Question

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

Answer 1

$\left(x^2-y^2\right)dx+2xydy=0$

$\Rightarrow 2xydy=\left(y^2-x^2\right)dx$

$\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}$

putting $y=vx$

$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{v^2{x}^2-x^2}{2x\times vx}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{v^2-1}{2v}$

$\Rightarrow x\frac{dv}{dx}=\frac{v^2-1}{2v}-v$

$\Rightarrow x\frac{dv}{dx}=\frac{v^2-1-2v^2}{2v}$

$\Rightarrow x\frac{dv}{dx}=\frac{-v^2-1}{2v}$

$\Rightarrow \int \frac{2v}{v^2+1}=-\int \frac{dx}{x}$

pu8tting $v^2+1=t$

$2vdv=dt$

$\Rightarrow \int \frac{dt}{t}=-\int \frac{dx}{x}$

$\Rightarrow \log \left|t\right|=-\log \left|x\right|+\log c$

$\Rightarrow \log \left(v^2+1\right)=-\log \left|x\right|+\log c$

$\Rightarrow \log \left(\frac{y^2+x^2}{x^2}\right)=\log \left(\frac{c}{x}\right)$

$\Rightarrow \frac{x^2+y^2}{x^2}=\frac{c}{x}$

$\Rightarrow x^2+y^2=cx$