Question

For the following differential equation, find a particular solution satisfying the given condition:

\[x\left( x^2 - 1 \right)\frac{dy}{dx} = 1, y = 0\text{ when }x = 2\]

Answer 1

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

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

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

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

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

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