Why is there an S attached to D in partial fractions for Laplace transforms?

In summary, The student attempted to solve a nonhomogeneous differential equation using partial fractions, but was having difficulty understanding why an S was attached to the D in the denominator. If the denominator happened to be (S - 1)(S - 2), the student would have decomposited the fraction as A/(S - 1) + B/(S - 2). If either or both of the factors in the denominator were repeated, the student would add another term with a constant for each repeated factor, just as they did to get A/S + B/S2 + C/S3.
  • #1
e_brock123
14
0

Homework Statement


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) )


Homework Equations





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.
 
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  • #2
e_brock123 said:

Homework Statement


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) )


Homework Equations





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.
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
e_brock123 said:

Homework Statement


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) )
That expression can be simplified considerably.

625 is a perfect square, so the numerator can be factored.
 
  • #4
Oh wow thanks heaps for the help I understand it much better now Mark44.
 

1. What are partial fractions?

Partial fractions are a mathematical technique used to break down a complex rational expression into simpler fractions. This is done by finding the individual fractions that, when added together, equal the original expression.

2. Why are partial fractions used?

Partial fractions are used to make complicated algebraic expressions easier to work with. They can also help in solving integrals and differential equations. Additionally, partial fractions can be used to simplify and solve equations in fields such as physics and engineering.

3. How do you find the partial fraction decomposition?

The process for finding the partial fraction decomposition involves several steps. First, you need to factor the denominator of the rational expression. Then, you set up an equation with unknown constants for each factor in the denominator. Finally, you solve for the unknown constants by equating coefficients on both sides of the equation.

4. Can any rational expression be decomposed into partial fractions?

No, not all rational expressions can be decomposed into partial fractions. The expression must have a proper fraction, which means that the degree of the numerator must be less than the degree of the denominator. Also, the denominator must be factorable into distinct linear factors or a combination of linear and quadratic factors.

5. What are some applications of partial fractions?

Partial fractions have many applications in mathematics and other fields. In mathematics, they are used to solve integrals, simplify algebraic expressions, and find solutions to differential equations. In physics and engineering, they can be used to model and solve problems related to circuits, fluid dynamics, and signal processing.

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