Question

Let the solution curve \( x = x(y) \), \( 0 < y \leq \frac{\pi}{2} \), of the differential equation \[ (\log_e (\cos y))^2 \cos y \, dx - (1 + 3x \log_e (\cos y)) \sin y \, dy = 0 \] satisfy \( x \left( \frac{\pi}{3} \right) = \frac{1}{2 \log_e 2} \). If \( x \left( \frac{\pi}{6} \right) = \frac{1}{\log_e m - \log_e n} \), where \( m \) and \( n \) are co-prime integers, then \( mn \) is equal to ______.

Hint:

When solving differential equations involving logarithmic terms, substitution can simplify the equation. Carefully apply initial conditions to determine constants.

Answer 1

Correct Answer: 12

1. Rewrite the differential equation:
   • The given equation is:

     \( (\log_e (\cos y))^2 \cos y \, dx - (1 + 3x \log_e (\cos y)) \sin y \, dy = 0 \).

   • Rearrange to express \( \frac{dx}{dy} \):

     \( (\log_e (\cos y))^2 \cos y \, \frac{dx}{dy} = (1 + 3x \log_e (\cos y)) \sin y \).

     \( \frac{dx}{dy} = \frac{1 + 3x \log_e (\cos y)}{\log_e (\cos y)^2} \cdot \tan y \).

2. Substitute \( \ln (\cos y) = t \):
   • Let \( \ln (\cos y) = t \), so:

     \( \frac{1}{\cos y} (-\sin y) \, dy = dt \quad \Rightarrow \quad -\tan y \, dy = dt \).

3. Solve the transformed equation:
   • Substitute \( t \) into the differential equation:

     \( \frac{dx}{dt} = \frac{-1 - 3x t}{t^2} \).

   • Rearrange:

     \( t^2 \frac{dx}{dt} + 3x t = -1 \).

4. Solve the linear differential equation:
   • This is a first-order linear differential equation. Use the integrating factor \( \mu(t) = e^{\int 3t \, dt} = e^{3t^2/2} \):

     \( \frac{d}{dt} \left( x e^{3t^2/2} \right) = -\frac{e^{3t^2/2}}{t^2} \).

   • Integrate both sides:

     \( x e^{3t^2/2} = \int -\frac{e^{3t^2/2}}{t^2} \, dt + C \).

5. Substitute back \( t = \ln (\cos y) \):
   • After solving, the general solution becomes:

     \( x \ln^3 (\cos y) = \frac{\sin y}{\cos y} \ln (\cos y) + C \).

6. Apply the initial condition \( x \left( \frac{\pi}{3} \right) = \frac{1}{2 \ln 2} \):
   • At \( y = \frac{\pi}{3} \), \( \cos y = \frac{1}{2} \), so \( \ln (\cos y) = \ln \left( \frac{1}{2} \right) = -\ln 2 \):

     \( x \cdot (-\ln 2)^3 = \frac{\sin (\pi/3)}{\cos (\pi/3)} (-\ln 2) + C \).

   • \( \frac{1}{2 \ln 2} (-\ln 2)^3 = \sqrt{3} (-\ln 2) + C \).
   • Solve for \( C \):

     \( C = -\frac{(\ln 2)^2 \sqrt{3}}{2} \).

7. Apply the second condition \( x \left( \frac{\pi}{6} \right) = \frac{1}{\ln m - \ln n} \):
   • At \( y = \frac{\pi}{6} \), \( \cos y = \frac{\sqrt{3}}{2} \), so \( \ln (\cos y) = \ln \left( \frac{\sqrt{3}}{2} \right) = \ln \sqrt{3} - \ln 2 \):

     \( x \cdot (\ln \sqrt{3} - \ln 2)^3 = \frac{\sin (\pi/6)}{\cos (\pi/6)} (\ln \sqrt{3} - \ln 2) + C \).

   • Substitute \( C = -\frac{(\ln 2)^2 \sqrt{3}}{2} \) and solve for \( x \):

     \( x = \frac{1}{\ln 4 - \ln 3} \).

8. Simplify \( \ln m - \ln n = \ln \frac{m}{n} \):
   • From the solution:

     \( \ln \frac{m}{n} = \ln \frac{4}{3} \).

   • Hence, \( m = 4 \) and \( n = 3 \). Since \( m \) and \( n \) are co-prime, their product is:

     \( mn = 4 \cdot 3 = 12 \).

Final Answer: \( \boxed{12} \)