Question

The solution of the differential equation $\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{1+x^{22}}$ 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)

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

Answer 2

Given Equation is $\frac{d y}{d x}+\frac{2 y x}{1+x^{2}}=\frac{1}{\left(1+x^{2}\right)^{2}}$
It is comparing with linear differential equation $\frac{d y}{d x}+p y=Q$, we get
$p =\frac{2 x }{1+ x ^{2}}$ and $Q =\frac{1}{\left(1+ x ^{2}\right)^{2}}$
Now, IF $=e^{P d x}=e^{\frac{2 x}{1+x^{2}} d x}$
$e^{\left(\log 1+x^{2}\right)}=1+x^{2}$
Solution of differential equation is
$y\left(1+x^{2}\right)=\int \frac{1}{\left(1+x^{2}\right)^{2}}\left(1+x^{2}\right) d x+c$
$\Rightarrow y \left(1+ x ^{2}\right)=\int \frac{1}{\left(1+ x ^{2}\right)} d x + c$
$\Rightarrow y \left(1+ x ^{2}\right)=\tan ^{-1} x + c$
$\Rightarrow y =\frac{\tan ^{-1} x }{1+ x ^{2}}+ c$