Question

The particular integral of \(\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = e^{-2x}\) is

• A. \(-xe^{-2x}\)
• B. \(xe^{-2x}\)
• C. \(-\frac{x}{2}e^{-2x}\)
• D. \(\frac{x}{2}e^{-2x}\)

Hint:

When finding the particular integral for \(e^{ax}\) using the operator method, always first check if 'a' is a root of the auxiliary equation. If it is (i.e., if \(f(a)=0\)), you know it's a case of failure and you must apply the rule of multiplying by x and differentiating the denominator.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to find the particular integral (PI) for a second-order non-homogeneous linear differential equation with constant coefficients.

Step 2: Key Formula or Approach:
We use the operator method. The particular integral \(y_p\) is given by \(y_p = \frac{1}{f(D)} R(x)\), where \(D = \frac{d}{dx}\), \(f(D)\) is the differential operator, and \(R(x)\) is the function on the right-hand side.
For \(R(x) = e^{ax}\), we evaluate \(\frac{1}{f(a)}e^{ax}\). If \(f(a)=0\) (case of failure), the rule is \(y_p = x \frac{1}{f'(a)}e^{ax}\).

Step 3: Detailed Explanation:
The differential equation is \((D^2 + 3D + 2)y = e^{-2x}\).
So, \(f(D) = D^2 + 3D + 2\) and \(R(x) = e^{-2x}\).
The particular integral is:
\[ y_p = \frac{1}{D^2 + 3D + 2} e^{-2x} \]
We have the form \(e^{ax}\) with \(a=-2\). Let's evaluate \(f(a) = f(-2)\):
\[ f(-2) = (-2)^2 + 3(-2) + 2 = 4 - 6 + 2 = 0 \]
This is a "case of failure" because the denominator becomes zero. This happens because \(a=-2\) is a root of the auxiliary equation \(m^2+3m+2=0\).
When failure occurs, we apply the rule: multiply by \(x\) and differentiate the denominator with respect to \(D\).
\[ f'(D) = \frac{d}{dD}(D^2 + 3D + 2) = 2D + 3 \]
Now, the particular integral is given by:
\[ y_p = x \frac{1}{f'(D)} e^{-2x} = x \frac{1}{2D + 3} e^{-2x} \]
Substitute \(D=-2\) into the new operator:
\[ y_p = x \frac{1}{2(-2) + 3} e^{-2x} = x \frac{1}{-4 + 3} e^{-2x} = x \frac{1}{-1} e^{-2x} \]
\[ y_p = -xe^{-2x} \]

Step 4: Final Answer:
The particular integral of the differential equation is \(-xe^{-2x}\).