Question

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

Answer 1

Given that, $x^2\frac{dy}{dx}=x^2+xy+y^2\Rightarrow \frac{dy}{dx}=1+\frac{y}{x}+\frac{y^2}{x^2}...\left(i\right)Letf(x,y)=1+\frac{y}{x}+\frac{y^2}{x^2}f(\lambda x,\lambda y)=1+\frac{\lambda y}{\lambda x}+\frac{{\lambda }^2{y}^2}{{\lambda }^2{x}^2}f(\lambda x,\lambda y)={\lambda }^0(1+\frac{y}{x}+\frac{y^2}{x^2})={\lambda }^{0}f(x,y)$
which is homogeneous expression of degree 0.
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
On substituting these values in Eq. (i), we get
$(v+x\frac{dv}{dx})=1+v+v^2\Rightarrow x\frac{dv}{dx}=1+v+v^2-v\Rightarrow x\frac{dv}{dx}=1+v^2\Rightarrow \frac{dv}{1+x^2}=\frac{dx}{x}$
On integrating both sides, we get
$ta{n}^{-1}v=log|x|+C\Rightarrow ta{n}^{-1}\left(\frac{y}{x}\right)=log|x|+C$