Question

Solve: $(x^2+y^2)dx+\left(2xy\right)dy=0$

Answer 1

Let $v=\frac{y}{x}$

$\therefore y=vx$ and $dy=xdv+vdx$

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

$(x^2+v^2{x}^2)dx-2v{x}^2(xdv+vdx)=0$

$\therefore (1+v^2)dx=2v(xdv+vdx)$

$\therefore (1-v^2)dx=2vxdv$

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

Integrating both sides, we have

$\log x=-\log (1-v^2)+c$ where c is an arbitrary constant

$\therefore \log x=\log \left(\frac{k}{1-v^2}\right)$ where k is a constant greater than zero

$\therefore k=x(1-v^2)$

$\therefore k=x(1-\frac{y^2}{x^2})$

i.e. $x^2-y^2=kx$