Inverse Laplace

Question

2S+12/S²+2S+5

Answer 1

There is a little bit of ambiguity in the way the function is written, but I'm guessing since this is tagged with partial fractions, the function is 2S+12/(S²+2S+5). Also, it's not common to find S by itself, so I'm wondering if that should be Y(s) or if it is truly just S. Regardless, let's break the problem down into two parts

First, L^-1[2S]. This is 2δ'(t), where δ is the dirac delta function. Otherwise, if this is meant to be L^-1[2Y(s)], then the inverse Laplace is 2y(t). There is no real algebra or steps to this part of the process, we just need to look at the Different Laplace transforms and work backwards.

Next, let's look at the section involving partial fractions. We have the fraction 12/(S²+2S+5), and the first step is to factor the denominator as much as possible. In this case, the denominator cannot be reduced further, so we will have to use some clever algebra to manipulate the function into a workable form, and in fact, we ultimately will not be using partial fraction decomposition.

This clever trick of algebra is completing the square for the denominator. To complete the square, we just want to consider the first 2 terms, which are S²+2S. Since the first term has the coefficient 1, then we take (b/2)² as the term needed. So, (2/2)² = 1, and thus S²+2S+1 = (S+1)² is a perfect square. Back to the problem, we can split the 5 into 1 and 4, giving S²+2S+1+4 = (S+1)²+4 = (S+1)²+2².

12/((S+1)²+2²) now is starting to look like the Laplace transform for sin(at). Recall that L[sin(at)] = a/(s²+a²), and furthermore that L[e^btf(t)] = F(s-b). Using these two facts, we can finish the problem. First, in our expression, the value of a is 2, and so we'll want a 2 on top. We can accomplish this by breaking the 12 into 2*6. Second, the value of b is -1. Now, we take the inverse Laplace in this form: L^-1[6*(2/((s+1)²+2²) = 6e^-tsin(2t)

So, the final solution is either:

2y(t)+6e^-tsin(2t)

or

2δ'(t)+6e^-tsin(2t)