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quasar_4
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Homework Statement
Determine the inverse Laplace transform of 1/((s^2 +1)*(s-1)).
The answer is 1/2*(e^x - cos(x) - sin(x)).
Homework Equations
We get a table of known inverse Laplace transforms.
The Attempt at a Solution
I tried to break this up using partial fractions, i.e., A/(s^2 +1 ) + B/(s-1). Then I solved for the coefficients A and B. Now for B I got B=1/2, which gives us 1/2*(s-1) which is just 1/2*e^x. So there's part of the answer already.
I am stuck on the other part. For A (maybe I am not solving for A correctly) I got A=1/(i-1). But then I have this fraction 1/((i-1)*(s^2 +1)). The 1/(s^2+1) is just sin(x). So I don't see why my answer wouldn't just be
1/2*e^x + (1/(i-1))*sin(x). I even tried expanding out the product (i-1)(s^2+1) and looking for a way to complete the square, but I must have missed it, whatever it was.
Where does the other half come from? And the other cos(x)? I know the answer above is right (both from the professor and Maple) but I don't see how they got it. Any help would be wonderful!