Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Differential Equations

Question:

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

Options:

$-x$

$\frac{x}{1+x^2}$

$\sqrt{1-x^2}$

$\frac{1}{2} \log \left(1-x^2\right)$

Correct Answer:

$\sqrt{1-x^2}$

Explanation:

We have,

$\left(1-x^2\right) \frac{d y}{d x}-x y=1 \Rightarrow \frac{d y}{d x}-\left(\frac{x}{1-x^2}\right) y=\frac{1}{1-x^2}$

It is a linear differential equation with Integrating factor given by

Integrating factor = $e^{-\int \frac{x}{1-x^2} d x}=e^{\frac{1}{2} \log \left(1-x^2\right)}=\sqrt{1-x^2}$