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प्रश्न
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
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उत्तर
We have,
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
\[ \Rightarrow \frac{y}{1 + y^2}dy = \frac{x}{1 - x^2}dx\]
Integrating both sides ,
\[\int\frac{y}{1 + y^2}dy = \int\frac{x}{1 - x^2}dx\]
\[\text{ Substituting }1 + y^2 = t\text{ and }1 - x^2 = u \]
\[2ydy = dt\text{ and }- 2x dx = du\]
\[ \therefore \frac{1}{2}\int\frac{1}{t}dt = \frac{- 1}{2}\int\frac{1}{u}du\]
\[ \Rightarrow \frac{1}{2}\log \left| t \right| = - \frac{1}{2}\log \left| u \right| + \log C\]
\[ \Rightarrow \frac{1}{2}\log \left| 1 + y^2 \right| = - \frac{1}{2}\log \left| 1 - x^2 \right| + \log C\]
\[ \Rightarrow \frac{1}{2}\left[ \log \left| 1 + y^2 \right| + \log \left| 1 - x^2 \right| \right] = \log C\]
\[ \Rightarrow \log \left( \left| 1 + y^2 \right|\left| 1 - x^2 \right| \right) = 2 \log C\]
\[ \Rightarrow \left( 1 + y^2 \right)\left( 1 - x^2 \right) = C^2 \]
\[ \Rightarrow \left( 1 + y^2 \right)\left( 1 - x^2 \right) = C_1 , ...........\left(\text{where }C_1 = C^2\right) \]
\[\text{ Hence, }\left( 1 + y^2 \right)\left( 1 - x^2 \right) = C_1\text{ is the required solution.}\]
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