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

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

Answers (1)

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}=tand 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}

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