#### Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 47

Answer: $x+y+x y=1$

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

Given: $\left(1+x^{2}\right) \frac{d y}{d x}+\left(1+y^{2}\right)=0, y=1 \text { when } x=0$

Solution:

\begin{aligned} &\left(1+x^{2}\right) \frac{d y}{d x}+\left(1+y^{2}\right)=0 \\\\ &\left(1+x^{2}\right) \frac{d y}{d x}=-\left(1+y^{2}\right) \\\\ &\Rightarrow \frac{d y}{1+y^{2}}=-\frac{1}{1+x^{2}} d x \end{aligned}

Integrating both sides

\begin{aligned} &\Rightarrow \int \frac{1}{\left(1+y^{2}\right)} d y=-\int \frac{1}{\left(1+x^{2}\right)} \mathrm{dx} \\\\ &\tan ^{-1} y=-\tan ^{-1} x+c \end{aligned}            ................(1)

Given that $y=1 \text { when } x=0$

$\therefore \tan ^{-1}(1)=-\tan ^{-1}(0)+c \Rightarrow \frac{\pi}{4}=0+c \Rightarrow c=\frac{\pi}{4}$

Put in (1) we get

\begin{aligned} &\Rightarrow \tan ^{-1} y=-\tan ^{-1} x+\frac{\pi}{4} \\\\ &{\left[\therefore \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)\right]} \\\\ &\Rightarrow \tan ^{-1} x+\tan ^{-1} y=\frac{\pi}{4} \end{aligned}

\begin{aligned} &\Rightarrow \tan ^{-1}\left(\frac{x+y}{1-x y}\right)=\frac{\pi}{4} \\\\ &\Rightarrow\left(\frac{x+y}{1-x y}\right)=\tan \frac{\pi}{4} \\\\ \end{aligned}

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