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प्रश्न
Find a particular solution of the following differential equation:- \[\left( 1 + x^2 \right)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}; y = 0,\text{ when }x = 1\]
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उत्तर
We have,
\[\left( 1 + x^2 \right)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}\]
\[ \Rightarrow \frac{dy}{dx} + \frac{2x}{\left( 1 + x^2 \right)}y = \frac{1}{\left( 1 + x^2 \right)^2}\]
\[\text{Comparing with }\frac{dy}{dx} + Py = Q,\text{ we get}\]
\[P = \frac{2x}{\left( 1 + x^2 \right)} \]
\[Q = \frac{1}{\left( 1 + x^2 \right)^2}\]
Now,
\[I . F . = e^{\int\frac{2x}{\left( 1 + x^2 \right)}dx} \]
\[ = e^{\log \left| 1 + x^2 \right|} \]
\[ = 1 + x^2 \]
So, the solution is given by
\[y \times I . F . = \int Q \times I . F . dx + C\]
\[ \Rightarrow y\left( 1 + x^2 \right) = \int\frac{1}{\left( 1 + x^2 \right)} dx + C\]
\[ \Rightarrow y\left( 1 + x^2 \right) = \tan^{- 1} x + C . . . . . \left( 1 \right)\]
Now,
When x = 1, y = 0
\[ \therefore 0\left( 1 + 1 \right) = \tan^{- 1} 1 + C\]
\[ \Rightarrow C = - 1\]
\[ \Rightarrow C = - \frac{\pi}{4}\]
Putting the value of `C` in (1), we get
\[y\left( 1 + x^2 \right) = \tan^{- 1} x - \frac{\pi}{4}\]
