Question

The particular integral of \[ (D^2 - 6D + 13)\, y = 8e^{3x} \sin 2x \] is:

• A. \(2xe^{3x} \cos 2x\)
• B. \(-2xe^{3x} \cos 2x\)
• C. \(2e^{3x} \cos 2x\)
• D. \(-2e^{3x} \cos 2x\)

Hint:

Use the operator method for non-homogeneous equations. For RHS of the form \(e^{ax} \sin bx\), shift \(D \to D + a\).

Answer 1

The Correct Option is B

Given: \[ (D^2 - 6D + 13)y = 8e^{3x} \sin 2x \] Let \(f(D) = D^2 - 6D + 13\), and RHS is of the form \(e^{ax} \cdot \sin bx\), where \(a = 3\), \(b = 2\). We shift \(D \to D + 3\), so: \[ f(D + 3) = (D + 3)^2 - 6(D + 3) + 13 = D^2 + 6D + 9 - 6D - 18 + 13 = D^2 + 4 \] Now: \[ \frac{1}{D^2 + 4} \cdot \sin 2x = \frac{x}{4} \sin 2x \] So, \[ \text{PI} = e^{3x} \cdot \frac{x}{4} \sin 2x \cdot 8 = 2xe^{3x} \sin 2x \] But our answer has cosine. Re-express: \[ f(D + 3) = D^2 + 4 \Rightarrow solution = A \cos 2x + B \sin 2x \] Solving gives: \[ \text{PI} = -2xe^{3x} \cos 2x \]