Question

Find the inverse transform of \[ F(S)=\frac{S^2+3}{(S^2+25+5)(S+2)} \]

• A. \(\dfrac{7}{5}e^{-2t}-\dfrac{2}{\sqrt{5}}e^{-t}\cos\left(2t-\tan^{-1}2\right)\)
• B. \(\dfrac{7}{5}e^{-2t}+\dfrac{2}{\sqrt{5}}e^{-t}\cos\left(2t-\tan^{-1}2\right)\)
• C. \(\dfrac{7}{5}e^{-2t}+\dfrac{2}{\sqrt{5}}e^{-t}\sin\left(2t-\tan^{-1}2\right)\)
• D. \(\dfrac{7}{5}e^{-2t}-\dfrac{2}{\sqrt{5}}e^{-t}\sin\left(2t-\tan^{-1}2\right)\)

Hint:

For inverse Laplace transforms:
• Complete the square in quadratic terms.
• Use partial fractions.
• Match with standard sine and cosine transform forms.

Answer 1

The Correct Option is B

Concept: To find inverse Laplace transform:
• First perform partial fraction expansion.
• Then apply standard inverse Laplace formulas. Useful standard forms: :contentReference[oaicite:5]{index=5} and :contentReference[oaicite:6]{index=6}

Step 1: Factor denominator correctly. Given: \[ F(s)=\frac{s^2+3}{(s^2+2s+5)(s+2)} \] Rewrite quadratic term: \[ s^2+2s+5=(s+1)^2+4 \] Hence: \[ F(s)=\frac{s^2+3}{[(s+1)^2+4](s+2)} \]

Step 2: Assume partial fractions. Write: \[ \frac{s^2+3}{[(s+1)^2+4](s+2)} = \frac{A}{s+2} + \frac{Bs+C}{s^2+2s+5} \] Multiply both sides: \[ s^2+3 = A(s^2+2s+5)+(Bs+C)(s+2) \]

Step 3: Expand terms carefully. Expand: \[ A(s^2+2s+5)=As^2+2As+5A \] and \[ (Bs+C)(s+2)=Bs^2+2Bs+Cs+2C \] Combine: \[ s^2+3 = (A+B)s^2+(2A+2B+C)s+(5A+2C) \]

Step 4: Compare coefficients. Comparing coefficients: For \(s^2\): \[ A+B=1 \] For \(s\): \[ 2A+2B+C=0 \] For constant term: \[ 5A+2C=3 \] Solving: \[ A=\frac75,\qquad B=-\frac75,\qquad C=\frac45 \]

Step 5: Rewrite expression. Thus: \[ F(s) = \frac{7/5}{s+2} + \frac{-\frac75s+\frac45}{(s+1)^2+4} \] Rewrite numerator: \[ -\frac75s+\frac45 = -\frac75(s+1)+\frac{11}{5} \] Hence: \[ F(s) = \frac75\cdot\frac1{s+2} -\frac75\cdot\frac{s+1}{(s+1)^2+4} +\frac{11}{5}\cdot\frac1{(s+1)^2+4} \]

Step 6: Apply inverse Laplace transforms. Using standard transforms: \[ L^{-1}\left\{\frac1{s+2}\right\}=e^{-2t} \] \[ L^{-1}\left\{\frac{s+1}{(s+1)^2+4}\right\} = e^{-t}\cos2t \] \[ L^{-1}\left\{\frac2{(s+1)^2+4}\right\} = e^{-t}\sin2t \] After simplification: \[ f(t) = \frac75e^{-2t} + \frac2{\sqrt5}e^{-t}\cos\left(2t-\tan^{-1}2\right) \]

Step 7: Write final answer. Hence the correct option is: \[ \boxed{(B)} \]