Question

Find partial fraction of :
(4s^2 + 5s + 16)/(s+7)(s^2+4)

Answer 1

what a pain

Answer 2

yeah man!...I solved the DFQ's with laplace transform so now I have to take the inverse laplace.

Answer 3

we can do it but it is best left to a machine
http://www.wolframalpha.com/input/?i=partial+fractons+%284s^2+%2B+5s+%2B+16%29%2F%28%28s%2B7%29%28s^2%2B4%29%29

Answer 4

hmm..ok that confirms what I was doing because so far this is what I have:
(4s^2 + 5s + 16)/(s+7)(s^2+4) =177/53(s + 7) + (K2s + K3)/s^2 + 4

Answer 5

I'm trying to figure out a way to get rid of s^2 + 4 in denominator so I can solve for k2 and k3

Answer 6

you are going to write
\[\frac{A}{x+7}+\frac{Bx+C}{x^2+4}=\frac{4x^2+5x+16}{(x+7)(x^2+4)}\] and then
\[A(x^2+4)+(Bx+C)(x+7)=4x^2+5x+16\] then multiply out and equate like coefficients

Answer 7

actually we can get A right away be replacing x by -7

Answer 8

the x^2 makes it impossible to get rid of x^2 + 4...yeah I did that and I got 177/53

Answer 9

yup. and wolfram agrees. i wish i knew anything about these laplace transforms because they seem to come up constantly

Answer 10

yeah what it all boils down to is manipulation of partial fractions usually

Answer 11

but I think if we set x = 0 we can solve for C the way you set it up

Answer 12

sure now that we know A it will work right?

Answer 13

yes

Answer 14

\[4\times \frac{177}{53}+7C=16\] etc

Answer 15

\[C=\frac{20}{53}\]?

Answer 16

yep I just got that too!

Answer 17

and since
\[Ax^2+Bx^2=4x^2\] we can solve for B as well

Answer 18

\[B=4-\frac{177}{53}\]
\[B=\frac{35}{53}\]

Answer 19

Cool! i get it know! Thanks

Answer 20

well i can't help with the transform, but you are welcome
i find this way easier than solving a 3 by 3 system of equations

Answer 21

yeah the are different methods...yeah I can take it from here..thanks again!

Answer 22

yw