Question

Find a particular solution of the following differential equation:- \[\left( 1 + x^2 \right)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}; y = 0,\text{ when }x = 1\]

Answer 1

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

\[y\left( 1 + x^2 \right) = \tan^{- 1} x - \frac{\pi}{4}\]