Question

The integrating factor of the differential equation $\frac{d y}{d x}+\left(3 x^{2} \tan ^{-1} y-x^{3}\right)\left(1+y^{2}\right)=0$is

• A. $e^{x^2}$
• B. $e^{x^3}$
• C. $e^{3x^2}$
• D. $e^{3x^3}$

Answer 1

Answer: (B)

$e^{x^3}$

Answer 2

Given, $\frac{d y}{d x}=-\left(3 x^{2} \tan ^{-1} y-x^{3}\right)\left(1+y^{2}\right)$
$\Rightarrow \frac{d y}{d x}=x^{3}\left(1+y^{2}\right)-3 x^{2}\left(\tan ^{-1} y\right)\left(1+y^{2}\right)$
$\Rightarrow \frac{1}{\left(1+y^{2}\right)} \cdot \frac{d y}{d x}=x^{3}-3 x^{2} \tan ^{-1} y$
$\Rightarrow \frac{1}{1+y^{2}} \cdot \frac{d y}{d x}+3 x^{2} \tan ^{-1} y=x^{3}$
Put $\tan ^{-1} y=t$
$\Rightarrow \frac{1}{1+y^{2}} \cdot \frac{d y}{d x}=\frac{d t}{d x}$
$\therefore \frac{d t}{d x}+3 t x^{2}=x^{3}$
which is linear differential equation in $t$.
Now, I $F=\theta^{\int 3 x^{2} d x}=e^{x^{3}}$