Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are continuous functions of $x$.
• Integrating Factor (I.F.): For a first-order linear differential equation, the integrating factor is given by $I.F. = e^{\int P(x) dx}$. The solution is then $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$.
• Integration by Parts: $\int u \, dv = uv - \int v \, du$

Step-by-Step Solution

Step 1: Calculate the Determinant of Matrix A

We are given the differential equation $\frac{dy}{dx} - |A| = 0$, which means $\frac{dy}{dx} = |A|$. We need to find the determinant of the matrix $A$: A = \left[ {\matrix{ y & {\sin x} & 1 \cr 0 & { - 1} & 1 \cr 2 & 0 & {{1 \over x}} \cr } } \right] We expand the determinant along the first row: |A| = y \cdot \left| {\matrix{ { - 1} & 1 \cr 0 & {{1 \over x}} \cr } } \right| - \sin x \cdot \left| {\matrix{ 0 & 1 \cr 2 & {{1 \over x}} \cr } } \right| + 1 \cdot \left| {\matrix{ 0 & { - 1} \cr 2 & 0 \cr } } \right| Now we calculate the $2 \times 2$ determinants:

• \left| {\matrix{ { - 1} & 1 \cr 0 & {{1 \over x}} \cr } } \right| = (-1) \cdot \frac{1}{x} - (1)(0) = -\frac{1}{x}
• \left| {\matrix{ 0 & 1 \cr 2 & {{1 \over x}} \cr } } \right| = (0) \cdot \frac{1}{x} - (1)(2) = -2
• \left| {\matrix{ 0 & { - 1} \cr 2 & 0 \cr } } \right| = (0)(0) - (-1)(2) = 2 Substituting these values back into the expression for $|A|$: $|A| = y \left( - \frac{1}{x} \right) - \sin x (-2) + 1 (2)$ $|A| = - \frac{y}{x} + 2\sin x + 2$

Step 2: Formulate the Differential Equation

We substitute the expression for $|A|$ back into the equation $\frac{dy}{dx} = |A|$: $\frac{dy}{dx} = - \frac{y}{x} + 2\sin x + 2$

Step 3: Rewrite the Equation in Standard Linear Form

We rearrange the differential equation into the standard linear form: $\frac{dy}{dx} + P(x)y = Q(x)$. $\frac{dy}{dx} + \frac{y}{x} = 2\sin x + 2$ Comparing this with the standard form, we identify $P(x)$ and $Q(x)$: $P(x) = \frac{1}{x}$ $Q(x) = 2\sin x + 2$

Step 4: Calculate the Integrating Factor (I.F.)

Using the formula for the integrating factor, $I.F. = e^{\int P(x)dx}$: $I.F. = e^{\int \frac{1}{x}dx}$ The integral of $\frac{1}{x}$ is $\ln|x|$. Since $x > 0$, we can drop the absolute value. $I.F. = e^{\ln x} = x$

Step 5: Solve the Linear Differential Equation

We apply the formula for the general solution: $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$. Substitute the calculated $I.F.$ and identified $Q(x)$: $y \cdot x = \int (2\sin x + 2) \cdot x \, dx + C$ $xy = \int (2x\sin x + 2x) \, dx + C$ We split the integral into two simpler integrals: $\int (2x\sin x + 2x) \, dx = \int 2x\sin x \, dx + \int 2x \, dx$ The second integral is straightforward: $\int 2x \, dx = x^2$ For the first integral, $\int 2x\sin x \, dx$, we use integration by parts. Let $u = 2x$ and $dv = \sin x \, dx$. Then, $du = 2 \, dx$ and $v = \int \sin x \, dx = -\cos x$. Applying the integration by parts formula: $\int 2x\sin x \, dx = (2x)(-\cos x) - \int (-\cos x)(2 \, dx)$ $= -2x\cos x + 2\int \cos x \, dx$ $= -2x\cos x + 2\sin x$ Combine the results of both integrals: $\int (2x\sin x + 2x) \, dx = -2x\cos x + 2\sin x + x^2$ Substitute this back into our general solution equation: $xy = x^2 - 2x\cos x + 2\sin x + C$

Step 6: Use the Initial Condition to Find the Constant C

We are given the initial condition $y(\pi) = \pi + 2$. Substituting $x = \pi$ and $y = \pi + 2$: $(\pi)(\pi + 2) = (\pi)^2 - 2(\pi)\cos(\pi) + 2\sin(\pi) + C$ Recall that $\cos(\pi) = -1$ and $\sin(\pi) = 0$. $\pi^2 + 2\pi = \pi^2 - 2\pi(-1) + 2(0) + C$ $\pi^2 + 2\pi = \pi^2 + 2\pi + C$ $C = 0$

Step 7: Write the Particular Solution

Since $C=0$, the particular solution is: $xy = x^2 - 2x\cos x + 2\sin x$

Step 8: Evaluate $y\left( {{\pi \over 2}} \right)$

We substitute $x = \frac{\pi}{2}$ into our particular solution: $\left( \frac{\pi}{2} \right) y\left( \frac{\pi}{2} \right) = \left( \frac{\pi}{2} \right)^2 - 2\left( \frac{\pi}{2} \right)\cos\left( \frac{\pi}{2} \right) + 2\sin\left( \frac{\pi}{2} \right)$ Recall that $\cos\left( \frac{\pi}{2} \right) = 0$ and $\sin\left( \frac{\pi}{2} \right) = 1$. $\frac{\pi}{2} y\left( \frac{\pi}{2} \right) = \frac{\pi^2}{4} - \pi(0) + 2(1)$ $\frac{\pi}{2} y\left( \frac{\pi}{2} \right) = \frac{\pi^2}{4} + 2$ Multiply both sides by $\frac{2}{\pi}$: $y\left( \frac{\pi}{2} \right) = \left( \frac{\pi^2}{4} + 2 \right) \cdot \frac{2}{\pi}$ $y\left( \frac{\pi}{2} \right) = \frac{\pi}{2} + \frac{4}{\pi}$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when calculating determinants and applying integration by parts. A small sign error can propagate through the entire solution.
• Integration by Parts: Remember the integration by parts formula $\int u \, dv = uv - \int v \, du$. Choose $u$ and $dv$ strategically to simplify the integral.
• Integrating Factor: Ensure the differential equation is in the standard linear form before calculating the integrating factor.

Summary

We solved the given differential equation by first calculating the determinant of matrix $A$, then rewriting the equation in standard linear form. We found the integrating factor, solved the equation, and used the given initial condition to find the particular solution. Finally, we evaluated the solution at $x = \frac{\pi}{2}$ to find the desired value.

The final answer is \boxed{{\pi \over 2} + {4 \over \pi}}, which corresponds to option (A).