Question

Particular integral of \(\left(\dfrac{d^3}{dx^3}+4\dfrac{d}{dx}\right)y=\sin 2x\) is

• A. \(-\dfrac{x}{8}\cos 2x\)
• B. \(-\dfrac{x}{8}\sin 2x\)
• C. \(\dfrac{x}{12}\cos 2x\)
• D. \(-\dfrac{x}{12}\sin 2x\)

Hint:

When \(F(ia)=0\), use the resonance rule involving \(F'(ia)\) and multiply by \(x\).

Answer 1

The Correct Option is B

Concept:
For \[ F(D)y=\sin ax \] if \(F(ia)=0\), then the usual formula fails and we use the repeated/resonance rule. Here, \[ F(D)=D^3+4D \]

Step 1: Factorize \(F(D)\). \[ F(D)=D^3+4D \] \[ F(D)=D(D^2+4) \] For \(\sin 2x\), the operator \(D^2+4\) gives resonance because \[ D^2(\sin2x)=-4\sin2x \] So, \[ (D^2+4)\sin2x=0 \]

Step 2: Use complex method. Write \[ \sin2x=\Im(e^{2ix}) \] For \(e^{2ix}\), replace \(D\) by \(2i\). \[ F(D)=D^3+4D \] \[ F(2i)=(2i)^3+4(2i) \] \[ =8i^3+8i \] \[ =-8i+8i=0 \] So resonance occurs.

Step 3: Use derivative of \(F\). \[ F'(D)=3D^2+4 \] \[ F'(2i)=3(2i)^2+4 \] \[ =3(-4)+4 \] \[ =-12+4 \] \[ =-8 \] Therefore, \[ \text{P.I. for }e^{2ix}=\frac{x e^{2ix}}{F'(2i)} \] \[ =\frac{x e^{2ix}}{-8} \]

Step 4: Take imaginary part. \[ \frac{x e^{2ix}}{-8} = -\frac{x}{8}(\cos2x+i\sin2x) \] The imaginary part is \[ -\frac{x}{8}\sin2x \]

Step 5: Final answer. \[ \boxed{-\frac{x}{8}\sin2x} \]