Step 1: Understanding the Concept:
This problem asks for the particular integral (P.I.) of a second-order linear non-homogeneous differential equation. The P.I. is a specific solution that satisfies the non-homogeneous equation. The method to find it depends on the form of the function on the right-hand side.
Step 2: Key Formula or Approach:
The particular integral for an equation of the form $F(D)y = e^{ax}$, where $D = \frac{d}{dx}$, is given by $P.I. = \frac{1}{F(D)}e^{ax}$.
Case 1: If $F(a) \neq 0$, then $P.I. = \frac{1}{F(a)}e^{ax}$.
Case 2: If $F(a) = 0$, and the root 'a' has multiplicity k, then $P.I. = \frac{x^k}{F^{(k)}(a)}e^{ax}$, where $F^{(k)}(a)$ is the k-th derivative of $F(D)$ with respect to D, evaluated at $D=a$. A simpler way is $P.I. = \frac{x^k}{k! G(a)}e^{ax}$ where $F(D) = (D-a)^k G(D)$ and $G(a) \neq 0$.
Step 3: Detailed Explanation:
The differential equation is $(D^2 + 3D + 2)y = e^{-2x}$.
The operator is $F(D) = D^2 + 3D + 2$. The right-hand side is of the form $e^{ax}$ with $a = -2$.
First, let's find the roots of the auxiliary equation $m^2 + 3m + 2 = 0$.
\[ (m+1)(m+2) = 0 \]
The roots are $m=-1$ and $m=-2$.
Now, let's check $F(a) = F(-2)$:
\[ F(-2) = (-2)^2 + 3(-2) + 2 = 4 - 6 + 2 = 0 \]
Since $F(a) = 0$, this is a case of failure (Case 2). The root $a=-2$ is a non-repeated root of the auxiliary equation.
The formula for a single root failure is $P.I. = x \frac{1}{F'(D)}e^{ax}$.
Find the derivative of the operator: $F'(D) = \frac{d}{dD}(D^2 + 3D + 2) = 2D + 3$.
Now evaluate $F'(a) = F'(-2)$:
\[ F'(-2) = 2(-2) + 3 = -4 + 3 = -1 \]
The particular integral is:
\[ P.I. = x \frac{1}{F'(a)} e^{ax} = x \frac{1}{-1} e^{-2x} = -xe^{-2x} \]
Step 4: Final Answer:
The particular integral of the given differential equation is $-xe^{-2x}$. Therefore, option (A) is correct.