Inverse Laplace- Partial Fractions with exponential

ns5032
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Homework Statement


[e^(-2s)] / (s^2+s-2)
Find the inverse Laplace transform.


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The Attempt at a Solution


I know that I can factor the denominator into (s+2)(s-1). Then I tried to use partial fractions to split up the denominator, but I don't know how to do that with an exponential on the top. Thanks for any help!
 
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just write it as exp(whatever)*(1/whatever). Then do partial fractions to get exp(whatever)*(?/a + ?/b). You will see that the exponential will be easy to 'invert' back into the time domain as it corresponds to unit step functions (i believe).
 
Im stuck in the same boat, but trying to get the partial fraction for "(e^[-s] -e^[-2s])/[(s^2)(s+1)]"

I wasn't too sure what EngageEngage meant.
 
i am having the same problems! never knew there was any issue with partial fractions involving exponential components.

my question was to get partial fraction from 3e^-2s/(s(s+5)), so i brought down the exponential function to get 3 different fractions with 1/e^2s, 1/s and 1/(s+5).

But it still doesn't work out.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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