Question

The particular integral of \((D^2-1)y=4\cosh x\) is

• A. \(xe^x-e^{-x}\)
• B. \(x(e^{-x}-e^x)\)
• C. \(e^x-xe^x\)
• D. \(x(e^x-e^{-x})\)

Hint:

When \(F(a)=0\) for \(e^{ax}\), use the resonance rule \(\frac{x e^{ax}}{F'(a)}\).

Answer 1

The Correct Option is D

Concept:
We need the particular integral of \[ (D^2-1)y=4\cosh x \] Recall that \[ \cosh x=\frac{e^x+e^{-x}}{2} \] Therefore, \[ 4\cosh x=2(e^x+e^{-x}) \]

Step 1: Write the particular integral. \[ \text{P.I.}=\frac{1}{D^2-1}\left[4\cosh x\right] \] \[ =\frac{1}{D^2-1}\left[2e^x+2e^{-x}\right] \] \[ =2\frac{1}{D^2-1}e^x+2\frac{1}{D^2-1}e^{-x} \]

Step 2: Observe resonance. Here, \[ D^2-1=(D-1)(D+1) \] For \(e^x\), \(a=1\), and \[ F(1)=1^2-1=0 \] For \(e^{-x}\), \(a=-1\), and \[ F(-1)=(-1)^2-1=0 \] So both terms are resonance cases.

Step 3: Find P.I. for \(2e^x\). For \[ F(D)=D^2-1 \] \[ F'(D)=2D \] For \(a=1\), \[ F'(1)=2 \] Therefore, \[ \frac{1}{F(D)}2e^x = 2\cdot \frac{x e^x}{F'(1)} \] \[ =2\cdot \frac{x e^x}{2} \] \[ =xe^x \]

Step 4: Find P.I. for \(2e^{-x}\). For \(a=-1\), \[ F'(-1)=2(-1)=-2 \] Therefore, \[ \frac{1}{F(D)}2e^{-x} = 2\cdot \frac{x e^{-x}}{F'(-1)} \] \[ =2\cdot \frac{x e^{-x}}{-2} \] \[ =-xe^{-x} \]

Step 5: Add both particular integrals. \[ \text{P.I.}=xe^x-xe^{-x} \] \[ \text{P.I.}=x(e^x-e^{-x}) \]

Step 6: Final answer. \[ \boxed{x(e^x-e^{-x})} \]