Question

Integrating factor of differential Equation $\frac{dy}{dx}{\cos }^{2}x+y=1$ is ___________.

• A. $e^{\cot x}$
• B. $e^{\tan x}$
• C. $e^{\sec x}$
• D. $e^{cosecx}$

Answer 1

The correct option is B $e^{\tan x}$

$\frac{dy}{dx}{\cos }^{2}x+y=1$
$\therefore \frac{dy}{dx}+\frac{y}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}$
$\therefore \frac{dy}{dx}+\left({\sec }^{2}x\right)y={\sec }^{2}x$
Now comparing with $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$
$\therefore P\left(x\right)={\sec }^{2}x$
$\therefore$ Integrating factors $=e^{\int P\left(x\right)dx}$
$=e^{\int {\sec }^{2}xdx}$
$=e^{tanx}$
$\therefore$ Integrating factors $=(I.F.)e^{\tan x}$.