# Find a Particular Solution of the Following Differential Equation:- ( 1 + X 2 ) D Y D X + 2 X Y = 1 1 + X 2 ; Y = 0 , When X = 1 - Mathematics

Sum

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

#### Solution

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

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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 67.1 | Page 147