# Thread: help with partial fractions

1. Thanks mate. Is their anyway I can get your help again in the future? Im working through the laplace transform unit thats why I needed help with partial fractions because these were to do with solving differential equations using laplace transforms. Ive started step functions and just dont know what is going on with it, I missed the lecture on it but even the worked examples dont make sense.

2. Sure, but you should post new questions in a separate thread. The general rule is no more than 2 questions per thread.

Also, if you have LT questions (I guessed that was the source, considering all the s's floating around), you should post them in the Differential Equations forum, not here in Pre-Algebra and Algebra.

3. Can you tell me:

if you have $(s-1)^2$ = $\dfrac{1}{s^2} -1$

how do you combine the right hand side?

same with

$(s+4)^2$ = $\dfrac{1}{s-3} -1 + s + 8$

4. In all cases, you have to get a common denominator. In the first case, that common denominator is s^2. In the second case, the common denominator is (s-3).

5. not sure what you mean by getting a common denominator

6. I'll give you more details. This is a 2nd order diff equation I had to solve and worked through it to this point:

$(s^2 -2s + 2)Y(s)$ = $\dfrac{1}{s^2}$

because its an irreducible quadratic I completed the square to give

$(s-1)^2 + 1$ = $\dfrac{1}{s^2}$

Now from here I dont know what to do to put it into its partial fractions. You say get the common denominator but I dont know what you mean. I looked at the wikipedia page. Same as the 2nd problem.

7. I think you do know, operationally, what the common denominator is. You've been using it several times in this thread! But maybe you haven't made the connection in your mind between what you're doing and what it's called. You can compute least common multiples. I've seen you do it. The common denominator is just the least common multiple of all the denominators in a sum or difference of fractions. Therefore, you know how to find the common denominator.

Incidentally, your solution method is a bit incomplete, I think. If you've got

$(s^2 -2s + 2)Y(s)=\dfrac{1}{s^{2}},$ then you need to solve for $Y(s)$ thus:

$Y(s)=\dfrac{1}{s^{2}(s^{2}-2s+2)}.$

Now you do partial fractions. What's your initial guess going to be?

8. When you complete the square for $Y(s)=\dfrac{1}{s^{2}(s^{2}-2s+2)}.$. Wouldnt it become $Y(s)=\dfrac{1}{s^{2}(s-1)^2+1)}.$?

9. Yes, but why do you need to complete the square? Just break the fraction apart. What's your initial guess?

10. not a clue how it can be split. The answers show theirs 3 parts to it.

11. Re-examine post # 36. I tell you there how to do it.

12. so it would give $\dfrac{1}{(s)^2(s^2-2s+2)}=\dfrac{A}{s}+\dfrac{B}{s^2}+\dfrac{Cs+D}{s^ 2-2s+2}$

just the answers say it has 3 parts not 4

13. What do you mean by "parts"? Do you mean fractions? Or constants to be found?

14. There should be 3 parts because when you take their laplace transforms you get

(-1/2e^-tcost) + (1/2) + (1/2t)

Its a matter now of how to put $\dfrac{1}{(s)^2(s^2-2s+2)}$ into 3 parts A, B and C.

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