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प्रश्न
For the following differential equation, find a particular solution satisfying the given condition:
\[x\left( x^2 - 1 \right)\frac{dy}{dx} = 1, y = 0\text{ when }x = 2\]
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उत्तर
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
\[y = \frac{1}{2}\log \left| x - 1 \right| + \frac{1}{2}\log \left| x + 1 \right| - \log \left| x \right| + \log \left| 2 \right| - \frac{1}{2}\log \left| 3 \right|\]
\[ \Rightarrow 2y = \log \left| x - 1 \right| + \log \left| x + 1 \right| - 2\log \left| x \right| + 2\log \left| 2 \right| - \log \left| 3 \right|\]
\[ \Rightarrow 2y = \log \left| x - 1 \right| + \log \left| x + 1 \right| - \log \left| x^2 \right| + \log \left| 4 \right| - \log \left| 3 \right|\]
\[ \Rightarrow 2y = \log\frac{\left( x - 1 \right)\left( x + 1 \right)}{x^2} - \left( \log\left| 3 \right| - \log\left| 4 \right| \right)\]
\[ \Rightarrow y = \frac{1}{2}\log\frac{\left( x^2 - 1 \right)}{x^2} - \frac{1}{2}\log \left( \frac{3}{4} \right)\]
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