Question

The solution of the differential equation $xdy-ydx=\left(\sqrt{x^2+y^2}\right)dx$ is?

• A. $y-\sqrt{x^2+y^2}=C{x}^2$
• B. $y+\sqrt{x^2+y^2}=C{x}^2$
• C. $y+\sqrt{x^2+y^2}+C{x}^2=0$
• D. None of the above

Answer 1

The correct option is B $y+\sqrt{x^2+y^2}=C{x}^2$

$xdy-ydx=\sqrt{x^2+y^2}dx$
$xdy=(y+\sqrt{x^2+y^2})dx$
$\frac{dy}{dx}=\frac{y+\sqrt{x^2+y^2}}{x}$
It is homogeneous equation.
Put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$v+x\frac{dv}{dx}=\frac{vx+x\sqrt{1+v^2}}{x}=\frac{v+\sqrt{1+v^2}}{1}$
$\Rightarrow x\frac{dv}{dx}=\sqrt{1+v^2}$
$=\Rightarrow \int \frac{dv}{\sqrt{1+v^2}}=\int \frac{dx}{x}$
$\Rightarrow log(v+\sqrt{1+v^2})=logx+logC$
$\Rightarrow log\left(\frac{y}{x}+\frac{\sqrt{x^2+y^2}}{x}\right)=logx+logC$
$\Rightarrow log(y+\sqrt{x^2+y^2})-logx=logx+logC$
$\Rightarrow log(y+\sqrt{x^2+y^2})=log\left(x^{2}C\right)$
$\Rightarrow y+\sqrt{x^2+y^2}=x^{2}C$.