Question

Solve : $\frac{dy}{dx}=\frac{y}{x}+\frac{\sqrt{x^2+y^2}}{x},x>0.$

Answer 1

$\text{We have}$ $\frac{dy}{dx}=\frac{y}{x}+\frac{\sqrt{x^2+y^2}}{x},x>\mathrm{0....}\left(i\right)\text{(Consider)}f(x,y)=\frac{y}{x}+\frac{\sqrt{x^2+y^2}}{x}.\text{Put}x=\lambda x,y=\lambda y\Rightarrow f(\lambda x,\lambda y)=\frac{\lambda y}{\lambda x}+\frac{\sqrt{(}{\lambda }^2{x}^2+{\lambda }^2{y}^2)}{\lambda x}=\frac{y}{x}+\frac{\sqrt{(}{x}^2+y^2)}{x}=f(x,y).\text{So, f is homogeneous}\text{Let y=}vx\to \frac{dy}{dx}=v+x\frac{dv}{dx}=v+x\frac{dv}{dx}in\left(i\right)\therefore v+x\frac{dv}{dx}=\frac{vx}{x}+\frac{\sqrt{x^2+v^2{x}^2}}{x}\Rightarrow x\frac{dv}{dx}=\sqrt{1+v^2}\Rightarrow \int \frac{dv}{\sqrt{1+v^2}}=\int \frac{dx}{x}\Rightarrow log|v+\sqrt{1+v^2}|=log|x|+logC\Rightarrow log\left|\frac{y+\sqrt{x^2+y^2}}{x}\right|=log|Cx|\therefore y+\sqrt{x^2+y^2}=C{x}^2\text{is the required solution}.$