Question

Show that the given differential equation is homogeneous and then solve it.

$xdy-ydx=\sqrt{x^2+y^2}dx$

Answer 1

Given, $xdy-ydx=\sqrt{x^2+y^2}dx$
$\Rightarrow \frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}$ ...(i)
Here, the numerator and denominator both have polynomial of degree 1. So, the given differential equation is homogeneous.
So, put y=vx
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx},$then Eq. (i) becomes
$v+x\frac{dv}{dx}=\frac{vx+\sqrt{x^2+x^2{v}^2}}{x}$
$\Rightarrow v+x\frac{dv}{dx}=\frac{x[v+\sqrt{1+v^2}]}{x}\Rightarrow v+x\frac{dv}{dx}=v+\sqrt{1+v^2}\Rightarrow \frac{1}{\sqrt{1+v^2}}dv=\frac{1}{x}dx$
On integrating both sides, we get
$\int \frac{1}{\sqrt{1+v^2}}dv=\int \frac{dx}{x}\Rightarrow log(v+\sqrt{1+v^2})=logx+C$ $[\because \int \frac{dx}{\sqrt{x^2}+a^2}=log|x+\sqrt{x^2+a^2}|+C]$
$\Rightarrow log[\frac{y}{x}+\sqrt{1+{\left(\frac{y}{x}\right)}^2}]=logx+logC$ $(Putv=\frac{y}{x})$
$\Rightarrow log[\frac{y}{x}+\sqrt{\frac{x^2+y^2}{x^2}}]=log\left(xC\right)$ $[\because logm+logn=logmn]$
$\Rightarrow \frac{y}{x}+\frac{x^2+y^2}{x}=Cx$
$\Rightarrow y+\sqrt{x^2+y^2}=C{x}^2$
which is the required solutions.