Step 1: Solve for the particular integral of each expression.
- B. \(\frac{1}{D^2+D+1}\cos x\): For cosine functions, we substitute \(D^2 \to -a^2\), so \(D^2 \to -1^2 = -1\).
\[ P.I. = \frac{1}{-1+D+1}\cos x = \frac{1}{D}\cos x = \int \cos x \,dx = \sin x \]
This matches II.
- C. \(\frac{1}{(D-1)^2e^x\):} This is a case of failure, as substituting \(D=1\) makes the denominator zero. We use the formula \(\frac{1}{f(D)}e^{ax}V = e^{ax}\frac{1}{f(D+a)}V\).
\[ P.I. = e^x \frac{1}{((D+1)-1)^2}(1) = e^x \frac{1}{D^2}(1) = e^x \iint 1 \,dx\,dx = e^x \frac{x^2}{2} \]
This matches III.
- D. \(\frac{1}{D^3-3D^2+4D-2}e^x\): Substitute \(D=1\): \(1-3+4-2=0\). It's a failure case. We use the rule: If \(f(a)=0\), P.I. is \(x \frac{1}{f'(a)}e^{ax}\).
Let \(f(D) = D^3-3D^2+4D-2\). Then \(f'(D) = 3D^2-6D+4\).
\(f'(1) = 3(1)^2 - 6(1) + 4 = 3-6+4=1\).
\[ P.I. = x \frac{1}{1}e^x = xe^x \]
This matches I.
- A. \(\frac{1}{(D-1)x^2\):} We use binomial expansion: \(\frac{1}{D-1} = -(1-D)^{-1} = -(1+D+D^2+...)\).
\[ P.I. = -(1+D+D^2)(x^2) = -(x^2 + D(x^2) + D^2(x^2)) = -(x^2 + 2x + 2) \]
This matches IV.
Step 2: Formulate the correct matching sequence.
The matches are: A\(\rightarrow\)IV, B\(\rightarrow\)II, C\(\rightarrow\)III, D\(\rightarrow\)I. This corresponds to option (3).