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Step-by-Step Solution
Step 1: Rewrite the Differential Equation in Standard Form
The given differential equation is
(x \cos x)\,dy + \bigl(x\,y\,\sin x + y\,\cos x - 1\bigr)\,dx = 0,\quad 0 < x < \frac{\pi}{2}.
We want it in the form
\frac{dy}{dx} + P(x)\,y = Q(x).
First, divide the entire equation by x \cos x (assuming x\neq0 and \cos x \neq 0 ):
\frac{dy}{dx}
= - \frac{x\,y\,\sin x + y\,\cos x - 1}{x \cos x}.
Distribute the negative sign:
\frac{dy}{dx}
= -\frac{x\,y\,\sin x + y\,\cos x}{x \cos x}
+ \frac{1}{x \cos x}.
Factor out y in the first part:
\frac{dy}{dx}
= -\Bigl(\frac{x \sin x + \cos x}{x \cos x}\Bigr)\,y
+ \frac{1}{x \cos x}.
Notice that
\frac{x \sin x + \cos x}{x \cos x}
= \frac{x \sin x}{x \cos x} + \frac{\cos x}{x \cos x}
= \tan x + \frac{1}{x}.
Therefore, the differential equation becomes
\frac{dy}{dx} + \Bigl(\tan x + \frac{1}{x}\Bigr)\,y
= \frac{1}{x \cos x}.
Step 2: Find the Integrating Factor
For the standard linear differential equation
\frac{dy}{dx} + P(x)\,y = Q(x),
the integrating factor (IF) is given by
\text{IF} = e^{\int P(x)\,dx}.
Here,
P(x) = \tan x + \frac{1}{x}.
Thus,
\text{IF} = e^{\int (\tan x + \frac{1}{x})\,dx}.
We compute these integrals separately:
\int \tan x\,dx = -\ln|\cos x| \quad (\text{or } \ln|\sec x|),
and
\int \frac{1}{x}\,dx = \ln|x|.
So
\int \bigl(\tan x + \frac{1}{x}\bigr)\,dx
= \ln(\sec x) + \ln(x)
= \ln\bigl(x\sec x\bigr).
Hence the integrating factor is
\text{IF}
= e^{\ln(x \sec x)}
= x \sec x.
Step 3: Solve for y(x)
Once the integrating factor is found, the general solution is
y \cdot \bigl(\text{IF}\bigr)
= \int \bigl(Q(x)\,\text{IF}\bigr)\,dx + C,
where C is a constant of integration. In our case,
Q(x) = \frac{1}{x \cos x},
\quad
\text{IF} = x \sec x.
Therefore,
y \cdot (x \sec x)
= \int \Bigl(\frac{1}{x \cos x} \times x \sec x\Bigr)\,dx + C
= \int \sec^2 x\,dx + C.
We know
\int \sec^2 x\,dx = \tan x.
Hence,
x\,y\,\sec x = \tan x + C.
Solving for y :
y = \frac{\tan x + C}{x \sec x}
= \frac{\sin x + C\,\cos x}{x}.
Step 4: Determine the Constant C Using the Given Initial Condition
We have
\frac{\pi}{3} \,y\Bigl(\tfrac{\pi}{3}\Bigr) = \sqrt{3}.
Hence,
y\Bigl(\tfrac{\pi}{3}\Bigr) = \frac{3\sqrt{3}}{\pi}.
Substitute x = \tfrac{\pi}{3} into
y = \frac{\sin x + C\,\cos x}{x}.
This gives
y\Bigl(\tfrac{\pi}{3}\Bigr)
= \frac{\sin\bigl(\tfrac{\pi}{3}\bigr) + C\,\cos\bigl(\tfrac{\pi}{3}\bigr)}{\tfrac{\pi}{3}}
= \frac{3\sqrt{3}}{\pi}.
Multiply both sides by \tfrac{\pi}{3} :
\sin\bigl(\tfrac{\pi}{3}\bigr) + C \,\cos\bigl(\tfrac{\pi}{3}\bigr)
= \sqrt{3}.
We know \sin\bigl(\tfrac{\pi}{3}\bigr) = \tfrac{\sqrt{3}}{2} and \cos\bigl(\tfrac{\pi}{3}\bigr)= \tfrac{1}{2} . So,
\frac{\sqrt{3}}{2} + \frac{C}{2} = \sqrt{3}.
Hence,
\frac{C}{2} = \sqrt{3} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}
\quad\Longrightarrow\quad
C = \sqrt{3}.
Step 5: Write the Particular Solution
Substituting C = \sqrt{3} back into
y = \frac{\sin x + C\,\cos x}{x},
we get
y(x) = \frac{\sin x + \sqrt{3}\,\cos x}{x}.
Step 6: Compute y'(x) and y''(x)
We need to evaluate
\left|\frac{\pi}{6}\,y''\Bigl(\tfrac{\pi}{6}\Bigr) + 2\,y'\Bigl(\tfrac{\pi}{6}\Bigr)\right|.
First, find y'(x) :
y'(x)
= \frac{d}{dx}\Bigl[\frac{\sin x + \sqrt{3}\,\cos x}{x}\Bigr].
Treat it as (\sin x + \sqrt{3}\cos x)\cdot x^{-1} and apply the product rule:
y'(x)
= x^{-1}\,\frac{d}{dx}(\sin x + \sqrt{3}\cos x)
+ (\sin x + \sqrt{3}\cos x)\,\frac{d}{dx}(x^{-1}).
Note:
\frac{d}{dx}(\sin x + \sqrt{3}\cos x) = \cos x - \sqrt{3}\,\sin x,
and
\frac{d}{dx}\bigl(x^{-1}\bigr) = -x^{-2}.
Therefore,
y'(x)
= \frac{\cos x - \sqrt{3}\,\sin x}{x}
- \frac{\sin x + \sqrt{3}\,\cos x}{x^2}.
We would then differentiate y'(x) to obtain y''(x) , but since we only need the specific combination
\frac{\pi}{6}\,y''\bigl(\tfrac{\pi}{6}\bigr) + 2\,y'\bigl(\tfrac{\pi}{6}\bigr) ,
we focus on evaluating at x = \tfrac{\pi}{6} and using the known trigonometric values.
Step 7: Evaluate the Expression at x = \tfrac{\pi}{6}
Through the detailed (but somewhat lengthy) calculation of y'(x) and y''(x) at x = \tfrac{\pi}{6} and combining them according to
\frac{\pi}{6}\,y''\Bigl(\tfrac{\pi}{6}\Bigr) + 2\,y'\Bigl(\tfrac{\pi}{6}\Bigr),
one arrives at the numerical value
\left|\frac{\pi}{6}\,y''\Bigl(\tfrac{\pi}{6}\Bigr) + 2\,y'\Bigl(\tfrac{\pi}{6}\Bigr)\right| = 2.
(A careful term-by-term differentiation and substitution confirms this result.)
Final Answer
Hence, the required absolute value is
\boxed{2}.
