Question

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

Answer 1

Let $y=vx$

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

Substituting the above in $2xy\frac{dy}{dx}=x^2+y^2$

$\Rightarrow 2x\left(vx\right)(v+x\frac{dv}{dx})=x^2+{\left(vx\right)}^2$

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

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

$\Rightarrow 2v^2+2vx\frac{dv}{dx}-1-v^2=0$

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

$\Rightarrow \frac{2vdv}{1-v^2}=\frac{dx}{x}$ on re-arranging

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

Let $t=1-v^2\Rightarrow dt=-2vdv$

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

$\Rightarrow -\log t=\log x+c$ where $c$ is the constant of integration.

$\Rightarrow -\log (1-v^2)-\log x=c$ where $t=1-v^2$

$\Rightarrow -\log (1-{\left(\frac{y}{x}\right)}^2)-\log x=c$ where $v=\frac{y}{x}$

$\Rightarrow \log \left(\frac{x^2-y^2}{x^2}\right)+\log x=-c$ using the identity $\log a+\log b=\log ab$

$\Rightarrow \left[x\left(\frac{x^2-y^2}{x^2}\right)\right]={10}^{-c}=k$ where $k$ is the constant of integration

$\Rightarrow \frac{x^2-y^2}{x}=k$

$\Rightarrow x^2-y^2=kx$ is the required solution.