Question

The particular integral of \( (D^2 + 4)y = \cos 3x + \sin 2x \) is:

• A. \( \frac{1}{20}(\cos 3x + 5x \cos 2x) \)
• B. \( \frac{1}{20}(4 \cos 3x + 5x \cos 2x) \)
• C. \( \frac{-1}{20}(4 \cos 3x + 5x \cos 2x) \)
• D. \( \frac{1}{20}(4 \cos 3x + x \cos 2x) \)

Hint:

For non-homogeneous linear differential equations with trigonometric inputs, remember to handle resonance cases (when the operator evaluates to zero) by multiplying the input by \( x \) and proceeding with inverse differential operators.

Answer 1

The Correct Option is C

We are given the differential equation: \[ (D^2 + 4)y = \cos 3x + \sin 2x \] We need to find the particular integral (P.I.). Break into two parts: \[ \text{P.I.} = \frac{1}{D^2 + 4}(\cos 3x + \sin 2x) = \frac{1}{D^2 + 4}(\cos 3x) + \frac{1}{D^2 + 4}(\sin 2x) \] Part 1: \( \frac{1}{D^2 + 4}(\cos 3x) \) Since \( D^2(\cos 3x) = -9 \cos 3x \), the operator becomes: \[ = \frac{1}{-9 + 4} \cos 3x = \frac{1}{-5} \cos 3x = -\frac{1}{5} \cos 3x \] Part 2: \( \frac{1}{D^2 + 4}(\sin 2x) \) Here, \( D^2(\sin 2x) = -4 \sin 2x \). Hence, \[ D^2 + 4 = 0 \quad \Rightarrow \text{Resonance Case (failure)} \] So we multiply by \( x \) and differentiate: \[ \frac{1}{(D^2 + 4)}(\sin 2x) = x \cdot \frac{1}{2D}(\sin 2x) \] Now \( D(\sin 2x) = 2 \cos 2x \), so: \[ \frac{1}{2D}(\sin 2x) = \frac{1}{2 \cdot 2 \cos 2x} = \frac{1}{4} \cos 2x \Rightarrow x \cdot \frac{1}{4} \cos 2x = \frac{x}{4} \cos 2x \] Final Result: \[ \text{P.I.} = -\frac{1}{5} \cos 3x + \frac{x}{4} \cos 2x = \frac{-4 \cos 3x + 5x \cos 2x}{20} = {\frac{-1}{20}(4 \cos 3x + 5x \cos 2x)} \]