Partial Fraction Expansion with repeated roots

In summary, when dealing with repeated roots in partial fractions, it is important to multiply both sides by the denominator of the coefficient you want and then let the root equal the constant on the other side of the equation. In this case, to find the coefficients a and b, we can set s=-2 and solve for b, and then use another value for s to solve for a. This technique can be applied to other problems with repeated roots as well.
  • #1
jhmz
9
0
im trying to work out how to use partial fractions on a fraction with repeated roots. I am learning about laplace transforms at the moment, i don't remeber the lecturer specifically going through how to solve the transforms that have repeated roots and can't find it anywhere in the lecture material however he has organised the study questions into distinct/repeated/purely imaginary/complex roots.

Anywho I am stuck on the first question.

2y'' + 8y' + 8y = 0
y(0) = 1
y'(0) = 0

which transforms to:
[tex]Y_{(s)}=\frac{s+4}{(s+2)^2}[/tex]

Now the partial fraction I've got so far from looking at websites such as:
http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/PartialFraction/RootsRepeat.html

looks like:
[tex]Y_{(s)}=\frac{s+4}{(s+2)^2}=\frac{a}{s+2}+\frac{b}{(s+2)^2}[/tex]

Now from here it says to multiply both sides by the denominator of the coefficient you want and let s=-2 in the case of the denominator of a, s+2 as doing this cancels everything on the left hand side which looks like:

[tex]a=(s+2)\frac{s+4}{(s+2)^2}[/tex] where s = -2

but then it equals zero
and i don't get what i am doing wrong
 
Last edited by a moderator:
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  • #2
I think in this case it might be better not to use that shortcut. Once you have

[tex]\frac{s+4}{(s+2)^2} = \frac{a}{s+2} + \frac{b}{(s+2)^2}[/tex]

then you can just do

[tex]\frac{s+4}{(s+2)^2} = \frac{a}{(s+2)} \frac{(s+2)}{(s+2)} + \frac{b}{(s+2)^2}[/tex]

[tex]\frac{s+4}{(s+2)^2} = \frac{as+2a}{(s+2)^2} + \frac{b}{(s+2)^2}[/tex]

[tex]s+4 = as + 2a + b[/tex]

from which it's pretty straightforward to get a and b.

In general, a useful algebra technique is to look for creative ways to multiply by 1.
 
  • #3
That is quite creative and works out logically...just manipulating what is there, i would have never thought of that. Thanks for your help and will surely be keeping that advice in mind

=)
 
  • #4
I'm having a bit of trouble with repeated roots myself but still not sure what to do with the s+4=as+2a+b to find a & b! Could anyone please explain? Thanks
 
  • #5
CremeBrulee said:
I'm having a bit of trouble with repeated roots myself but still not sure what to do with the s+4=as+2a+b to find a & b! Could anyone please explain? Thanks

The left hand side and the right hand side must be equal, thus you need to set the factors for the s term and the constant equal to each other. This gives a (very easy) system of two equations in two unknowns (a and b):

[tex]1=a[/tex]

[tex]4=2a+b[/tex]

I assume you can solve this. Don't make it too difficult on yourself, the solution is often more easy than you might suspect especially in exercises from textbooks or the ones given in class.

Can you proceed from here?

coomast
 
  • #6
jhmz said:
im trying to work out how to use partial fractions on a fraction with repeated roots. I am learning about laplace transforms at the moment, i don't remeber the lecturer specifically going through how to solve the transforms that have repeated roots and can't find it anywhere in the lecture material however he has organised the study questions into distinct/repeated/purely imaginary/complex roots.

Anywho I am stuck on the first question.

2y'' + 8y' + 8y = 0
y(0) = 1
y'(0) = 0

which transforms to:
[tex]Y_{(s)}=\frac{s+4}{(s+2)^2}[/tex]

Now the partial fraction I've got so far from looking at websites such as:
http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/PartialFraction/RootsRepeat.html

looks like:
[tex]Y_{(s)}=\frac{s+4}{(s+2)^2}=\frac{a}{s+2}+\frac{b}{(s+2)^2}[/tex]

Now from here it says to multiply both sides by the denominator of the coefficient you want and let s=-2 in the case of the denominator of a, s+2 as doing this cancels everything on the left hand side which looks like:

[tex]a=(s+2)\frac{s+4}{(s+2)^2}[/tex] where s = -2

but then it equals zero
and i don't get what i am doing wrong
No, it is NOT 0 because there is also an [itex](s+2)^2[/itex] in the denominator. That "multiply both sides by the denominator of the coefficient you want" applies only to non-multiple roots.

Mutiplying both sides by [itex](s+2)^2[/itex] gives
[tex]s+ 4= a(s+2)+ b[/tex]
Now, taking s= -2 gives -2+ 4= a(0)+ b so b= 2.
Knowing that the equation is [itex]s+ 4= a(s+2)+ 2[/itex] for all s and you can get an equation to solve a by replacing s by any number other than -2. Simplest is to take s= 0 so that [itex]0+ 4= a(0+ 2)+ 2[/itex] so 2a= 2 and a= 1.
 
Last edited by a moderator:
  • #7
(s+4)/(s+2)^2 = (s+2+2)/(s+2)^2 =

(s+2)/(s+2)^2 + 2/(s+2)^2 =

1/(s+2) + 2/(s+2)^2
 

Related to Partial Fraction Expansion with repeated roots

What is partial fraction expansion with repeated roots?

Partial fraction expansion with repeated roots is a mathematical technique used to decompose a rational function with repeated roots in the denominator into simpler fractions. This is useful for solving integrals and other mathematical problems.

Why do we need to use partial fraction expansion with repeated roots?

Repeated roots in the denominator of a rational function can make it difficult to solve for the integral or evaluate the function. By using partial fraction expansion, we can break down the complex fraction into simpler fractions that are easier to integrate or evaluate.

How do we perform partial fraction expansion with repeated roots?

To perform partial fraction expansion with repeated roots, you first need to factor the denominator of the rational function. Then, you set up a system of equations using the coefficients of the fractions in the decomposition. This system of equations can then be solved to find the values of the coefficients.

What are the different types of repeated roots in partial fraction expansion?

There are two types of repeated roots in partial fraction expansion: simple repeated roots and multiple repeated roots. Simple repeated roots occur when a factor in the denominator appears more than once. Multiple repeated roots occur when a factor in the denominator appears with a power greater than one.

What are some common mistakes to avoid when performing partial fraction expansion with repeated roots?

One common mistake is forgetting to include all the terms in the decomposition. It's important to make sure all the terms in the original rational function are accounted for in the decomposition. Another mistake is to incorrectly set up the system of equations, which can result in incorrect coefficients. It's important to double check the set up of the equations before solving for the coefficients.

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