Question

The particular integral of \(\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}+y=e^x\sin x\) is

• A. \(e^x\cos x\)
• B. \(-e^x\sin x\)
• C. \(-e^x\cos x\)
• D. \(e^x\sin x\)

Hint:

For \(e^{ax}V\), use the shifting rule: replace \(D\) by \(D+a\).

Answer 1

The Correct Option is B

Concept:
For equations of the form \[ F(D)y=e^{ax}V \] we use the shifting rule: \[ \frac{1}{F(D)}e^{ax}V=e^{ax}\frac{1}{F(D+a)}V \]

Step 1: Identify \(F(D)\). The differential equation is \[ (D^2-2D+1)y=e^x\sin x \] So, \[ F(D)=D^2-2D+1 \] \[ F(D)=(D-1)^2 \]

Step 2: Apply shifting rule. Since the right hand side is \[ e^x\sin x \] we replace \(D\) by \(D+1\): \[ \text{P.I.}=e^x\frac{1}{F(D+1)}\sin x \] Now, \[ F(D+1)=((D+1)-1)^2 \] \[ F(D+1)=D^2 \] Therefore, \[ \text{P.I.}=e^x\frac{1}{D^2}\sin x \]

Step 3: Evaluate \(\dfrac{1}{D^2}\sin x\). Since \[ D^2(\sin x)=-\sin x \] we get \[ \frac{1}{D^2}\sin x=-\sin x \]

Step 4: Find the particular integral. \[ \text{P.I.}=e^x(-\sin x) \] \[ \text{P.I.}=-e^x\sin x \]

Step 5: Final answer. \[ \boxed{-e^x\sin x} \]