Question

The general solution of the differential equation
$\frac{dy}{dx}=\left(x^3-2x{\tan }^{-1}y\right)\left(1+y^2\right)$ is

• A. $2{\tan }^{-1}x=y^2-1+2C{e}^{-x^2}$
• B. $2{\tan }^{-1}y=x^2-1+2C{e}^{-x^2}$
• C. $2{\tan }^{-1}y=y^2-1+2C{e}^{-x^2}$
• D. $2{\tan }^{-1}x=x^2-1+2C{e}^{-x^2}$

Answer 1

Answer: (B)

$\frac{1}{1+y^2}\frac{dy}{dx}+2x\left({\tan }^{-1}y\right)=x^3$
Put ${\tan }^{-1}y=z$
$\therefore \frac{1}{1+y^2}\frac{dy}{dx}=\frac{dz}{dx}$
$\frac{dz}{dx}+(2x)z=x^3$
$\Rightarrow z{e}^{x^2}=\frac{1}{2}\int 2e^{x^2}{x}^{3}dx+C$
$\Rightarrow 2e^{x^2}\left({\tan }^{-1}y\right)=x^2{e}^{x^2}-e^{x^2}+2C$
$\Rightarrow 2{\tan }^{-1}y=x^2-1+2C{e}^{-x^2}$