#### Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 18

Answer: $\sqrt{1+y^{2}}+\sqrt{1+x^{2}}+\frac{1}{2} l \operatorname{og}\left|\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right|+c=0$

Hint: Separate the terms of x and y and then integrate them.

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

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

\begin{aligned} &\Rightarrow x y \frac{d y}{d x}=-\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}} \\\\ &\Rightarrow x y \frac{d y}{d x}=-\sqrt{\left(1+x^{2}\right)\left(1+y^{2}\right)} \\\\ &\Rightarrow \frac{y d y}{\sqrt{1+y^{2}}}=\frac{-\sqrt{1+x^{2}}}{x} \end{aligned}

Integrating both sides

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

$\text { Let } 1+y^{2}=t \Rightarrow 2 y d y=d t \Rightarrow y d y=\frac{d t}{2}$

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

$\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{t}} d t=-\int \frac{m}{m^{2}-1} d m \cdot m$

$\Rightarrow \frac{1}{2} \frac{t^{\frac{1}{2}}}{\frac{1}{2}}+\int \frac{m^{2}}{m^{2}-1} d m=0$

\begin{aligned} &\Rightarrow \sqrt{t}+\int \frac{m^{2}-1+1}{m^{2}-1} d m=0 \\\\ &\Rightarrow \sqrt{t}+\int \frac{m^{2}-1}{m^{2}-1} d m+\int \frac{1}{m^{2}-1} d m=0 \\\\ &\Rightarrow \sqrt{t}+m+\frac{1}{2} \log \frac{(m-1)}{m+1}=0 \end{aligned}       $\cdots \cdot \int \frac{1}{m^{2}-1} d m=\frac{1}{2} \log \frac{(m-1)}{m+1}$

\begin{aligned} &\text { Where } m=\sqrt{1+x^{2}}, 1+y^{2}=t \\\\ &\sqrt{1+y^{2}}+\sqrt{1+x^{2}}+\frac{1}{2} l \operatorname{og}\left|\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right|+c=0 \end{aligned}