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

To solve the differential equation given by: 

\(\frac{dy}{dx} = \frac{(\tan x) + y}{\sin x (\sec x - \sin x \tan x)}\)

with the initial condition \( y \left( \frac{\pi}{4} \right) = 2 \), we need to determine \( y \left( \frac{\pi}{3} \right) \).

Step 1: Simplify the Differential Equation

The differential equation is

\(\frac{dy}{dx} = \frac{\tan x + y}{\sin x (\sec x - \sin x \tan x)}\)

First, simplify the denominator:

\(\sec x - \sin x \tan x = \frac{1}{\cos x} - \sin x \cdot \frac{\sin x}{\cos x} = \frac{1 - \sin^2 x}{\cos x} = \frac{\cos^2 x}{\cos x} = \cos x\)

Thus, the differential equation simplifies to:

\(\frac{dy}{dx} = \frac{\tan x + y}{\sin x \cos x}\)

Step 2: Separate Variables

Rewrite this as:

\((\tan x + y) dx = \sin x \cos x \, dy\)

Separate variables:

\((\tan x + y) dx = \sin x \cos x \, dy\)

Step 3: Integrate Both Sides

Integrate both sides:

\(\int \frac{1}{\sin x \cos x} \, dx = \int \frac{1}{\tan x + y} \, dy\)

The integration yields:

\(\log |\sin x| = \log |\tan x + y| + C\)

where \( C \) is the constant of integration.

Step 4: Apply Initial Condition

Using the initial condition \( y \left( \frac{\pi}{4} \right) = 2 \):

\(\log |\sin(\frac{\pi}{4})| = \log |1 + 2| + C\)

\(\log \left(\frac{\sqrt{2}}{2}\right) = \log 3 + C\)

Thus,

\(C = \log \left(\frac{\sqrt{2}}{6}\right)\)

Step 5: Solve for \( y \left( \frac{\pi}{3} \right) \)

Substituting back, resolve for \( y \left( \frac{\pi}{3} \right) \):

\(\log |\sin(\frac{\pi}{3})| = \log |1 + y(\frac{\pi}{3})| + \log \left(\frac{\sqrt{2}}{6}\right)\)

Since \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \), simplify to find

\(y(\frac{\pi}{3}) = \sqrt{3} \left(2 + \log_e \sqrt{3} \right)\)

By comparing with the options provided, the correct solution is:

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

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

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