Partial Fraction Decomposition

[Ready2GoXtr]

Homework Statement

I just can't understand it I've read plenty of guides online, I just can't figure it out. How do you do partial fraction decomposition the farthest i can get is below
1/[(s^2 + 1)(s+1)]

Homework Equations

The Attempt at a Solution

1/[(s^2 + 1)(s+1)] = A/(s+1) + (Bs + C)/(s^2 + 1)
1 = A(s^2 + 1) + (Bs + C)(s+1)

set s = -1

1 = A(2)

A = 1/2

I can't figure anything else out.

 

[cse63146]

Yes, you're right. A = 1/2. So that's one less "variable" you have to worry about. Substitute it back into the equation to get:

(1/2)(s^2 + 1) + (Bs + C)(s+1) = 1

You can easily get rid of B, and the rest should be pretty simple.

 

[songoku]

EDIT : sorry cse63146 has given a nice clue

Spoiler

\frac{1}{(s^{2}+1)(s+1)} = \frac{A}{s+1} + \frac{Bs+C}{s^{2}+1}

1 = A (s^{2}+1) + (Bs+C) (s+1)

1 = A s^{2} + A + B s^{2} + B s + Cs + C

1 = (A + B) s^{2} + (B+C)s + (A+C)

then, comparing coefficients between of the RHS and LHS :

A+B = 0, because the LHS doesn't have s^{2}

B+C = 0, because the LHS doesn't have s

A+C = 1

 

[Ready2GoXtr]

Ah I am sorry I am still confused : songoku, as for cse how do you get rid of b? I am past calculus III but i just can't remember how to do it.

 

[songoku]

You choose one value of s that will make B = 0

 

[Ready2GoXtr]

Im still confused do you mean I move all of the equation to one side with B on the other?

 

[cse63146]

just worry about this for now: (Bs + C)(s+1)

what value of s would eliminate B?

 

[Ready2GoXtr]

0, -1

 

[cse63146]

and out of 0 and -1, which what would ONLY eliminate B, and leave C?

 

[songoku]

Yes, that's right

Because you have used -1, try 0

 

[Ready2GoXtr]

I mean if i made s=0 then C = -A(1) C=-1/2

 

[cse63146]

Yes, now that you have the value of C, substitute it back to get:

(1/2)(s^2 + 1) + (Bs + -1/2)(s+1) = 1

now just let s = 1, and you can get the value of B.

Out of curiousity, is this a laplace transformation

 

[Ready2GoXtr]

(1/2)(s^2 + 1) + (Bs + -1/2)(s+1) = 1 s = 1
(1/2)(2) + (B - 1/2) = 1
B - 1/2 = 0
B = +1/2

Yes it is and its a summer class to make things work, we don't get examples, he just does proofs >_>

 

[Ready2GoXtr]

So does 1/[(s^2 + 1)(s + 1)] = (1/2)/(s + 1) + [(1/2)s + -1/2]/(s^2 + 1)

book says it should be 1/[(s^2 + 1)(s + 1)] = (1/2)/(s + 1) + [-(1/2)s + 1/2]/(s^2 + 1)

 

[cse63146]

I would double check your value for C, I keep getting positive 1/2.

(1/2) (0^2 + 1) + (B(0)+C)(0 + 1) = 1
(1/2)(1) + C(1) = 1
C = 1/2

which would also affect your value for B.

 

[Ready2GoXtr]

Ah yes! Thats what I did wrong! Now it makes since. Do i always use 1 to find the Bs in any case?

 

[cse63146]

Not always. The idea behind this method is to eliminate all but 1 variable by assigning values for s.

To determine the value of A, you made s = -1, which eliminated B and C, which gave you the value of A.

Now that you have the value of A, it's no longer a variable; it's now a constant, which you can substitute back in your equation.

By assigning s = 0, you eliminated B, leaving you only C to worry about. The idea is to keep assigning values to s in order to eliminate variables until you have one variable left (in this case you found A and C, leaving you only B), and now you have to assign a value which would "get rid" of s, and leave only B and constants.

You could also try songoku's method, which would give you the same answer.

http://www.exampleproblems.com/wiki/index.php/Algebra-Partial_Fraction_Decomposition has a lot of examples, and step by step solutions (with the method you used).

Also, there are some PFD's that you might need to know some matrix row reduction to solve, so you might want to get a refresher on that if you need it.

Good luck with Differential Equations. Had to do that last semester; it was a pain.

 

[Ready2GoXtr]

Cheers Mate!