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Find the inverse Laplace transform of e^(-5s)/(s^2+3s+2).
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this looks like a heaviside function
How do I do it?
separate the laplace and you have e^-(5s)*(1)/(s^2+3s+2)
f(t) will be the inverse laplace transform of (1)/(s^2+3s+2)
do you know the heaviside formula?
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No.
ok the eqaution is L{f(t-a)U(t-a)}=e^(-as)F(s)
i mean equation* now you see the equation, we will work backwards to find the LHS of the equation
our U(t-a) will be U(t-5) because and you can see that from the formula e^(-5s) where a=-5
now we have to work on finding f(t) from (1)/(s^2+3s+2)
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you have the table in front of you
Okay, I get it.
@Idealist, Heaviside function is the same as unit step function.
How would you do this problem?
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