Subtopic: Higher order linear ODEs with constant coefficients
Question Statement:
The particular integral (PI) of the differential equation $\frac{d^2 y}{dx^2} + 4y = \sin(2x)$ is:
Options:
(A) $-\frac{x}{4} \cos(2x)$
(B) $\frac{x}{4} \cos(2x)$
(C) $-\frac{x}{2} \cos(2x)$
(D) $\frac{x}{2} \cos(2x)$
Answer Key:
A
Step-by-Step Solution & Explanation
Here $f(D) = D^2 + 4$. When evaluating for $\sin(2x)$, replacing $D^2$ with $-2^2 = -4$ gives $f(-4) = 0$ (case of failure / resonance).
Using the derivative rule for case of failure:
$PI = \frac{1}{D^2 + 4} \sin(2x) = x \cdot \frac{1}{\frac{d}{dD}(D^2 + 4)} \sin(2x) = x \cdot \frac{1}{2D} \sin(2x)$
$\frac{1}{D} \sin(2x) = \int \sin(2x) \, dx = -\frac{\cos(2x)}{2}$
$PI = x \left(-\frac{\cos(2x)}{4}\right) = -\frac{x}{4}\cos(2x)$
Thus, Option A is correct.