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:

When the coefficient P involves $\tan x$, you do not need to memorize $\int \tan x\,dx = \ln|\sec x|$ -- just substitute $u=\cos x$ directly, since $\tan x\,dx$ becomes $-\frac{du}{u}$, and integrate from there.

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: \[ \text{I.F.} = \cos x \]

Answer 2

Concept:

• Integrate $-\tan x\,dx$ using a direct $u = \cos x$ substitution instead of recalling the standard result $\int \tan x\, dx = \ln|\sec x|$.

Step 1: Write the equation in standard linear form.
Comparing $\frac{dy}{dx} - y\tan x = e^x$ with $\frac{dy}{dx} + Py = Q$ gives $P = -\tan x$.

Step 2: Set up the integral for the Integrating Factor.
$\text{I.F.} = e^{\int P\,dx} = e^{-\int \tan x\,dx} = e^{-\int \frac{\sin x}{\cos x}\,dx}$

Step 3: Integrate by direct substitution.
Let $u = \cos x$, so $du = -\sin x\,dx$, meaning $\sin x\,dx = -du$. Then:
$\int \frac{\sin x}{\cos x}\,dx = \int \frac{-du}{u} = -\ln|u| = -\ln|\cos x|$
So $-\int \tan x\,dx = \ln|\cos x|$.

Step 4: Simplify the exponential.
$\text{I.F.} = e^{\ln|\cos x|}$. Since the exponential and natural log cancel directly:
$\text{I.F.} = \cos x$

Final Answer: $\cos x$