Question

The general solution of the differential equation $\sqrt{1+x^2+y^2+x^2{y}^2}+xy\frac{dy}{dx}=0$ is (where $C$ is a constant of integration)

• A. $\sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}{\log }_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+C$
• B. $\sqrt{1+y^2}-\sqrt{1+x^2}=\frac{1}{2}{\log }_e\left(\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2}-1}\right)+C$
• C. $\sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}{\log }_e\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)+C$
• D. $\sqrt{1+y^2}-\sqrt{1+x^2}=\frac{1}{2}{\log }_e\left(\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right)+C$

Answer 1

Answer: (A)

$\sqrt{1+x^2+y^2+x^2{y}^2}+xy\frac{dy}{dx}=0$
$\Rightarrow \sqrt{(1+x{)}^2\left(1+y^2\right)}+xy\frac{dy}{dx}=0$
$\Rightarrow \sqrt{1+x^2}\sqrt{1+y^2}=-xy\frac{dy}{dx}$
$\Rightarrow \int \frac{ydy}{\sqrt{1+y^2}}=-\int \frac{\sqrt{1+x^2}}{x}dx.....(1)$
Now put $1+x^2=u^2$ and $1+y^2=v^2$
$2xdx=2udu$ and $2ydy=2vdv$
$\Rightarrow xdx=udu$ and $ydy=vdv$
substitude these values in equation (1)
$\int \frac{vdv}{v}=-\int \frac{u^2\cdot du}{u^2-1}$
$\Rightarrow \int dv=-\int \frac{u^2-1+1}{u^2-1}du$
$\Rightarrow v=-\int \left(1+\frac{1}{u^2-1}\right)du$
$\Rightarrow v=-u-\frac{1}{2}{\log }_e\left|\frac{u-1}{u+1}\right|+c$
$\Rightarrow \sqrt{1+y^2}=-\sqrt{1+x^2}+\frac{1}{2}{\log }_e\left|\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2-1}}\right|+c$
$\Rightarrow \sqrt{1+y^2}+\sqrt{1+x^2}=\frac{1}{2}{\log }_e\left|\frac{\sqrt{1+x^2}+1}{\sqrt{1+x^2-1}}\right|+c$