# Partial Fraction Expansion with repeated roots

1. Oct 6, 2008

### jhmz

im trying to work out how to use partial fractions on a fraction with repeated roots. Im learning about laplace transforms at the moment, i dont remeber the lecturer specifically going through how to solve the transforms that have repeated roots and cant find it anywhere in the lecture material however he has organised the study questions into distinct/repeated/purely imaginary/complex roots.

Anywho im stuck on the first question.

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

which transforms to:
$$Y_{(s)}=\frac{s+4}{(s+2)^2}$$

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

looks like:
$$Y_{(s)}=\frac{s+4}{(s+2)^2}=\frac{a}{s+2}+\frac{b}{(s+2)^2}$$

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:

$$a=(s+2)\frac{s+4}{(s+2)^2}$$ where s = -2

but then it equals zero
and i dont get what i am doing wrong

2. Oct 6, 2008

### Ben Niehoff

I think in this case it might be better not to use that shortcut. Once you have

$$\frac{s+4}{(s+2)^2} = \frac{a}{s+2} + \frac{b}{(s+2)^2}$$

then you can just do

$$\frac{s+4}{(s+2)^2} = \frac{a}{(s+2)} \frac{(s+2)}{(s+2)} + \frac{b}{(s+2)^2}$$

$$\frac{s+4}{(s+2)^2} = \frac{as+2a}{(s+2)^2} + \frac{b}{(s+2)^2}$$

$$s+4 = as + 2a + b$$

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. Oct 6, 2008

### jhmz

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. Aug 23, 2009

### CremeBrulee

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. Aug 23, 2009

### coomast

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):

$$1=a$$

$$4=2a+b$$

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. Aug 23, 2009

### HallsofIvy

Staff Emeritus
No, it is NOT 0 because there is also an $(s+2)^2$ 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 $(s+2)^2$ gives
$$s+ 4= a(s+2)+ b$$
Now, taking s= -2 gives -2+ 4= a(0)+ b so b= 2.
Knowing that the equation is $s+ 4= a(s+2)+ 2$ 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 $0+ 4= a(0+ 2)+ 2$ so 2a= 2 and a= 1.

7. Aug 23, 2009

### Count Iblis

(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

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}