Check my answer -ordinary differential equation I'm trying to solve the ODE $x^2y'+2xy-y^3=0$ or in other words $y'+\frac{2}{x}y=\frac{1}{x^2}y^3$
According to wikipedia (at least in hebrew), this is what's called a Bernoulli equation, that we can solve by using the transform $z=y^{1-3}=y^{-2}$ and then our differential equation will be $z'-\frac{4}{x}z=-\frac{2}{x^2}$
Note: The exact phrasing is - if $y'+p(x)y=q(x)y^n$, then use the transformation $z=y^{1-n}$ and you will get the linear equation $z'+(1-n)p(x)z=(1-n)q(x)$. It is not immediately apparent to me why this is true.
At any rate, I used the transformation, define $u=e^{\int \frac{-4}{x}dx}=e^{-4 \ln x}=x^{-4}$
$uz=\frac{z}{x^4}\int -\frac{2}{x^6}dx=-2\int x^{-6}=\frac{2}{5}x^{-5}+C$ and so $z=\frac{2}{5}x^{-1}+Cx^4$
Since $z=\frac{1}{y^2}$, we get that $y^2=\frac{1}{z}$ and so $y=\frac{1}{\sqrt{\frac{2}{5}x^{-1}+Cx^4}}$ or $y=-\frac{1}{\sqrt{\frac{2}{5}x^{-1}+Cx^4}}$
Is this correct? If so, why is the thing in the "note" correct? If it isn't, where did I go wrong?

Your solution looks correct. Also, your equation for $z$ looks consistent with the 'note':
$$z' + (1-3)\frac{2}{x} z = (1-3)\frac{1}{x^2}$$

Added:
If $z = y^{1-n}$, then $z' = (1-n)y^{-n}y'$.
Thus
$$y' + p(x)y = q(x)y^n$$
$$\Leftrightarrow (1-n)y^{-n} [ y' + p(x)y ] = (1-n)y^{-n} [q(x)y^n]$$ 
$$\Leftrightarrow (1-n)y^{-n}y' + (1-n)p(x)y^{1-n} = (1-n)q(x)$$
$$\Leftrightarrow z' + (1-n)p(x)z = (1-n)q(x)$$
Personally, I never use this last form for Bernoulli equations. I just work through them starting with the substitution.
