# The general solution of the differential equation

Question:

The general solution of the differential equation

$\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0$ is :

(where $\mathrm{C}$ is a constant of integration)

1. $\sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right)+C$

2. $\sqrt{1+y^{2}}-\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right)+C$

3. $\sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\frac{1}{2} \log _{c}\left(\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)+C$

4. $\sqrt{1+y^{2}}-\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)+C$

Correct Option: 1

Solution:

$\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0$

$\Rightarrow \sqrt{(1+x)^{2}\left(1+y^{2}\right)}+x y \frac{d y}{d x}=0$

$\Rightarrow \sqrt{1+x^{2}} \sqrt{1+y^{2}}=-x y \frac{d y}{d x}$

$\Rightarrow \int \frac{y d y}{\sqrt{1+y^{2}}}=-\int \frac{\sqrt{1+x^{2}}}{x} d x$ ..........(1)

Now put $1+x^{2}=u^{2}$ and $1+y^{2}=v^{2}$

$2 x d x=2 u d u$ and $2 y d y=2 v d v$

$\Rightarrow x d x=u d u$ and $y d y=v d v$

substitude these values in equation (1)

$\int \frac{v d v}{v}=-\int \frac{u^{2} \cdot d u}{u^{2}-1}$

$\Rightarrow \int \mathrm{dv}=-\int \frac{\mathrm{u}^{2}-1+1}{\mathrm{u}^{2}-1} \mathrm{du}$

$\Rightarrow \mathrm{v}=-\int\left(1+\frac{1}{\mathrm{u}^{2}-1}\right) \mathrm{du}$

$\Rightarrow \mathrm{v}=-\mathrm{u}-\frac{1}{2} \log _{\mathrm{e}}\left|\frac{\mathrm{u}-1}{\mathrm{u}+1}\right|+\mathrm{c}$

$\Rightarrow \sqrt{1+y^{2}}=-\sqrt{1+x^{2}}+\frac{1}{2} \log _{e}\left|\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}-1}}\right|+c$

$\Rightarrow \sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left|\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}-1}}\right|+c$