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. $ylog(1+x^2)=C+{\tan }^{-1}x$
• 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
$y.\left(1+x^2\right)=\int \frac{1}{{\left(1+x^2\right)}^2}.\left(1+x^2\right)dx+C$
$\Rightarrow y\left(1+x^2\right)=\int \frac{1}{1+x^2}dx+C$
$\Rightarrow y\left(1+x^2\right)={\tan }^{-1}x+C$