Can I Simplify This Diff EQ by Using Laplace Transform and Partial Fractions?

In summary, the conversation discusses solving a differential equation using the Laplace transform. The focus is on finding the partial fraction part, specifically determining the values of C and D. By equating coefficients and using a quick formula, the values of C and D can be found.
  • #1
kdinser
337
2
This is actually part of a larger problem of solving a diff EQ by use of a laplce transform, but it's the partial fraction part that I'm stuck on.

[tex]\frac{1}{s^2(s^2+w^2)} = \frac{A}{s}+\frac{B}{s^2}+\frac{C*s+D}{s^2+w^2}[/tex]

Good so far?

If I then multiply through by the common denominator I get

EQ.2[tex]1 = A*s(s^2+w^2) + B(s^2+w^2) + s^2(C*s+D)[/tex]

So then I set s=0

[tex]1=B*w^2 \longrightarrow B=1/w^2 [/tex]

Then using the quadratic equation, I find that the roots of s^2+w^2 are +/- jw

letting s=jw I end up with

[tex]1 = -jC*w^3 - Dw^2[/tex]

this is where I get stuck, I'm not sure how to find C and D at this point.

Note: I just noticed a mistake in EQ.2,
 
Last edited:
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  • #2
you can equate the various bits, ie the coefficients of the quadratic linear and constant terms. you cut out a bit of that with s=0 which is a good trick but u need the other way to get all the constants (if only in terms of w)

ie [tex]{s^2}(C*s + D) = 0[/tex]

i think
 
  • #3
Thanks for the reply, but I'm not really sure what you are saying.
 
  • #4
You can get two equations for [itex]C[/itex] and [itex]D[/itex] by letting [itex]s=\pm i\omega[/itex]. You can then solve for [itex]C[/itex] and [itex]D[/itex].

By the way if you want to know a quick way to do this you can say the following:

[tex]
\frac{1}{xy} = \frac{1}{x-y}\left\{\frac{1}{y}-\frac{1}{x}\right\}
[/itex]

So if I choose [itex]x=s^2+\omega^2[/itex], and [itex]y=s^2[/itex] then your initial fraction can be cast directly into the form on the RHS of your equation.
 
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What is partial fraction decomposition?

Partial fraction decomposition is a method used to break down a rational function into smaller, simpler fractions. This is used to make integration and simplification of complex algebraic expressions easier.

Why do we use partial fraction decomposition?

Partial fraction decomposition is used to simplify complex rational expressions and make them easier to integrate or evaluate. It also helps to identify the roots and poles of a function.

What are the steps for performing partial fraction decomposition?

The steps for performing partial fraction decomposition are:

  1. Factor the denominator of the rational function.
  2. Equate the original expression with the sum of the partial fractions.
  3. Identify the coefficients of each term by choosing appropriate values for the unknown constants.
  4. Use algebraic manipulation to solve for the unknown constants.

Can any rational function be decomposed into partial fractions?

Yes, every rational function can be decomposed into partial fractions. However, some cases may require the use of complex numbers.

What is the difference between proper and improper partial fractions?

A proper partial fraction has a numerator with a lower degree than the denominator, while an improper partial fraction has a numerator with a greater or equal degree to the denominator. Proper partial fractions can be easily integrated, while improper partial fractions require additional steps.

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