# Thread: help with partial fractions

1. I let s = -1:

1 + 2(-1) = -3*(-1-2)(-1)^2 + 5/4*(-1-1)(-1)^2 + C*(-1-1)(-1-2)(-1)^2 + 1/2*(-1-1)(-1-2)

-1 = 9 -5/2 + 3 + 6C
-21/2 = 6C
-7/4 = C

2. Hmm. You should have this equation:

$1+2s=A(s-2)s^{2}+B(s-1)s^{2}+C(s-1)(s-2)s+D(s-1)(s-2).$

You can see that there is only an $s$ multiplying C, not a $s^{2}.$

Carry this correction all the way through to see how that changes things.

3. I used $\dfrac{1+2s}{(s-1)(s-2)s^{2}}=\dfrac{A}{s-1}+\dfrac{B}{s-2}+\dfrac{C}{s}+\dfrac{D}{s^{2}}.$

Is this not the way?

Yeah the answers are now correct. Nice one.

4. I got s^2 at the end of C using $\dfrac{1+2s}{(s-1)(s-2)(s)^2}=\dfrac{A}{s-1}\,\dfrac{(s-2)(s)^2}{(s-2)(s)^2}+\dfrac{B}{s-2}\,\dfrac{(s-1)(s)^2}{x(s-1)(s)^2}

+\dfrac{C}{s}\,\dfrac{(s-1)(s-2)(s)^2}{(s-1)(s-2)(s)^2}+\dfrac{D}{s^2}\,\dfrac{(s-1)(s-2)}{(s-1)(s-2)}.$

5. No. The C term doesn't require both s's, since it already has one in its denominator.

Glad you got required result using the other method. Any more questions?

6. Given that $\dfrac{3s}{(s-2)(s+1)(s^2+4)}=\dfrac{A}{s-2}\,\dfrac{(s+1)(s^2+4)}{(s+1)(s^2+4)}+\dfrac{B}{s +1}\,\dfrac{(s-2)(2^2+4)}{(s-2)(2^2+4)}

+\dfrac{Cs+D}{s^2+4}\,\dfrac{(Cs+D)(s-2)(s+1)}{(Cs+D)(s-2)(s+1)}$

Can you confirm that C = -17/10 and D = -3/10
the answers say C = -9/20 D = -3/20

7. You should have

$\dfrac{3s}{(s-2)(s+1)(s^2+4)}=\dfrac{A}{s-2}\,\dfrac{(s+1)(s^2+4)}{(s+1)(s^2+4)}+\dfrac{B}{s +1}\,\dfrac{(s-2)(s^2+4)}{(s-2)(s^2+4)}
+\dfrac{Cs+D}{s^2+4}\,\dfrac{(s-2)(s+1)}{(s-2)(s+1)};$

You'll see that there should be an s in a coupled of places, not a 2. Also, no need to multiply by Cs+D top and bottom, since that's in the numerator, and plays no role in determining the common denominator.

What did you get for A and B?

8. Yeah sorry the extra (Cs+D) shouldnt of been there. For A = 1/4 and B = 1/5

9. That's correct. D you can get by setting s = 0. What do you get for D?

10. Yeah I got when S = 0

3(0) = 1/4*(0+1)((0)^2+4) + 1/5*(0-2)((0)^2+4) + (C(0) + D)*(0-2)(0+1)

0 = 1 - 8/5 - 2D

3/5 = -2D

-3/10 = D

What did you get?

11. Your D is correct. So, what do you get for C? I'd try s = 1 to get your C.

12. yeah I used s = 1 and got C = -17/10

13. Let me see your work.

14. s = 1

3(1) = 1/4*(1+1)((1)^2+4) + 1/5*(1-2)((1)^2+4) + (C(1) -3/10)*(1-2)(1+1)

3 = 5/2 -1 -2C + 3/5

9/10 = -2C

-9/20 = C

15. There you go. So you got D correct, and the book got C correct.

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