#### Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 21 maths textbook solution

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

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

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

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

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

Integrating both sides

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

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

Put

\begin{aligned} &1-x^{2}=t \\\\ &-2 x d x=d t \\\\ &-\log |y|+\log |y-1|=-\frac{1}{2} \int \frac{d t}{t} \end{aligned}

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