Question

Find $L\{t\,u(t-2)\}$.

• A. $\frac{e^{-2s}}{s}$
• B. $\frac{e^{-2s}(2s+1)}{s}$
• C. $\frac{e^{-2s}(2s+1)}{s^2}$
• D. $\frac{e^{-2s}(2s+1)}{s^3}$

Hint:

Always split expressions before applying Laplace shifting.

Answer 1

The Correct Option is C

Concept: Use shifting property: \[ L\{f(t-a)u(t-a)\}=e^{-as}F(s) \]

Step 1: Rewrite function.
\[ t u(t-2)=(t-2+2)u(t-2) \] \[ =(t-2)u(t-2)+2u(t-2) \]

Step 2: Apply Laplace transform.
\[ L\{(t-2)u(t-2)\}=e^{-2s}\frac{1}{s^2} \] \[ L\{2u(t-2)\}=2\cdot \frac{e^{-2s}}{s} \]

Step 3: Combine results.
\[ L = e^{-2s}\left(\frac{1}{s^2}+\frac{2}{s}\right) \] \[ = \frac{e^{-2s}(2s+1)}{s^2} \]