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

Answer: (A)

$\sqrt{3} \left( 2 + \log_e \sqrt{3} \right)$

Answer 2

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

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 $x = \frac{\pi}{4}$, $y = 2$:

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