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Step-by-Step Solution
Step 1: Understand the Problem
We are given a function $y = y(x)$ satisfying the differential equation
$
\frac{dy}{dx} - |A| = 0, \quad x > 0,
$
where
$
A = \begin{pmatrix}
y & \sin x & 1\\
0 & -1 & 1\\
2 & 0 & \frac{1}{x}
\end{pmatrix},
$
and the initial condition $y(\pi) = \pi + 2.$ We need to find the value of $y\left(\frac{\pi}{2}\right).$
Step 2: Determine the Determinant $|A|$
From the given solution reference, the determinant of $A$ is
$
|A| = -\frac{y}{x} + 2\sin x + 2.
$
This expression will appear on the right-hand side of the differential equation.
Step 3: Form the Differential Equation
Since
$
\frac{dy}{dx} - |A| = 0,
$
substituting the expression for $|A|$ gives
$
\frac{dy}{dx} = -\frac{y}{x} + 2\sin x + 2.
$
Rearrange to standard linear form:
$
\frac{dy}{dx} + \frac{y}{x} = 2\sin x + 2.
$
Step 4: Identify the Integrating Factor (IF)
The above is a first-order linear differential equation of the form
$
\frac{dy}{dx} + P(x)\,y = Q(x),
$
where $P(x) = \frac{1}{x}$ and $Q(x) = 2\sin x + 2.$
The integrating factor (IF) is found by
$
\text{IF} = e^{\int P(x)\,dx} = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x.
$
Step 5: Multiply Through by the Integrating Factor
Multiply both sides of
$
\frac{dy}{dx} + \frac{y}{x} = 2\sin x + 2
$
by $x$:
$
x\,\frac{dy}{dx} + y = x\,(2\sin x + 2).
$
The left-hand side is the derivative of $(x\,y)$:
$
\frac{d}{dx}(x\,y) = x(2\sin x + 2).
$
Step 6: Integrate Both Sides
Integrate with respect to $x$:
$
x\,y = \int x(2\sin x + 2)\,dx.
$
Break it into two integrals:
$
\int x(2\sin x + 2)\,dx = \int 2x \sin x\,dx + \int 2x\,dx.
$
$
\int 2x \sin x\,dx
$
Use integration by parts:
Let $u = 2x \implies du = 2\,dx$ and $dv = \sin x\,dx \implies v = -\cos x.$
$
\int 2x \sin x\,dx = 2x(-\cos x) - \int (-\cos x)\,2\,dx
= -2x\cos x + 2 \int \cos x\,dx
= -2x\cos x + 2\sin x.
$
$
\int 2x\,dx = x^2.
$
Summing these results:
$
\int x(2\sin x + 2)\,dx
= x^2 - 2x\cos x + 2\sin x + C.
$
Hence
$
x\,y = x^2 - 2x\cos x + 2\sin x + C.
$
Step 7: Use the Initial Condition to Find $C$
We are given that $y(\pi) = \pi + 2.$ Substitute $x = \pi$ and $y(\pi) = \pi + 2$ into
$
x\,y = x^2 - 2x\cos x + 2\sin x + C.
$
Left side:
$
\pi \cdot (\pi + 2) = \pi^2 + 2\pi.
$
Right side:
$
\pi^2 - 2\pi\cos(\pi) + 2\sin(\pi) + C
= \pi^2 - 2\pi(-1) + 2 \cdot 0 + C
= \pi^2 + 2\pi + C.
$
Equating both sides:
$
\pi^2 + 2\pi = \pi^2 + 2\pi + C
\implies C = 0.
$
Step 8: Write the General Solution
With $C = 0,$ the solution simplifies to
$
x\,y = x^2 - 2x\cos x + 2\sin x.
$
Thus,
$
y = \frac{x^2 - 2x\cos x + 2\sin x}{x}.
$
Step 9: Evaluate $y\left(\frac{\pi}{2}\right)$
Substitute $x = \frac{\pi}{2}$:
$
\frac{\pi}{2} \, y\left(\frac{\pi}{2}\right)
= \left(\frac{\pi}{2}\right)^2 - 2 \cdot \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) + 2 \sin\left(\frac{\pi}{2}\right).
$
Calculate each term:
$
\left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4}.
$
$
\cos\left(\frac{\pi}{2}\right) = 0
\implies - 2 \cdot \frac{\pi}{2} \cdot 0 = 0.
$
$
\sin\left(\frac{\pi}{2}\right) = 1
\implies 2 \cdot 1 = 2.
$
So the right-hand side becomes
$
\frac{\pi^2}{4} + 2.
$
Hence,
$
\frac{\pi}{2}\,y\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4} + 2
\;\;\Longrightarrow\;\;
y\left(\frac{\pi}{2}\right)
= \frac{\frac{\pi^2}{4} + 2}{\frac{\pi}{2}}
= \left(\frac{\pi^2}{4} + 2\right)\,\frac{2}{\pi}.
$
Simplify:
$
\left(\frac{\pi^2}{4}\right)\frac{2}{\pi} = \frac{\pi}{2},
\quad
2 \cdot \frac{2}{\pi} = \frac{4}{\pi}.
$
Adding these together gives
$
\frac{\pi}{2} + \frac{4}{\pi}.
$
Step 10: Final Answer
Therefore, the value of
$
y\left(\frac{\pi}{2}\right)
$
is
$
\frac{\pi}{2} + \frac{4}{\pi}.
$
