# For the Following Differential Equation, Find a Particular Solution Satisfying the Given Condition:- X ( X 2 − 1 ) D Y D X = 1 , Y = 0 When X = 2 - Mathematics

Sum

For the following differential equation, find a particular solution satisfying the given condition:- $x\left( x^2 - 1 \right)\frac{dy}{dx} = 1, y = 0\text{ when }x = 2$

#### Solution

We have,

$x\left( x^2 - 1 \right)\frac{dy}{dx} = 1$

$\Rightarrow \frac{dy}{dx} = \frac{1}{x\left( x^2 - 1 \right)}$

$\Rightarrow dy = \left\{ \frac{1}{x\left( x^2 - 1 \right)} \right\}dx$

Integrating both sides, we get

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

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

$\Rightarrow y = \int\left\{ \frac{1}{x\left( x + 1 \right)\left( x - 1 \right)} \right\}dx + C . . . . . . . . \left( 1 \right)$

$\text{Let }\frac{1}{x\left( x + 1 \right)\left( x - 1 \right)} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{x - 1}$

$\Rightarrow 1 = A\left( x + 1 \right)\left( x - 1 \right) + Bx\left( x - 1 \right) + Cx\left( x + 1 \right)$

$\Rightarrow 1 = A\left( x^2 - 1 \right) + B\left( x^2 - x \right) + C\left( x^2 + x \right)$

$\Rightarrow 1 = x^2 \left( A + B + C \right) + x\left( - B + C \right) - A$

Comparing both sides, we get

$- A = 1 . . . . . . . . . (2)$

$- B + C = 0 . . . . . . . . .(3)$

$A + B + C = 0 . . . . . . . . (4)$

Solving (2), (3) and (4), we get

$A = - 1$

$B = \frac{1}{2}$

$C = \frac{1}{2}$

$\therefore \frac{1}{x\left( x + 1 \right)\left( x - 1 \right)} = \frac{- 1}{x} + \frac{1}{2\left( x + 1 \right)} + \frac{1}{2\left( x - 1 \right)}$

Now, (1) becomes

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

$\Rightarrow y = - \int\frac{1}{x}dx + \frac{1}{2}\int\frac{1}{x - 1}dx + \frac{1}{2}\int\frac{1}{x - 1}dx$

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

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

Given:- y(2) = 0

$\therefore 0 = \frac{1}{2}\log \left| 2 - 1 \right| + \frac{1}{2}\log \left| 2 + 1 \right| - \log \left| 2 \right| + C$

$\Rightarrow C = \log \left| 2 \right| - \frac{1}{2}\log \left| 3 \right|$

Substituting the value of C, we get

$y = \frac{1}{2}\log \left| x - 1 \right| + \frac{1}{2}\log \left| x + 1 \right| - \log \left| x \right| + \log \left| 2 \right| - \frac{1}{2}\log \left| 3 \right|$

$\Rightarrow 2y = \log \left| x - 1 \right| + \log \left| x + 1 \right| - 2\log \left| x \right| + 2\log \left| 2 \right| - \log \left| 3 \right|$

$\Rightarrow 2y = \log \left| x - 1 \right| + \log \left| x + 1 \right| - \log \left| x^2 \right| + \log \left| 4 \right| - \log \left| 3 \right|$

$\Rightarrow 2y = \log\frac{\left( x - 1 \right)\left( x + 1 \right)}{x^2} - \left( \log\left| 3 \right| - \log\left| 4 \right| \right)$

$\Rightarrow y = \frac{1}{2}\log\frac{\left( x^2 - 1 \right)}{x^2} - \frac{1}{2}\log \left( \frac{3}{4} \right)$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 65.1 | Page 146