Question

The general solution of the differential equation $[2\sqrt{xy}-x]dy+ydx=0,x>0$ is :
(where $c$ is integration constant)

• A. $\ln x+\sqrt{\frac{y}{x}}=c$
• B. $\ln y-\sqrt{\frac{x}{y}}=c$
• C. $\ln y+\sqrt{\frac{x}{y}}=c$
• D. $\ln y+\sqrt{\frac{y}{x}}=c$

Answer 1

The correct option is C $\ln y+\sqrt{\frac{x}{y}}=c$

$x>0\Rightarrow y>0$
Now
$[2\sqrt{xy}-x]dy+ydx=0\Rightarrow \frac{dy}{dx}=\frac{y}{x-2\sqrt{xy}}\cdots \left(i\right)$
This is a homogeneous D.E.

Put $y=vx$ so that
$\frac{dy}{dx}=v+x\frac{dv}{dx}.$

Then, equation $\left(i\right)$ becomes : $v+x\frac{dv}{dx}=\frac{vx}{x-2\sqrt{v{x}^2}}\Rightarrow x\frac{dv}{dx}=\frac{v}{1-2\sqrt{v}}-v$
$\Rightarrow x\frac{dv}{dx}=\frac{2v^{\frac{3}{2}}}{1-2\sqrt{v}}\Rightarrow \int \frac{1-2v^{\frac{1}{2}}}{v^{\frac{3}{2}}}dv=2\int \frac{dx}{x}\Rightarrow \int \frac{dv}{v^{\frac{3}{2}}}-2\int \frac{dv}{v}=2\int \frac{dx}{x}\Rightarrow \frac{-2}{\sqrt{v}}-2\ln |v|=2\ln x+2k$
$\Rightarrow \frac{1}{\sqrt{v}}+\ln \left(|v|x\right)+k=0\Rightarrow \ln y+\sqrt{\frac{x}{y}}=c$ where $c=-k=$ constant.