Question

Find the inverse Laplace transform....
\[1) \space \frac{ 3s+5 }{ s^2-4s+13 }\]\[2) \space \frac{ 5 }{ (2s+7)^3 }\]\[3) \space \frac{ 1 }{ (1-e^{-3s})s^2 }\]

Answer 1

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

Answer 2

Let's complete the square on the first one.
\[\large \frac{3s+5}{(s^2-4s+4)+9}\]\[\large \frac{3s+5}{(s-2)^2+3^2} = \quad \frac{3(s-2)+6+5}{(s-2)^2+3^2}\]
\[\large 3\frac{(s-2)}{(s-2)^2+3^2}+\frac{11}{9} \cdot \frac{9}{(s-2)^2+3^2}\]

Answer 3

Taking the inverse Laplace on the first term ... I think will give us...
\[\huge 3e^{2t}\cos3t\]

Answer 4

Good job. You don't have to go further.

Answer 5

I forgot about how to change the numerator

Answer 6

This was a test question on the test I just took

Answer 7

They all were

Answer 8

ah i see c: