Question

The solution of the differential equation $\frac{dy}{dx}+\frac{2yx}{1+x^2}=\frac{1}{1+x^{22}}$ 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

The correct option is A $y(1+x^2)=c+{\tan }^{-1}x$

Given Equation is $\frac{dy}{dx}+\frac{2yx}{1+x^2}=\frac{1}{(1+x^2{)}^2}$

It is comparing with linear differential equation

$\frac{dy}{dx}+py=Q$, we get

$p=\frac{2x}{1+x^2}$ and $Q=\frac{1}{(1+x^2{)}^2}$

Now, IF $=e^{Pdx}=e^{\frac{2x}{1+x^2}dx}$

$e^{(log1+x^2)}=1+x^2$

Solution of differential equation is

$y(1+x^2)=\int \frac{1}{(1+x^2{)}^2}(1+x^2)dx+c$

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

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

$\Rightarrow y=\frac{{\tan }^{-1}x}{1+x^2}+c$