#### Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.7 question 38 sub question (ii) maths

Answer: $\left(1+y^{2}\right)=\frac{c^{1}}{\left(1-x^{2}\right)}$

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

Given: $y\left(1-x^{2}\right) \frac{d y}{d x}=x\left(1+y^{2}\right)$

Solution: $y\left(1-x^{2}\right) \frac{d y}{d x}=x\left(1+y^{2}\right)$

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

Integrating both sides

$\int \frac{y d y}{\left(1+y^{2}\right)}=\int \frac{x}{\left(1-x^{2}\right)} d x$

$\frac{1}{2} \int \frac{2 y}{\left(1+y^{2}\right)} d y=\frac{-1}{2} \int \frac{-2 x}{\left(1-x^{2}\right)} d x$

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

\begin{aligned} &\frac{1}{2}\left[\log \left|1+y^{2}\right|+\log \left|1-x^{2}\right|\right]=\log c \\\\ &\log \left(1+y^{2}\right)\left(1-x^{2}\right)=\log c^{2} \\\\ &\left(1+y^{2}\right)\left(1-x^{2}\right)=c^{2} \\\\ &\left(1+y^{2}\right)=\frac{c^{1}}{1-x^{2}} \end{aligned}