Question

Let \( y = y(x) \) be the solution of the differential equation \[ \cos(x \log(\cos x))^2 \, dy + (\sin x - 3 \sin x \log(\cos x)) \, dx = 0, \quad x \in \left( 0, \frac{\pi}{2} \right) \] If \( y\left( \frac{\pi}{4} \right) = -1 \), then \( y\left( \frac{\pi}{6} \right) \) is equal to:

• A. \( \frac{1}{\ln 3 - \ln 4} \)
• B. \( 2 \log 3 - \log 4 \)
• C. \( -1 \log 4 \)
• D. \( 1 \log 3 - \log 4 \)

Hint:

To solve differential equations, check if separation of variables or an integrating factor is useful.

Answer 1

The Correct Option is A

The differential equation to be solved is:

\( \cos x (\ln(\cos x))^2 \, dy + (\sin x - 3y \sin x \ln(\cos x)) \, dx = 0 \)

Rearranging and dividing by \( \cos x (\ln(\cos x))^2 \) yields:

\( \frac{dy}{dx} - \frac{3 \tan x}{\ln(\cos x)} y = -\frac{\tan x}{(\ln(\cos x))^2} \)

Using the identity \( \ln(\cos x) = -\ln(\sec x) \), the equation transforms to:

\( \frac{dy}{dx} + \frac{3 \tan x}{\ln(\sec x)} y = -\frac{\tan x}{(\ln(\sec x))^2} \)

1. Integrating Factor Calculation:
The integrating factor (I.F.) is determined by:

\( I.F. = e^{\int \frac{3 \tan x}{\ln(\sec x)} \, dx} \)

The integral \( \int \frac{\tan x}{\ln(\sec x)} \, dx \) evaluates to \( \ln(\ln(\sec x)) \).

Therefore:

\( I.F. = e^{3 \ln(\ln(\sec x))} = (\ln(\sec x))^3 \)

2. Solving the Differential Equation:
Multiplying the equation by the integrating factor \( (\ln(\sec x))^3 \):

\( y \cdot (\ln(\sec x))^3 = -\int \frac{\tan x}{(\ln(\sec x))^2} \cdot (\ln(\sec x))^3 \, dx \)

Simplifying the integral yields:

\( y \cdot (\ln(\sec x))^3 = -\int \tan x \cdot \ln(\sec x) \, dx \)

Applying the substitution \( u = \ln(\sec x) \), which implies \( du = \tan x \, dx \):

\( y \cdot (\ln(\sec x))^3 = -\int u \, du = -\frac{u^2}{2} + C = -\frac{(\ln(\sec x))^2}{2} + C \)

3. Applying the Initial Condition:
Given \( x = \frac{\pi}{4} \) and \( y = -\frac{1}{\ln 2} \), substituting these values:

\( -\frac{1}{\ln 2} \cdot (\ln(\sqrt{2}))^3 = -\frac{1}{2} \cdot (\ln(\sqrt{2}))^2 + C \)

Using \( \ln(\sqrt{2}) = \frac{1}{2} \ln 2 \):

\( -\frac{1}{\ln 2} \cdot \left(\frac{1}{2} \ln 2\right)^3 = -\frac{1}{2} \cdot \left(\frac{1}{2} \ln 2\right)^2 + C \)

Simplification leads to:

\( -\frac{(\ln 2)^3}{8 \ln 2} = -\frac{(\ln 2)^2}{8} + C \)

\( -\frac{(\ln 2)^2}{8} = -\frac{(\ln 2)^2}{8} + C \)

This results in \( C = 0 \).

4. Final Solution:
The solution is:

\( y \cdot (\ln(\sec x))^3 = -\frac{1}{2} (\ln(\sec x))^2 \)

Dividing by \( (\ln(\sec x))^3 \):

\( y = -\frac{1}{2 \ln(\sec x)} \)

Substituting \( \ln(\sec x) = -\ln(\cos x) \):

\( y = \frac{1}{2 \ln(\cos x)} \)

5. Evaluating \( y \) at \( x = \frac{\pi}{6} \):
Substituting \( x = \frac{\pi}{6} \):

\( y = \frac{1}{2 \ln(\cos \frac{\pi}{6})} \)

Since \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \):

\( y = \frac{1}{2 \ln\left(\frac{\sqrt{3}}{2}\right)} \)

Using \( \ln\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{2} \ln 3 - \ln 2 \):

\( y = \frac{1}{2 \left(\frac{1}{2} \ln 3 - \ln 2\right)} = \frac{1}{\ln 3 - \ln 4} \)

Final Answer:
The value of \( y \) at \( x = \frac{\pi}{6} \) is \( \frac{1}{\ln 3 - \ln 4} \).