# Partial fractions question.

1. Nov 8, 2012

### e_brock123

1. The problem statement, all variables and given/known data
Hi I just have a problem in regards to setting up your partial fractions when doing nonhomogeneous differential equations using Laplace transforms.
I’m at the stage of getting the inverse Laplace of: (1-625S^4)/(S^3 (25S^2+1) )

2. Relevant equations

3. The attempt at a solution
So I’m going to simplify that equation by using partial fractions and I get:
A/S+B/S^2 +C/S^3 +(DS+E)/(25S^2+1)
I roughly understand repeated linear factors but however I’m having trouble grasping why there is an S attached to the D. If you could explain to me why this occurred that would be very helpful thanks in advance.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 8, 2012

### Staff: Mentor

Because that's the way it works.

If the denominator happened to be (S - 1)(S - 2), you would decompose the fraction as A/(S - 1) + B/(S - 2). If either or both of the factors in the denominator were repeated, you would just add another term with a constant for each repeated factor, just as you did to get A/S + B/S2 + C/S3. In each case, the polynomial in the numerator is one degree less than the degree of the factor that is repeated. (IOW, the factor that is repeated is S, and it's degree 1, so each numerator will be a polynomial of degree 0 - a constant.)

With irreducible quadratic factors, you want a polynomial of degree one less than the quadratic, so the numerator will be a first-degree polynomial, which will have an S term and a constant. If there are repeated irreducible quadtratic factors, then each one will have a numerator that is first degree.

3. Nov 8, 2012

### SammyS

Staff Emeritus
That expression can be simplified considerably.

625 is a perfect square, so the numerator can be factored.

4. Nov 8, 2012

### e_brock123

Oh wow thanks heaps for the help I understand it much better now Mark44.