#### Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 31 Maths Textbook Solution.

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

Hint: you must know the rules of solving differential equation. First rearrange the values and then solve.

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

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

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

Integrating both sides, we get,

\begin{aligned} &\int \frac{1}{\mathrm{y}(\mathrm{y}-1)} \mathrm{dy}=\int \frac{\mathrm{x}}{1-\mathrm{x}^{2}} \mathrm{dx} \ldots . .(\mathrm{I}) \\ &\text { L.H.S }:-\frac{1}{\mathrm{y}(\mathrm{y}-1)}=\frac{\mathrm{A}}{\mathrm{y}}+\frac{\mathrm{B}}{\mathrm{y}-1} \\ &1=\mathrm{A}(\mathrm{y}-1)+\mathrm{By} \end{aligned}

Substituting  $y=1$

$1=A\left ( 0-1 \right )+B\left ( 0 \right )$

$1=B$

Again substituting, $y=0$

$1=A\left ( 0-1 \right )+B(0)$

$1=-A$

$A=-1$

Substituting values of A and B in $\frac{1}{y\left ( y-1 \right )}=\frac{A}{y}+\frac{B}{y-1}$, we get,

\begin{aligned} &\frac{1}{y(y-1)}=\frac{-1}{y}+\frac{1}{y-1} \\ &\int \frac{1}{y(y-1)} d y=\int \frac{-1}{y} d y+\int \frac{1}{y-1} d y \\ &=-\log |y|+\log |y-1|+C_{1} \end{aligned}

Now considering R.H.S of (II), we have,

$\int \frac{x}{1-x^{2}}dx$

Here, putting $1-x^{2}=t$and differentiate both sides, we get,

\begin{aligned} &-2 x d x=d t \\ &x d x=\frac{-d t}{2} \\ &\because \int \frac{x}{1-x^{2}} d x \end{aligned}

\begin{aligned} &\Rightarrow \frac{-1}{2} \int \frac{1}{t} d t \\ &\Rightarrow \frac{-1}{2} \log |t|+C_{2} \quad\left[\because \int \frac{d t}{t}=\log |t|+C\right] \\ &\Rightarrow \frac{-1}{2} \log \left|1-x^{2}\right|+C_{2}\left[\text { where } t=1-x^{2}\right] \end{aligned}

Now substituting the values of $\int \frac{1}{y\left ( y-1 \right )}dy$ and $\int \frac{x}{1-x^{2}}dx$ in (I)

\begin{aligned} &\Rightarrow-\log |y|+\log |y-1|+C_{1}=-\frac{1}{2} \log \left|1-x^{2}\right|+c_{2} \\ &\Rightarrow-\log |y|+\log |y-1|=-\frac{1}{2} \log \left|1-x^{2}\right|+C \quad\left[\because \text { where } C_{2}-C 1=C\right] \end{aligned}