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Step-by-Step Solution
Step 1: Express the Differential Equation in a Standard Form
The given differential equation is:
(\ln(\cos y))^2 \cos y \, dx \;-\; \bigl(1 \;+\; 3x\,\ln(\cos y)\bigr)\,\sin y \, dy \;=\; 0.
Rewrite it as
\frac{dx}{dy} \;=\; \frac{\bigl(1 + 3x\,\ln(\cos y)\bigr)\,\sin y}{(\ln(\cos y))^2 \cos y}.
Factor out \tan y :
\frac{dx}{dy} \;=\; \tan y \left(\frac{1}{(\ln(\cos y))^2} \;+\; \frac{3x}{\ln(\cos y)}\right).
This is a linear differential equation in x and can be written in the form:
\frac{dx}{dy} \;-\; \frac{3\,\tan y}{\ln(\cos y)} x
\;=\; \frac{\tan y}{(\ln(\cos y))^2}.
Step 2: Identify and Compute the Integrating Factor
For a linear differential equation of the form
\frac{dx}{dy} + P(y)\,x = Q(y),
the integrating factor (I.F.) is
\mathrm{I.F.} = e^{\int P(y)\,dy}.
In our case:
P(y) = -\,\frac{3\,\tan y}{\ln(\cos y)}.
Therefore the integrating factor becomes:
\mathrm{I.F.}
= e^{-\int \frac{3\,\tan y}{\ln(\cos y)} \, dy}.
Recognize that \frac{d}{dy}\bigl(\ln(\cos y)\bigr) = - \tan y. Hence:
-\int \frac{3\,\tan y}{\ln(\cos y)} \, dy
= 3 \int \frac{\tan y}{\ln(\cos y)} \,(-1)\,d(\ln(\cos y))
= 3 \ln(\ln(\cos y)).
More directly, we often use a known result that leads to:
\mathrm{I.F.} = (\ln(\cos y))^3.
Step 3: Multiply the Original Equation by the Integrating Factor
Our differential equation
\frac{dx}{dy} - \frac{3\,\tan y}{\ln(\cos y)}\,x = \frac{\tan y}{(\ln(\cos y))^2}
is multiplied by (\ln(\cos y))^3 :
(\ln(\cos y))^3 \,\frac{dx}{dy}
\;-\; \frac{3\,\tan y}{\ln(\cos y}}\,(\ln(\cos y))^3\,x
\;=\;
(\ln(\cos y))^3 \,\frac{\tan y}{(\ln(\cos y))^2}.
The left-hand side becomes the derivative of
x \,(\ln(\cos y))^3,
while the right-hand side simplifies to
\tan y \,\ln(\cos y).
Thus the equation becomes:
\frac{d}{dy}\Bigl[x \,(\ln(\cos y))^3\Bigr]
= \tan y \,\ln(\cos y).
Step 4: Integrate Both Sides
Integrate with respect to y :
x \,(\ln(\cos y))^3
= \int \tan y \,\ln(\cos y)\,dy + C.
Let u = \ln(\cos y) . Then du = -\tan y \, dy, which means \tan y \, dy = -du. So
\int \tan y\,\ln(\cos y)\,dy
= \int \ln(\cos y)\,\tan y \,dy
= \int u\,(-\,du)
= -\int u \,du
= -\frac{u^2}{2}
= -\frac{(\ln(\cos y))^2}{2}.
Hence,
x \,(\ln(\cos y))^3
= -\,\frac{(\ln(\cos y))^2}{2} + C.
Step 5: Use the Given Initial Condition at y = \frac{\pi}{3}
We are given
x\Bigl(\tfrac{\pi}{3}\Bigr) = \frac{1}{2\,\ln(2)}.
Also,
\cos\Bigl(\tfrac{\pi}{3}\Bigr) = \frac{1}{2} \quad\Rightarrow\quad \ln\Bigl(\tfrac{1}{2}\Bigr) = -\ln(2).
Substitute y = \frac{\pi}{3} into
x \,(\ln(\cos y))^3
= -\,\frac{(\ln(\cos y))^2}{2} + C.
The left-hand side becomes:
\biggl(\frac{1}{2\,\ln(2)}\biggr)
\Bigl[-\ln(2)\Bigr]^3
= \frac{1}{2\,\ln(2)}\cdot\bigl[-(\ln(2))^3\bigr]
= -\frac{(\ln(2))^2}{2}.
On the right-hand side:
-\frac{\bigl[-\ln(2)\bigr]^2}{2} + C
= -\frac{(\ln(2))^2}{2} + C.
Equating both sides:
-\frac{(\ln(2))^2}{2}
= -\frac{(\ln(2))^2}{2} + C
\;\;\Longrightarrow\;\; C=0.
Step 6: Write the General Solution After Finding C=0
With C=0, the solution simplifies to
x \,(\ln(\cos y))^3
= -\frac{(\ln(\cos y))^2}{2}.
Thus
x
= -\frac{1}{2}
\cdot \frac{(\ln(\cos y))^2}{(\ln(\cos y))^3}
= -\frac{1}{2\,\ln(\cos y)}.
Step 7: Evaluate x at y = \frac{\pi}{6}
We now find
x\Bigl(\tfrac{\pi}{6}\Bigr).
Note that
\cos\Bigl(\tfrac{\pi}{6}\Bigr) = \frac{\sqrt{3}}{2}
\quad\Longrightarrow\quad \ln\Bigl(\frac{\sqrt{3}}{2}\Bigr)
= \frac{1}{2}\ln(3) - \ln(2).
Substitute into the expression for x :
x\Bigl(\tfrac{\pi}{6}\Bigr)
= -\,\frac{1}{2\,\ln\Bigl(\frac{\sqrt{3}}{2}\Bigr)}
= -\,\frac{1}{2\Bigl(\frac{1}{2}\ln(3) - \ln(2)\Bigr)}.
Simplify the denominator:
\frac{1}{2}\ln(3) - \ln(2)
= \frac{\ln(3) - 2\,\ln(2)}{2}
= \frac{\ln(3) - \ln(4)}{2}
= \frac{\ln\bigl(\frac{3}{4}\bigr)}{2}.
Hence,
x\Bigl(\tfrac{\pi}{6}\Bigr)
= -\,\frac{1}{ \ln\Bigl(\frac{\sqrt{3}}{2}\Bigr) }
= \frac{1}{ \ln(4) - \ln(3) }
= \frac{1}{ \ln\!\bigl(\tfrac{4}{3}\bigr) }.
This matches the given form
\frac{1}{ \ln(m) - \ln(n) }
= \frac{1}{ \ln\!\bigl(\tfrac{m}{n}\bigr) }.
Step 8: Identify m and n , Then Compute m\,n
From
x\Bigl(\tfrac{\pi}{6}\Bigr)
= \frac{1}{ \ln\!\bigl(\tfrac{4}{3}\bigr) },
we see that m = 4 and n = 3. Because m and n are coprime, their product is
m n = 4 \times 3 = 12.
Hence, the required value is
\boxed{12}.
