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Find the inverse Laplace transform for the following:

a. $F(s) = \dfrac{{2s - 3}}{{{s^2} - 3s + 2}}$.

I used partial fractions and I ended up getting

$$y(t) = {e^t} + {e^{2t}}.$$

b. $F(s) = \dfrac{{{e^{ - 2s}}}}{{s - 9}}$.

I ended up getting $e^{9t-9} h(t-2)$ and $0$ being $\neq 2$.

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1 Answer 1

Hints:

a). is correct

b.) is incorrect - think about the $t$ term and that being $(t-2)$ in the exponential (you got the Heaviside portion correct, so it is close).

It should be:

$$\displaystyle e^{9 (t-2)} H(t-2)$$

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Oh so it would be e^ (9t-2) h(t-2)...I have that written down lol –  user1988 May 4 '13 at 5:51
    
You are my lifesaver. Thank you again –  user1988 May 4 '13 at 5:54
    
Oh I forgot my parenthesis. I leave little mistakes out like that. And those are important. Thank you –  user1988 May 4 '13 at 5:57
    
+1 nice hints. Correct hints Amzoti. –  Babak S. May 4 '13 at 6:20
    
@BabakS. thank you, great to see you on! Regards –  Amzoti May 4 '13 at 6:21

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