Question

Solution of differential equation $(y+x\sqrt{xy}(x+y\left)\right)dx+\left(y\sqrt{xy}\right(x+y)-x)dy=0$ is

• A. $\frac{x^2+y^2}{2}+{\cot }^{-1}\sqrt{\frac{x}{y}}=\lambda$
• B. $\frac{x^2+y}{2}-2{\cot }^{-1}\sqrt{\frac{x}{y}}=\lambda$
• C. $\frac{x^2+y^2}{2}+{\tan }^{-1}\sqrt{\frac{y}{x}}=\lambda$
• D. $\frac{x^2+y^2}{2}+{\tan }^{-1}\sqrt{\frac{y}{x}}=\lambda$

Answer 1

The correct option is D $\frac{x^2+y^2}{2}+{\tan }^{-1}\sqrt{\frac{y}{x}}=\lambda$

From given differential equations $(ydx-xdy)+x\left(\sqrt{xy}\right(x+y)dx+y\sqrt{xy}(x+y)dy=0$
$\Rightarrow ydx-xdy+\sqrt{xy}(x+y)(xdx+ydy)=0$
$\Rightarrow \frac{ydx-xdy}{(x+y)\sqrt{xy}}+xdx+ydy=0$
$\Rightarrow \frac{\frac{ydx-xdy}{y^2}}{(\frac{x}{y}+1)\sqrt{\frac{x}{y}}}+\frac{1}{2}(2xdx+2dy)=0$
$\Rightarrow \frac{d\left(\frac{x}{y}\right)}{sqrt\frac{x}{y}[1+{\left(\sqrt{\frac{x}{y}}\right)}^2]}+\frac{1}{2}(x^2+y^2)=0$
$\Rightarrow \int \frac{2vdv}{v(1+v^2)}+\frac{1}{2}\int d(x^2+y^2)=0$
$\Rightarrow 2{\tan }^{-1}\left(\frac{\sqrt{x}}{y}\right)+\frac{1}{2}(x^2+y^2)=\lambda$
Put $\Rightarrow \sqrt{\frac{x}{y}}=v\Rightarrow \frac{x}{y}=v^2$
$\Rightarrow d\left(\frac{x}{y}\right)=2vdv$
$=\frac{x^2+y^2}{2}+\pi -2{\cot }^{-1}\sqrt{\frac{x}{y}}=\lambda$
$=\frac{x^2+y^2}{2}+2{\cot }^{-1}\sqrt{\frac{x}{y}}=\lambda -\pi$