+Cx^2$
$dy=(2x^5+Cx^2)dx$
Integrating the last equation,
$y=\frac{x^6}{3}+C_1x^3+C_2$
The final solution is:
$y=\frac{x^6}{3}+C_1x^3+C_2$
