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.