<s>[INST] Solve the initial value problem $x^2y'+y(x-y)=0$ $x^2y'+y(x-y)=0$
$y(1)=-1$
Can someone point me in the right direction on this? I started doing this:
$y'+p(x)y=q(x)y^n$
So 
$p(x)=x-y, q(x)=1/y^2, n=2$
$x^2y'+y(x-y)=(1/y^2)y^2=1$
Then ran out of ideas...
(I don't want you give give me the answer, just steer me the right way please!)
 [/INST] $$y'+\frac{y}{x}-\frac{y^2}{x^2}=0$$
$$xy'+y=\frac{y^2}{x}$$
$$(xy)'=\frac{(xy)^2}{x^3}$$
Substitute $z=xy$:
$$z'=\frac{z^2}{x^3}$$
integrate:
$$\int \frac {dz}{z^2}=\int\frac{dx}{x^3}=\frac{-1}{2x^2}+K$$
$$ \frac {1}{z}=\frac{1}{2x^2}+K$$
$$ \frac {1}{xy}=\frac{1}{2x^2}+K$$
$$ x=y(\frac 1 2 +Kx^2)$$
$$ y=\frac x {(\frac 1 2 +Kx^2)}=\frac {2x} {(1 +Kx^2)}$$
$$....$$
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