Question

use properties of the Laplace transform and the table of Laplace transforms to determine L[f]
\(f(t)=2(t-5)u_5(t)\)

Answer 1

\(\text{Step 1}\)
\(\text{Given: }\)
\(f(t)=2(t-5)u_5(t)\)
\(L\left[f(t)\right]-?\)
\(\text{Step 2}\)
\(\text{Formula : } L\left[f(t-c)u_c(t)\right]=e^{-cs}L\left[f(t)\right]\)
\(\text{Step 3}\)
\(\text{Solution : } f(t)=2(t-5)u_5(t)\)
\(\text{Take Laplace Transform to both sides: }\)
\(L\left[f(t)\right]=L\left[2(t-5)u_5(t)\right]\)
\(L\left[f(t)\right]=2L\left[(t-5)u_5(t)\right]\)
\(L\left[f(t)\right]=2e^{-5s}L(t) \Rightarrow \left[L\left[f(t-c)u_c(t)\right]=e^{-cs}L\left[f(t)\right]\right]\)
\(L\left[f(t)\right]=2e^{-5s}\bigg(\frac{1!}{s^{1+1}}\bigg) \Rightarrow \left[L(t^n)=\frac{n!}{s^{n+1}}\right]\)
\(L\left[f(t)\right]=2e^{-5s}\bigg(\frac{1}{s^2}\bigg)\)
\(\begin{array}{|c|c|} \hline L\left[f(t)\right]=2\frac{e^{-5s}}{s^2} \\ \hline \end{array}\)
\(\text{Step 4}\)
\(\text{Answer : } L\left[f(t)\right]=2\frac{e^{-5s}}{s^2}\)