Question

The solution of the differential equation $\frac{dy}{dx} + \frac{2yx}{1+x^{2}} = \frac{1}{\left(1+x^{2}\right)^{2}} $ is

• A. $y(1+x^2) = C + \tan^{-1} x $
• B. $\frac{y}{1 + x^2} = C + \tan^{-1} x$
• C. $y \, log (1+ x^2) = C + \tan^{-1} x $
• D. $y (1 + x^2) = C + \sin^{-1} x $

Answer 1

Answer: (A)

$\frac{dy}{dx} + \frac{2yx}{1+x^{2}} = \frac{1}{\left(1+x^{2}\right)^{2}} $
which is a linear differential equation.
Here, $P = \frac{2x}{1+x^{2}}, Q = \frac{1}{\left(1+x^{2}\right)^{2}} $
Now, $IF - e^{\int Pdx}$
$ = e^{\int \frac{2x}{1+x^{2}}dx} = e^{\log\left(1+x^2\right) } $
$= \left(1+x^{2}\right) $
$\therefore $ Solution of differential equation is
$y. \left(1+x^{2}\right) = \int \frac{1}{\left(1+x^{2}\right)^{2}}. \left(1+x^{2}\right)dx + C $
$\Rightarrow y\left(1+x^{2}\right) = \int \frac{1}{1+x^{2}} dx + C$
$ \Rightarrow y \left(1+x^{2}\right) = \tan^{-1}x + C $