Question

Solution of $(y+x\sqrt{xy}(x+y)+y)dx+(y\sqrt{xy}(x+y)-x)dy=0$. is

• A. $x^2+y^2=2{\tan }^{-1}\sqrt{\frac{y}{x}}+c$
• B. $x^2+y^2=4{\tan }^{-1}\sqrt{\frac{y}{x}}+c$
• C. $x^2+y^2={\tan }^{-1}\sqrt{\frac{y}{x}}+c$
• D. None of these

Answer 1

The correct option is B $x^2+y^2=4{\tan }^{-1}\sqrt{\frac{y}{x}}+c$

The given differential equation is

$(y+x\sqrt{xy}(x+y))dx+(y\sqrt{xy}(x+y)-x)dy=0$

$\Rightarrow ydx-xdy+\sqrt{xy}(x+y)(xdx+ydy)=0$

$\Rightarrow xdx+ydy=\frac{xdy-ydx}{(x+y)\sqrt{xy}}$

$\Rightarrow xdx+ydy=\frac{xdy-ydx}{(x+y)\sqrt{xy}}$

$\Rightarrow \frac{1}{2}d(x^2+y^2)=2d\left({\tan }^{-1}\sqrt{\frac{y}{x}}\right)$

$\Rightarrow x^2+y^2=4{\tan }^{-1}\sqrt{\frac{y}{x}}+c.$