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

4. (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:

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

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

Let $x=\tan \theta \Rightarrow d x=\sec ^{2} \theta d \theta$

$\Rightarrow \int \frac{\sec ^{3} \theta d \theta}{\tan \theta}=-\int \frac{2 y}{2 \sqrt{1+y^{2}}} d y$

$\Rightarrow \int \frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cdot \cos ^{2} \theta} d \theta=-\sqrt{1+y^{2}}$

$\Rightarrow \int(\tan \theta \cdot \sec \theta+\operatorname{cosec} \theta) d \theta=-\sqrt{1+y^{2}}$

$\Rightarrow \sec \theta+\log _{e}|\operatorname{cosec} \theta-\cot \theta|=-\sqrt{1+y^{2}}+C$

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

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