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

Answer 1

Answer: (B)

$x \,d y-y d x=\sqrt{x^{2}+y^{2}} d x$
$x\, d y=\left(y+\sqrt{x^{2}+y^{2}}\right) d x$
$\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}$
It is homogeneous equation.
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$v+x \frac{d v}{d x}=\frac{v x+x \sqrt{1+v^{2}}}{x}=\frac{v+\sqrt{1+v^{2}}}{1}$
$\Rightarrow x \frac{d v}{d x}=\sqrt{1+v^{2}}$
$\Rightarrow \int \frac{d v}{\sqrt{1+v^{2}}}=\int \frac{d x}{x}$
$\Rightarrow \log \left(v+\sqrt{1+v^{2}}\right)=\log x+\log C$
$\Rightarrow \log \left(\frac{y}{x}+\frac{\sqrt{x^{2}+y^{2}}}{x}\right)=\log x+\log C$
$\Rightarrow \log \left(y+\sqrt{x^{2}+y^{2}}\right)-\log x=\log x+\log C$
$\Rightarrow \log \left(y+\sqrt{x^{2}+y^{2}}\right)=\log \left(x^{2} C\right)$
$\Rightarrow y+\sqrt{x^{2}+y^{2}}=x^{2} C$