Question

The general solution of the differential equation $[2\sqrt{xy}-x]dy+ydx=0$ is

• A. $\sqrt{xy}+y=c$
• B. $\sqrt{xy}-y=c$
• C. $y-\sqrt{xy}=c$
• D. none of these

Answer 1

The correct option is A $\sqrt{xy}+y=c$

$2\sqrt{xy}-x+y\frac{dx}{dy}=0$

$\Rightarrow$ $y\frac{dx}{dy}-x=-\sqrt{x}\sqrt{y}$

$\Rightarrow$ $\frac{dx}{dy}+\left(\frac{-1}{y}\right)x=\frac{-2\sqrt{x}}{\sqrt{y}}$

$\Rightarrow$ $\frac{1}{\sqrt{x}}\frac{dx}{dy}+\left(\frac{-1}{y}\right)\sqrt{x}=\frac{-2}{\sqrt{y}}$

Put $\sqrt{x}=t$

$\Rightarrow$ $\frac{1}{2\sqrt{x}}\frac{dx}{dy}=\frac{dt}{dy}$

$\Rightarrow$ $\frac{1}{\sqrt{x}}\frac{dx}{dy}=2\frac{dt}{dy}$

$\Rightarrow$ $2\frac{dt}{dy}+(-\frac{1}{y}).t=-\frac{2}{\sqrt{y}}$

$\Rightarrow$ $2\frac{dt}{dy}+(-\frac{1}{2y})t=-\frac{1}{\sqrt{y}}$

$\frac{dt}{dy}+(-\frac{1}{2y})t=-\frac{1}{\sqrt{y}}$

I.F$=e^{-\frac{1}{2}xy}=\sqrt{y}$

$\Rightarrow$ $\frac{d}{dy}(\sqrt{y}.t)=-1$

$\Rightarrow$ $t.\sqrt{y}=-y+c$

$\Rightarrow$ $t.\sqrt{y}=-y+c$

$\Rightarrow$$\sqrt{x}.\sqrt{y}=-y+c$

$\Rightarrow$ $\sqrt{xy}+y=c$