Target Exam

CUET

Subject

Section B1

Chapter

Differential Equations

Question:

The integrating factor of the differential equation $(1 - y^2) \frac{dx}{dy} + yx = ay, (-1 < y < 1)$ is:

Options:

$\frac{1}{y^2 - 1}$

$\frac{1}{\sqrt{y^2 - 1}}$

$\frac{1}{1 - y^2}$

$\frac{1}{\sqrt{1 - y^2}}$

Correct Answer:

$\frac{1}{\sqrt{1 - y^2}}$

Explanation:

The correct answer is Option (4) → $\frac{1}{\sqrt{1 - y^2}}$ ##

The given differential equation is

$(1 - y^2) \frac{dx}{dy} + yx = ay, (-1 < y < 1)$

Rewrite the differential equation:

$\frac{dx}{dy} + \frac{yx}{(1-y^2)} = \frac{ay}{(1-y^2)}$

The integrating factor is $e^{\int \frac{y}{1-y^2} dy}$

$= e^{-\frac{1}{2} \log |1-y^2|}$

$= \frac{1}{\sqrt{1-y^2}}$