Question

The particular integral of the differential equation \[ (D^2+1)y = \sin x\,\sin2x \] is

• A. \(\dfrac{x}{4}\sin x+\dfrac{1}{16}\cos3x\)
• B. \(\dfrac{x}{2}\sin x+\dfrac{1}{8}\cos3x\)
• C. \(\dfrac{x}{4}\cos x+\dfrac{1}{16}\sin3x\)
• D. \(\dfrac{x}{2}\cos x-\dfrac{1}{8}\cos3x\)

Hint:

Remember the identity \[ \boxed{ \sin A\sin B = \frac12[\cos(A-B)-\cos(A+B)]. } \] Use it before finding the particular integral.

Answer 1

The Correct Option is A

Step 1: Use the trigonometric identity. \[ \sin x\sin2x = \frac12(\cos x-\cos3x). \] Hence, \[ (D^2+1)y = \frac12\cos x - \frac12\cos3x. \]

Step 2: Find the particular integral. For the resonant term, \[ \frac{1}{D^2+1}\left(\frac12\cos x\right) = \frac{x}{4}\sin x. \] For the second term, \[ \frac{1}{D^2+1}\left(-\frac12\cos3x\right) = -\frac12\cdot\frac{\cos3x}{1-9} = \frac1{16}\cos3x. \] Therefore, \[ \boxed{ \text{P.I.} = \frac{x}{4}\sin x + \frac1{16}\cos3x. } \] Hence, \[ \boxed{(A)} \] is the correct answer.