Question

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

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

Answer 1

Answer: (B)

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

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

$\frac{dy}{dx}=\frac{\sqrt{x^2-y^2}+y}{x}$

$\Rightarrow F(x,y)=\frac{\sqrt{x^2+y^2}+y}{x}$

$F(kx\cdot ky)=\frac{\sqrt{k^2{x}^2+k^2{y}^2+ky}}{kx}=k^o\cdot F(x,y)$

$y=vx$ and $\frac{dy}{x}=v+x\frac{dy}{dx}$

$v+x\frac{dv}{dx}=\frac{\sqrt{x^2+v^2{x}^2}+vx}{x}$

$v+x\frac{dv}{dx}=\sqrt{1+v^2}+v$

$\frac{xdv}{dx}=\sqrt{1+v^2}$

$\frac{dv}{\sqrt{1+v^2}}=\frac{dx}{k}$

$log|v+\sqrt{1-v^2})l=logx+loge$

$v=\frac{y}{x}$

$log\left(\frac{y}{x}\right)+\sqrt{1+y^2/x^2}=logcx$

$log\frac{y+\sqrt{x^2+y^2}}{x}=logcx$

$log(y+\sqrt{x^2+y^2})=logc{x}^2$

$y+\sqrt{x^2+y^2}=c{x}^2$.