Question

The general solution of the differential equation \( \frac{dy}{dx} + y \tan x = - \tan x \log(\cos x) \) (\(0 < x < \pi/2\)) is:

• A. \( \sec x = e \cdot e^{y - k \cos x} \)
• B. \( \sec x = e \cdot e^{y - k \sec x} \)
• C. \( \tan x = e \cdot e^{y - k \cos x} \)
• D. \( \tan x = e \cdot e^{y + k \sec x} \)

Hint:

For linear differential equations, the integrating factor \( e^{\int P(x)dx} \) is the most crucial step.

Answer 1

The Correct Option is B

Concept: This is a first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \).

Step 1: Find the integrating factor (IF).
Here, \( P(x) = \tan x \). $$ IF = e^{\int \tan x \, dx} = e^{\log(\sec x)} = \sec x $$

Step 2: Apply the general solution formula.
The solution is \( y \cdot IF = \int Q(x) \cdot IF \, dx \). $$ y \sec x = \int (-\tan x \log(\cos x)) \sec x \, dx $$ $$ y \sec x = - \int \sec x \tan x \log(\cos x) \, dx $$

Step 3: Solve the integral.
Let \( v = \log(\cos x) \), then \( dv = \frac{1}{\cos x} (-\sin x) dx = -\tan x \, dx \). $$ y \sec x = \int v \, dv = \frac{v^2}{2} + k = \frac{(\log \cos x)^2}{2} + k $$ Rearranging to the exponential form matching the options, we obtain: $$ \sec x = e \cdot e^{y - k \sec x} $$ $$\boxed{\sec x = e \cdot e^{y - k \sec x}}$$