Question

Solution of the differential equation $xdy-ydx-\sqrt{x^2+y^2}dx=0$ is

• A. $y-\sqrt{x^2+y^2}=C{x}^2$
• B. $y+\sqrt{x^2+y^2}=C{x}^2$
• C. $x+\sqrt{x^2+y^2}=C{x}^2$
• D. $x-\sqrt{x^2+y^2}=C{x}^2$

Answer 1

Answer: (B)

The given differential equation is
$xdy-ydx-\sqrt{x^2+y^2}dx=0$
$\Rightarrow xdy=\left(y+\sqrt{x^2+y^2}\right)dx$
$\Rightarrow \frac{dy}{dx}=\frac{\left(y+\sqrt{x^2+y^2}\right)}{x}$
This is the linear differential equation.
Put y = vx, so that $\frac{dy}{dx}=v+x\frac{dy}{dx}.$Then
$v+x\frac{dy}{dx}=\frac{vx+\sqrt{x^2+v^2{x}^2}}{x}$
$\Rightarrow v+x\frac{dy}{dx}=v+\sqrt{1+v^2}$
$\frac{dv}{\sqrt{1+v^2}}=\frac{dx}{x}$
Integrating both sides, we get
$\int \frac{dv}{\sqrt{1+v^2}}=\int \frac{dx}{x}$
$\Rightarrow log\left(v+\sqrt{1+v^2}\right)=logx+logC$
$\Rightarrow v+\sqrt{1+v^2}=Cx$
$\Rightarrow \frac{y}{x}+\sqrt{1+\frac{y^2}{x^2}}=Cx\left[\because y=vx\right]$
$y+\sqrt{x^2+y^2}=C{x}^2$