Question

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

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

Answer 1

The correct option is B $e^{x^3}$

Let ${\tan }^{-1}y=t$
Differentiate both sides, we get
$\frac{1}{1+y^2}\frac{dy}{dx}=\frac{dt}{dx}$

Now, given equation
$\frac{dy}{dx}+(3x^2{\tan }^{-1}y-x^3)(1+y^2)=0$
$\Rightarrow \frac{1}{1+y^2}\frac{dy}{dx}+(3x^2{\tan }^{-1}y-x^3)=0$
Substitute the value in above equation.
$\frac{dt}{dx}+(3x^{2}t-x^3)=0$
$\Rightarrow \frac{dt}{dx}+3x^{2}t=x^3$
Therefore, the integrating factor is
$IF=e^{\int 3x^{2}dx}=e^{x^3}$