Is This the Correct Approach to Solving an Inverse Laplace Transform?

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Homework Statement



Could anyone please check my work, The answer is wrong. Correct Answer: e^2t(-t+e^t-1)

Homework Equations



Inverse Laplace

The Attempt at a Solution



Here's my solution, please point out my mistakes. One mistake i found after taking the picture is: Its -2te^-2t ( third last step), Then my answer is just e^-t+e^-2t

http://postimage.org/image/mfzr6aom5/
 
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Desii said:

Homework Statement



Could anyone please check my work, The answer is wrong. Correct Answer: e^2t(-t+e^t-1)

Homework Equations



Inverse Laplace

The Attempt at a Solution



Here's my solution, please point out my mistakes. One mistake i found after taking the picture is: Its -2te^-2t ( third last step), Then my answer is just e^-t+e^-2t

http://postimage.org/image/mfzr6aom5/

You have a couple of mistakes near the beginning.

You have
$$ \frac{1}{(s + 1)^2(s + 1)^2} = \frac{A}{s+ 1} + \frac{Bs + C}{(s + 2)^2}$$

The first mistake is minor and probably a typo - you wrote (s + 1)2 in the first fraction when I think you meant (s + 2)2.

The second mistake is one of understanding. To decompose a repeated linear factor such as (s + 2)2 you do it like this:
$$ \frac{1}{(s + 1)^2(s + 2)^2} = \frac{A}{s+ 1} + \frac{B}{s + 2} + \frac{C}{(s + 2)^2}$$
The way you did it is for irreducible quadratics (i.e., quadratics with complex factors), not quadratics that are perfect squares.

After that I stopped looking.
 
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Gotcha, Thanks mate. This solves the mystery :) and yes i meant (s+2)^2
 
Desii said:
Gotcha, Thanks mate. This solves the mystery :)
That's why they pay us the big bucks. Oh wait, they don't pay us at all!
Desii said:
and yes i meant (s+2)^2

Glad I could help...
 
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