Question

Let \( y = y(x) \) be the solution of the differential equation \[ \frac{dy}{dx} = \frac{(\tan x) + y}{\sin x (\sec x - \sin x \tan x)} \], \( x \in \left( 0, \frac{\pi}{2} \right) \) satisfying the condition \( y \left( \frac{\pi}{4} \right) = 2 \). Then, \( y \left( \frac{\pi}{3} \right) \) is

• A. \( \sqrt{3} \left( 2 + \log_e \sqrt{3} \right) \)
• B. \( \frac{\sqrt{3}}{2} \left( 2 + \log_e 3 \right) \)
• C. \( \sqrt{3} \left( 1 + 2 \log_e 3 \right) \)
• D. \( \sqrt{3} \left( 2 + \log_e 3 \right) \)

Answer 1

The Correct Option is A

\[ \frac{dy}{dx} - 2 \cos(2x) \cdot y = \sec^2 x \] \[ \frac{dy}{dx} + p \cdot y = Q \]

Integrating Factor (IF):

\[ IF = e^{\int p \, dx} = e^{-2 \int \csc(2x) \, dx} \]

Let \( 2x = t \):

\[ 2dx = dt \implies dx = \frac{dt}{2} \]

Calculating the Integrating Factor:

\[ e^{\int \csc(2x) dx} = e^{-\int \tan t \, dt} = e^{-\ln |\tan x|} = \frac{1}{|\tan x|} \]

So, the solution becomes:

\[ y(IF) = \int Q \cdot (IF) \, dx + c \] \[ y = \frac{1}{|\tan x|} \int \sec^2 x \cdot \frac{1}{|\tan x|} \, dx + c \] \[ y = \frac{1}{|\tan x|} \int \frac{dt}{|t|} + c \quad \text{(for \( \tan x = t \))} \] \[ y = \frac{1}{|\tan x|} \ln |t| + c \] \[ y = |\tan x| \left( \ln |\tan x| + c \right) \]

Setting specific values:

Put \( x = \frac{\pi}{4} \), \( y = 2 \):

\[ 2 = \ln 1 + c \implies c = 2 \] \[ y = |\tan x| \left( \ln |\tan x| + 2 \right) \] \[ y\left(\frac{\pi}{3}\right) = \sqrt{3} \left( \ln \sqrt{3} + 2 \right) \]