Question

Determine the exact expression for the Integrating Factor (I.F.) of the following first-order linear differential equation: \( \frac{dy}{dx} - y\tan x = e^x \)

• A. \( \sec x \)
• B. \( \cos x \)
• C. \( \sin x \)
• D. \( e^{-\tan x} \)

Hint:

Always ensure your linear differential equation is in standard form before identifying \( P \). The coefficient of the leading \( \frac{dy}{dx} \) term must equal exactly 1. If it doesn't, divide the entire equation by that term first.

Answer 1

The Correct Option is B

Concept: A first-order linear differential equation written in standard form is expressed as: \[ \frac{dy}{dx} + Py = Q \] Where \( P \) and \( Q \) are functions of \( x \) or constants. The Integrating Factor (I.F.) needed to solve this class of differential equations is given by the calculus formula: \[ \text{I.F.} = e^{\int P \, dx} \]

Step 1: Isolate the coefficient function \( P \) from the equation structure.
Comparing our given equation to the standard linear format shows: \[ P = -\tan x \] *(Note: It is crucial to include the negative sign attached to the tangent function).*

Step 2: Evaluate the indefinite integral of \( P \).
Set up and integrate the tangent function with respect to \( x \): \[ \int P \, dx = \int -\tan x \, dx = -\ln|\sec x| \] Using logarithmic properties, bring the negative sign inside as a reciprocal power: \[ -\ln|\sec x| = \ln\left| \frac{1}{\sec x} \right| = \ln|\cos x| \]

Step 3: Substitute the integrated term back into the exponential base.
Plug the evaluated integral into the final integrating factor template: \[ \text{I.F.} = e^{\ln|\cos x|} \] Since the exponential base \( e \) and natural logarithm \( \ln \) cancel each other out, the expression simplifies to: \[ \text{I.F.} = \cos x \]