December 22, 2014

Homework Help: Calculus - Integrals

Posted by David on Tuesday, March 25, 2008 at 9:43pm.

I made this question, just wanted to make sure it didn't drown under the tide...

I've set a problem up, something like this.

s^4: A+D=0
s^3: -2A+B-3D+E=0
s^2: 2A-B+C+3D-3E=0
s^1: -2A+B-D+3E=4
s^0: A+C-E=4

Count Iblis
-It's easier to substitute for s the values that make the denominators zero and then demand equalty. You then get simpler equations than comparing the coefficients of equal powers of s on both sides.-

Would anybody mind showing it out? I tried what Count Iblis said, but it didn't really work out well...

Then, Count Iblis said this...

Count Iblis
-Hi David,

If I give you the answer (I just verified that it is correct):

(s+1)/[(s^2+1)(s-1)^3] =

1/(s-1)^3 - 1/2 1/(s-1)^2 +

1/2 1/(s^2+1)

Can you verify that adding up the partial fractions yields the original fraction?

Can you extract your A, B, C, etc. from the solution and verify that it satisfies your equations?

Now, you have to admit that just writing down the equations by equating equal powers of s and solving them is not an effient way to solve the problem. It is perhaps a good algebra exercise, but I can assure you that people like me who have to do this for their work usually do not use this awkward method. only in somple cases when you have just two terms is the method useful.

There is a slight shortcut when you solve partial fractions by introducing the unknown constants: You add up the partial fractions so that you have a common denominator and then equate the two numerators as you've done.

You can insert s = 1, s = i and s = -i and demand that the two numerators are the same for these three cases. If you do that then you lose some terms so the equations becomes simpler Also, you can demand that the first and second derivatives of both sides are the same at s = 1.-

I'm sorta confused at this point. I just can't find the A, B, C, D, and E.

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