Mathematics - Differential Equation Question with Solution | TestHub

$\left(x^2-y^2\right)dx+2xydy=0$ $$\Rightarrow \frac{dy}{dx}=\frac{y^2-x^2}{2xy}$$ Let $y=vx$ $$\begin{array}{rl}& \frac{dy}{dx}=v+x\frac{dv}{dx}\\ \Rightarrow & v+x\frac{dv}{dx}=\frac{v^2{x}^2-x^2}{2v{x}^2}\\ \Rightarrow & v+x\frac{dv}{dx}=\frac{v^2-1}{2v}\\ \Rightarrow & x\frac{dv}{dx}=\frac{-v^2-1}{2v}\\ \Rightarrow & \frac{2vdv}{v^2+1}=-\frac{dx}{x}\end{array}$$ After integrating, we get $$\begin{array}{rl}& \ln \left|v^2+1\right|=-\ln |x|+\ln c\\ & \frac{y^2}{x^2}+1=\frac{c}{x}\end{array}$$ As curve passes through the point $(1,1)$, so $1+1=c\Rightarrow c=2$ $x^2+y^2-2x=0$, which is a circle of radius one.