Question

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

• A. \(\dfrac{3}{8}\cos 2x\)
• B. \(\dfrac{1}{4}\cos 2x\)
• C. \(\dfrac{1}{8}\cos 2x\)
• D. \(-\dfrac{1}{8}\sin 2x\)

Hint:

When \(F(D)\) contains odd powers of \(D\), use the complex exponential method for \(\sin ax\) and \(\cos ax\).

Answer 1

The Correct Option is D

Concept:
For a linear differential equation \[ F(D)y=\cos ax \] we use \[ \text{P.I.}=\frac{1}{F(D)}\cos ax \] If \(F(D)\) contains odd powers of \(D\), then we must be careful because \(D\) changes \(\cos ax\) into \(\sin ax\).

Step 1: Identify \(F(D)\). \[ F(D)=D^2-4D+4 \] \[ F(D)=(D-2)^2 \] We need \[ \text{P.I.}=\frac{1}{D^2-4D+4}\cos 2x \]

Step 2: Use complex method. Write \[ \cos 2x=\Re(e^{2ix}) \] So first calculate \[ \frac{1}{F(D)}e^{2ix} \] For \(e^{2ix}\), replace \(D\) by \(2i\): \[ F(2i)=(2i)^2-4(2i)+4 \] \[ F(2i)=-4-8i+4 \] \[ F(2i)=-8i \]

Step 3: Calculate particular integral. \[ \frac{1}{F(D)}e^{2ix}=\frac{e^{2ix}}{-8i} \] Since \[ \frac{1}{-8i}=\frac{i}{8} \] we get \[ \frac{e^{2ix}}{-8i}=\frac{i}{8}e^{2ix} \] Now, \[ e^{2ix}=\cos2x+i\sin2x \] So, \[ \frac{i}{8}e^{2ix} = \frac{i}{8}\cos2x+\frac{i^2}{8}\sin2x \] \[ = \frac{i}{8}\cos2x-\frac{1}{8}\sin2x \] The real part is \[ -\frac{1}{8}\sin2x \]

Step 4: Final answer. \[ \boxed{-\frac{1}{8}\sin 2x} \]