e^-5s/(s^2+3s+2)+e^-10s/(s(s^2+3s+2))+(1/2)/(s^2+3s+2)

How do you find the inverse laplace transform? Please help.

e^-5s/(s^2+3s+2)+e^-10s/(s(s^2+3s+2))+(1/2)/(s^2+3s+2)

How do you find the inverse laplace transform? Please help.

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Well you have to use the idea of partial fraction expansion to answer this and then use known laplace transform pairs.

Let's do it step by step:

We know that the inverse laplace transform of 1/(s-a) is e^at, for s>a. We also know that when you multiply by an exponential in the s domain it's a time shift.

Let's start from the first part, e^-5s/(s^2+3s+2).

We can rewrite this as: e^(-5s)/((s+1)(s+2)), which can also be written as: e^(-5s)*(A/(s+1)+B/(s+2)). Now this inverse laplace transform became very easy, since we know it will be in the following form:

A*e^(-(t-5))+B*e^(-2*(t-5)), all we have left to do now is find A and B. This can be easily done using the residue theory.

We find that A is 1 and B is -1.

If we go on to the second part and do the inverse laplace transform in the same method.

we have e^(-10s)/(s(s^2+3s+2)), again using partial fractions it becomes:

e^(-10s)*(A/s+B/(s+2)+C(s+1))

Therefore the inverse laplace transform of this (After finding the residues):

1/2+1/2*e^(-2*(t-10))-1*e^(-(t-10))

Going on to the last part, which is 1/2/(s^2+3s+2), we can find the inverse laplace transform to be:

1/2*e^(-t)-1/2*e^(-2t)

Add all of those together and that will be the inverse laplace transform of the above equation. Hope this helps, let me know if you have any questions.

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## Comments

How did you get the inverse laplace transform for (1/2)/(s^2+3s+2)? I know that (s+1)(s+2)=0, x=-1, -2. But from (1/2)(A/(s+2)+B/(s+1)), how do you solve for A and B? Can you please show step by step help?