Question

The inverse Laplace transform of $\frac{s+3}{s^2 - 4s + 13}$ is:}

• A. \(e^{2t} [\cos 2t + 3\sin 2t]\)
• B. \(\frac{e^{2t}}{3} [3\cos 3t + 5\sin 3t]\)
• C. \(e^{2t} [t + 3\sin 3t]\)
• D. \(\frac{e^{2t}}{5} [5\cos 3t + 3\sin 3t]\)

Hint:

When completing the square in the denominator $(s-a)^2 + \omega^2$, you immediately find the damping factor $e^{at}$ and frequency $\omega$. Here, $a=2$ and $\omega=3$, which instantly eliminates options (A) and (C).

Answer 1

The Correct Option is B

Concept: To find the inverse Laplace transform of a quadratic fractional function, we first complete the square for the polynomial in the denominator. Then, we apply the First Shifting Theorem, which states: \[ \mathcal{L}^{-1}\{F(s-a)\} = e^{at} \mathcal{L}^{-1}\{F(s)\} \] The standard baseline inverse transforms are: \[ \mathcal{L}^{-1}\left\{\frac{s}{s^2 + \omega^2}\right\} = \cos(\omega t), \quad \mathcal{L}^{-1}\left\{\frac{\omega}{s^2 + \omega^2}\right\} = \sin(\omega t) \]

Step 1: Complete the square of the denominator.
The given expression in $s$ is: \[ F(s) = \frac{s+3}{s^2 - 4s + 13} \] Focusing on the denominator $s^2 - 4s + 13$: \[ s^2 - 4s + 13 = (s^2 - 4s + 4) + 9 = (s-2)^2 + 3^2 \]

Step 2: Rewrite the numerator in terms of $(s-2)$.
To apply the shift consistently, look to convert the $s$ in the numerator to match $(s-2)$: \[ s + 3 = (s - 2) + 2 + 3 = (s - 2) + 5 \] Substituting these components back into the expression for $F(s)$: \[ F(s) = \frac{(s-2) + 5}{(s-2)^2 + 3^2} \]

Step 3: Separate the fraction into core transform components.
Splitting the numerator allows us to isolate the cosine and sine operational profiles: \[ F(s) = \frac{s-2}{(s-2)^2 + 3^2} + \frac{5}{(s-2)^2 + 3^2} \] To adjust the sine component perfectly to match $\frac{\omega}{s^2+\omega^2}$ where $\omega=3$, multiply and divide the second term by 3: \[ F(s) = \frac{s-2}{(s-2)^2 + 3^2} + \frac{5}{3} \cdot \frac{3}{(s-2)^2 + 3^2} \]

Step 4: Take the Inverse Laplace Transform.
Applying the first shifting property with $a = 2$: \[ \mathcal{L}^{-1}\{F(s)\} = e^{2t} \mathcal{L}^{-1}\left\{ \frac{s}{s^2 + 3^2} \right\} + \frac{5}{3} e^{2t} \mathcal{L}^{-1}\left\{ \frac{3}{s^2 + 3^2} \right\} \] \[ \mathcal{L}^{-1}\{F(s)\} = e^{2t} \cos 3t + \frac{5}{3} e^{2t} \sin 3t \]

Step 5: Factor out constants to match the option formatting.
Bringing out a common factor of $\frac{e^{2t}}{3}$: \[ \mathcal{L}^{-1}\{F(s)\} = \frac{e^{2t}}{3} \left[ 3\cos 3t + 5\sin 3t \right] \] This matches Option (B).