Answer
$y=x^6+\frac{Cx^3}{3}+C_2$
Work Step by Step
We are given:
$y''-2x^{-1}y'=18x^4$
Suppose that
$u=\frac{dy}{dx}$
Now, $\frac{du}{dx}=\frac{d^2x}{dx^2}$
Subsitute to the equation:
$\frac{du}{dx}-\frac{2u}{x}=18x^4$
Integrating factor:
$I(x)=e^{\int -\frac{2}{x}dx}=e^{-2\ln x}=x^{-2}$
The equation becomes:
$\frac{d}{dx}(ux^{-2})=18x^4 \times x^{-2}$
Integrating both sides:
$ux^{-2}=6x^5+C$1
$ u=6x^5+Cx^2$
$\frac{dy}{dx}=6x^5+Cx^2$
Integrating the last equation,
$y=x^6+\frac{Cx^3}{3}+C_2$
The final solution is:
$y=x^6+\frac{Cx^3}{3}+C_2$
