Step 1: Understanding the Concept:
This problem requires finding the particular integral for a differential equation where the right-hand side is a polynomial function of \(x\). The method involves using the binomial expansion of the operator \(\frac{1}{f(D)}\).
Step 2: Key Formula or Approach:
To evaluate \(\frac{1}{f(D)}x^m\), we first rewrite \(f(D)\) in the form \(k(1 \pm \phi(D))\) and then use the binomial expansion for \((1 \pm \phi(D))^{-n}\).
The relevant expansion is \((1-z)^{-2} = 1 + 2z + 3z^2 + \dots\).
We operate on the polynomial term by term, noting that \(D^k(x^m) = 0\) for \(k > m\).
Step 3: Detailed Explanation:
We need to evaluate:
\[ \frac{1}{(D-2)^2} x^2 \]
First, factor out \(-2\) from the denominator to get it into the form \((1-z)\):
\[ \frac{1}{(-2(1 - D/2))^2} x^2 = \frac{1}{4(1 - D/2)^2} x^2 \]
Now, use the binomial expansion for \((1 - D/2)^{-2}\):
\[ \frac{1}{4} (1 - D/2)^{-2} x^2 = \frac{1}{4} \left[ 1 + 2\left(\frac{D}{2}\right) + 3\left(\frac{D}{2}\right)^2 + \dots \right] x^2 \]
Since we are operating on \(x^2\), any derivative of order higher than 2 will be zero, so we only need terms up to \(D^2\).
\[ = \frac{1}{4} \left[ 1 + D + \frac{3}{4}D^2 \right] x^2 \]
Now, apply the operator to \(x^2\):
\[ = \frac{1}{4} \left[ 1 \cdot x^2 + D(x^2) + \frac{3}{4}D^2(x^2) \right] \]
Calculate the derivatives:
- \(D(x^2) = \frac{d}{dx}(x^2) = 2x\)
- \(D^2(x^2) = \frac{d^2}{dx^2}(x^2) = \frac{d}{dx}(2x) = 2\)
Substitute these back into the expression:
\[ = \frac{1}{4} \left[ x^2 + 2x + \frac{3}{4}(2) \right] \]
\[ = \frac{1}{4} \left( x^2 + 2x + \frac{3}{2} \right) \]
Step 4: Final Answer:
The result of the operation is \(\frac{1}{4}\left(x^2 + 2x + \frac{3}{2}\right)\), which corresponds to option (A).