Question

The solution of \( \frac{dy}{dx} + y \tan x = \sec x \), given \( y(0) = 0 \), is:

• A. \( y \sec x = \tan x \)
• B. \( y \tan x = \sec x \)
• C. \( \tan x = y \tan x \)
• D. \( x \sec x = \tan y \)
• E. \( y \cot x = \sec x \)

Hint:

Remember that \( \int \tan x \, dx = \ln|\sec x| \). This is a very common Integrating Factor step in calculus exams.

Answer 1

The Correct Option is A

Concept: This is a first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \). We solve this by finding the Integrating Factor \( \text{I.F.} = e^{\int P(x) \, dx} \). The general solution is then \( y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C \).

Step 1: Find the Integrating Factor.
Here, \( P(x) = \tan x \). \[ \text{I.F.} = e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x \]

Step 2: Find the general solution.
Multiply the entire equation by the I.F. and integrate: \[ y \cdot \sec x = \int \sec x \cdot \sec x \, dx \] \[ y \sec x = \int \sec^2 x \, dx \] \[ y \sec x = \tan x + C \]

Step 3: Apply the initial condition to find \( C \).
Given \( y = 0 \) when \( x = 0 \): \[ (0) \sec(0) = \tan(0) + C \] \[ 0 = 0 + C \implies C = 0 \] Substituting \( C = 0 \) back into the solution: \[ y \sec x = \tan x \]