I am not sure where I went wrong. I am very confident my algebra is fine and I set up my partial fractions exactly as I was supposed to. I am not sure what is wrong.
$(x^2-y^2)\frac{dy}{dx}=2xy$
of course $v=\frac{y}{x}$ and $y=xv$ and $y' = v+\frac{dv}{dx}$
using y = xv:
$v + x \frac{dv}{dx} = \frac{2x^2v}{x^2-x^2v^2}$
$v + x\frac{dv}{dx} = \frac{2v}{1-v^2}$
subtracting V over:
$x\frac{dv}{dx} = \frac{2v}{1-v^2}-\frac{v(1-v^2)}{1-v^2}$
$x\frac{dv}{dx} = \frac{2v-(v(1-v^2))}{1-v^2}$
$x\frac{dv}{dx} = \frac{2v-(v-v^2)}{1-v^2}$
$x\frac{dv}{dx} = \frac{v+v^3}{1-v^2}$
$\int \frac{1-v^2}{v(1+v^2)}dv = \int \frac{dx}{x}$
LHS: $\frac{A}{v}+\frac{Bx+c}{1+v^2}$
$A(1+v^2)+(Bv+C)v$
$1-v^2 = A + Av^2+ Bv^2 + Cv$
equating coefficients:
$-v^2 : -1 = B+A$
$0 = Cv \to C = 0$
$1 = A$ so $-1 = B + 1 \to B = -2$
$\int \frac{1}{v}dv + \frac{-2}{1+v^2} = \int \frac{dx}{x}$
$\int \frac{1}{v} dv + -2\int \frac{1}{1+v^2}dv = \int \frac{dx}{x}$
$ln \vert v \vert -2arctan{v} = ln \vert x \vert + C$
After using the answer:
After getting $ln \vert v \vert - ln \vert 1+v^2 \vert$
I cannot get to the answer
$y = C(x^2 + y^2)$
my attempt multiply both sides by $e$ and obtain:
$v - 1 - v^2 = x + C$
one gets absorbed with C:
$v-v^2= x + C$
$\frac{y}{x} - \frac{y^2}{x^2} = x+ C$
I've tried multiplying x over, Ive tried subtracting x nothing seems to work..
Your partial fraction set up is \begin{align*} \frac{1-v^2}{v(1+v^2)}&=\frac{A}{v}+\frac{B\color{red}{v}+C}{1+v^2} \end{align*}
Then you got the values as $A=1, B=-2$ and $C=0$. This means \begin{align*} \frac{1-v^2}{v(1+v^2)}&=\frac{1}{v}-\frac{2\color{red}{v}}{1+v^2} \end{align*} Now when you integrate you get \begin{align*} \int \frac{1-v^2}{v(1+v^2)} \, dv & =\int \frac{1}{v} \, dv-\int \frac{2\color{red}{v}}{1+v^2}\,dv\\ &=\ln |v |-\ln|1+v^2|+C. \end{align*}