Question

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

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

Answer 1

Answer: (B)

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