Solve the initial value problem $x^2y'+y(x-y)=0$

Question

$x^2y'+y(x-y)=0$

$y(1)=-1$

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!)

Answer 1

write $$y'+\frac{y}{x}-\frac{y^2}{x^2}=0$$ and Substitute $$\frac{y}{x}=u$$ if $x\neq 0$

Answer 2

$$y'+\frac{y}{x}-\frac{y^2}{x^2}=0$$

$$xy'+y=\frac{y^2}{x}$$

$$(xy)'=\frac{(xy)^2}{x^3}$$ Substitute $z=xy$: $$z'=\frac{z^2}{x^3}$$ integrate: $$\int \frac {dz}{z^2}=\int\frac{dx}{x^3}=\frac{-1}{2x^2}+K$$ $$ \frac {1}{z}=\frac{1}{2x^2}+K$$ $$ \frac {1}{xy}=\frac{1}{2x^2}+K$$ $$ x=y(\frac 1 2 +Kx^2)$$ $$ y=\frac x {(\frac 1 2 +Kx^2)}=\frac {2x} {(1 +Kx^2)}$$ $$....$$

Answer 3

A Bernoulli equation is of the form $p(x)y'+q(x)y = y^n$. These are solved with the substitution $v=y^{1-n}$, which converts it into a linear equation. Your equation is such with $n=2$.