Question

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

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

Answer 1

Answer: (B)

$y+\sqrt{x^{2}+y^{2}}=Cx^{2}$

Answer 2

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