The solution of differential equation d y d x + 2 x y 1 + x 2 = 1 ( 1 + x 2 ) 2 is:

Question

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{{(1 + x^{2})}^{2}}$ is:

Smart Achievers · Accepted Answer

(1)

Ans. $y (1 + x^{2}) = C + \tan^{- 1} x$

Explanation:

Given that,

$\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{{(1 + x^{2})}^{2}}$

Here, $P = \frac{2 x}{1 + x^{2}}$ and $Q = \frac{1}{{(1 + x^{2})}^{2}}$

which is a linear differential equation.

$∴$ $IF = e^{\int \frac{2 x}{1 + x^{2}} d x}$

Put $1 + x^{2} = t \Rightarrow 2 x d x = d t$

$∴$ $IF = e^{\int \frac{d t}{t}} = e^{\log t} = e^{\log (1 + x^{2})} = 1 + x^{2}$

The general solution is

$y (1 + x^{2}) = \int (1 + x^{2}) \frac{1}{{(1 + x^{2})}^{2}} d x + C$

$\Rightarrow$ $y (1 + x^{2}) = \int \frac{1}{1 + x^{2}} d x + C$

$\Rightarrow$ $y (1 + x^{2}) = \tan^{- 1} x + C$

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