Help Check DE Homework: x^2y' -2xy = 3y^4

[CINA]

Homework Statement

x^{2}\frac{dy}{dx}-2xy=3y^{4}

Homework Equations

Bernoulli's equation

The Attempt at a Solution

x^{2}\frac{dy}{dx}-2xy=3y^{4}

First I divide through by x^{2} and y^{n} to put it in standard form, and then to begin Bernoulli's equation process.

y^{-4}y'-\frac{2}{x}y^{-3}=\frac{3}{x^{2}}

u-substitution:

w=y^{1-n}=y^{-3}
w'=-3y^{-4}y'

Substitute in...
\frac{-1}{3}w'-\frac{2}{x}w=\frac{3}{x^{2}}

Put in Standard Form...

w'+\frac{6}{x}w=-\frac{9}{x^{2}}

Get the integrating factor...

p(x)=\frac{6}{x} \Rightarrow \int\frac{6}{x}dx = 6ln(x)\Rightarrow\mu=e^{6lnx}=x^{6}\int\frac{d}{dx}(w*x^{6})dx=\int\frac{-9}{x^{4}}dx

w=\frac{3}{x^{9}}+cx^{-6}

I'll stop here since I don't think the simplification with help much (w->y^-3)

Wolfram got this:

Clickity

Which kinda-sorta looks like my answer, except the powers are off. Where did I go wrong?

Using wolfram to simplify, with y^-3 plugged in for w

 

[jackmell]

Get the integrating factor...

p(x)=\frac{6}{x} \Rightarrow \int\frac{6}{x}dx = 6ln(x)\Rightarrow\mu=e^{6lnx}=x^{6}


\int\frac{d}{dx}(w*x^{6})dx=\int\frac{-9}{x^{4}}dx

That part is wrong.