Question

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

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

Answer 1

Answer: (B)

$e^{x^3}$

Given, $\frac{dy}{dx}=-(3x^{2}ta{n}^{-1}y-x^3)(1+y^2)$

$\Rightarrow \frac{dy}{dx}=x^3(1+y^2)-3x^{2}ta{n}^{-1}y(1+y^2)$

$\Rightarrow \frac{1}{(1+y^2)}\cdot \frac{dy}{dx}=x^3-3x^{2}ta{n}^{-1}y$

$\Rightarrow \frac{1}{1+y^2}\cdot \frac{dy}{dx}+3x^2{\tan }^{-1}y=x^3$

Put ${\tan }^{-1}y=t\Rightarrow \frac{1}{1+y^2}\cdot \frac{dy}{dx}=\frac{dt}{dx}$

$\therefore \frac{dt}{dx}+3t{x}^2=x^3$

Which is linear differential equation in it.

Now, $I=F=e^{{\int }^{3x^{2}dx}}=e^{x^3}$