Question

Question:
\[\mathcal{L}\left \{ (3t+1)\mathcal{U}(t-1) \right \}\]Properties:
\[\mathcal{L}\left \{ f(t-a)\mathcal{U}(t-a) \right \}=e^{-as}F(s)\]Procedure:
We see that\[f(t)=3t+1\]\[f(t-1)=3(t-1)+1=3t-3+1=3t-2.\]Therefore,\[\mathcal{L}\left \{ (3t-2+3)\mathcal{U}(t-1) \right \}=\mathcal{L}\left \{ (3t-2)\mathcal{U}(t-1) \right \}+\mathcal{L}\left \{ 3\mathcal{U}(t-1) \right \}\implies\]\[e^{-as}\left (\frac{3}{s^2}-\frac{2}{s}\right )+3\frac{e^{-as}}{s}\]Does this seem plausible?

Answer 1

yerp

Answer 2

I guess this could be further simplified to:\[e^{-s}\left ( \frac{3}{s^2}+\frac{1}{s} \right ).\]

Answer 3

dont forget the "a" in the exponent

Answer 4

http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx

Look at number 28 on the table.

Answer 5

Thank you so much, Joe.