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प्रश्न
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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
\[\int dy = \int\left\{ \frac{1}{x\left( x^2 - 1 \right)} \right\}dx\]
\[ \Rightarrow y = \int\left\{ \frac{1}{x\left( x^2 - 1 \right)} \right\}dx\]
\[ \Rightarrow y = \int\frac{1}{x\left( x - 1 \right)\left( x + 1 \right)}dx\]
\[\text{ Let }\frac{1}{x\left( x - 1 \right)\left( x + 1 \right)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}\]
\[ \Rightarrow 1 = A\left( x^2 - 1 \right) + B\left( x^2 + x \right) + C\left( x^2 - x \right)\]
\[ \Rightarrow 1 = \left( A + B + C \right) x^2 + \left( B - C \right)x - A\]
Equating the coefficients on both sides we get
\[A + B + C = 0 . . . . . \left( 1 \right)\]
\[B - C = 0 . . . . . \left( 2 \right)\]
\[A = - 1 . . . . . \left( 3 \right)\]
\[\text{ Solving }\left( 1 \right), \left( 2 \right)\text{ and }\left( 3 \right),\text{ we get }\]
\[A = - 1\]
\[B = \frac{1}{2}\]
\[C = \frac{1}{2}\]
\[ \therefore y = \frac{1}{2}\int\frac{1}{x - 1}dx - \int\frac{1}{x}dx + \frac{1}{2}\int\frac{1}{x + 1}dx\]
\[ = \frac{1}{2}\log\left| x - 1 \right| - \log\left| x \right| + \frac{1}{2}\log\left| x + 1 \right| + C\]
\[ = \frac{1}{2}\log\left| x - 1 \right| + \frac{1}{2}\log\left| x + 1 \right| - \log\left| x \right| + C\]
\[\text{ It is given that }y\left( 2 \right) = 0 . \]
\[ \therefore 0 = \frac{1}{2}\log\left| 2 - 1 \right| + \frac{1}{2}\log\left| 2 + 1 \right| - \log\left| 2 \right| + C\]
\[ \Rightarrow C = \log\left| 2 \right| - \frac{1}{2}\log\left| 3 \right|\]
Substituting the value of C, we get
\[y = \frac{1}{2}\log\left| x - 1 \right| + \frac{1}{2}\log\left| x + 1 \right| - \log\left| x \right| + \log\left| 2 \right| - \frac{1}{2}\log\left| 3 \right|\]
\[ \Rightarrow 2y = \log\left| x - 1 \right| + \log\left| x + 1 \right| - 2\log\left| x \right| + 2\log\left| 2 \right| - \log\left| 3 \right|\]
\[ \Rightarrow 2y = \log\left| x - 1 \right| + \log\left| x + 1 \right| - \log\left| x^2 \right| + \log 4 - \log 3\]
\[ \Rightarrow 2y = \log\frac{4\left( x - 1 \right)\left( x + 1 \right)}{3 x^2}\]
\[ \Rightarrow y = \frac{1}{2}\log\frac{4\left( x^2 - 1 \right)}{3 x^2}\]
\[\text{ Hence, } y = \frac{1}{2}\log\frac{4\left( x^2 - 1 \right)}{3 x^2}\text{ is the solution to the given differential equation }.\]
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