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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George C. | Humboldt State and Georgetown graduateHumboldt State and Georgetown graduate

Yaniv A. | Math, CS, EE, SAT tutorMath, CS, EE, SAT tutor

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?