Question

(x^2-y^2)dx+2xydy=0


so this is a problem of a differential equation I've been trying so hard to match with the given answer but failed every time I tried. So, is there anyone who can really can help me out to catch the mistakes that I'm making?
* The last line of my workout is just a dump guess.
The pictures are my workouts also the answer to this question is attached. Please read my solutions by this order : pic-1, pic-3 & pic-2, if necessary.

Answer 1

$(x^2 - y^2) \, dx + 2xy \, dy = 0$

Multiply both sides by $\frac1{x^2}$.

$\left(1 - \dfrac{y^2}{x^2}\right) \, dx + \dfrac{2y}x \, dy = 0$

Substitute $y=vx$, so $v=\frac yx$ and $dy=x\,dv+v\,dx$.

$(1-v^2) \, dx + 2v (x\,dv + v\,dx) = 0$

$(1 + v^2) \, dx + 2xv \, dv = 0$

Separate the variables.

$2xv\,dv = -(1 + v^2) \, dx$

$\dfrac{v}{1+v^2}\,dv = -\dfrac{dx}{2x}$

Integrate both sides

$\displaystyle \int \frac{v}{1+v^2}\,dv = -\frac12 \int \frac{dx}x$

On the left side, substitute $w=1+v^2$ and $dw=2v\,dv$.

$\displaystyle \frac12 \int \frac{dw}w = -\frac12 \int\frac{dx}x$

$\displaystyle \ln|w| = -\ln|x| + C$

Solve for $w$, then $v$, then $y$.

$e^{\ln|w|} = e^{-\ln|x| + C}$

$w = e^C e^{\ln|x^{-1}|}$

$w = Cx^{-1}$

$1 + v^2 = Cx^{-1}$

$1 + \dfrac{y^2}{x^2} = Cx^{-1}$

$\implies \boxed{x^2 + y^2 = Cx}$

Your mistake is in the first image, between third and second lines from the bottom. (It may not be the only one, it's the first one that matters.)

You incorrectly combine the fractions on the left side.

$\dfrac1{-2v} -\dfrac v{-2} = \dfrac1{-2v} - \dfrac{v^2}{-2v} = \dfrac{1-v^2}{-2v} = \dfrac{v^2-1}{2v}$