Question

Decompose the following into partial fractions... Help please!

Answer 1

\[6 / [(s+4)(s+7)^4] + 4/(s+4)\]

Answer 2

I think you get something like \[A/(s+4)+B/(s+7)+C/(s+7)^2+D/(s+7)^3+E/(s+7)^4+4/(s+4)\]?? But I dont know how to solve for A->E

Answer 3

what a colossal pain

Answer 4

\[\frac{4}{(s+4)(s+7)^4}\] is what you want?

Answer 5

A is easy enough, replace x by -7 and see what you get
\[-3A=6\]
\[A=-2\] but the rest are annoying

Answer 6

\[-3(x+7)^4+B(x+4)(x+7)^3+C(x+4)(x+7)^2\]
\[+D(x+4)(x+7)+E(x+4)=6\]

Answer 7

A colossal pain is right.... Why have you swapped 6 for 4 in the first term, a few comments up?

Answer 8

typo

Answer 9

Ah ok!

Answer 10

i think @eliassaab has a snap way to do it, otherwise you have to solve a large system of equations
easier to cheat

Answer 11

http://www.wolframalpha.com/input/?i=partial+fractions+6%2F%28%28x%2B4%29%28x%2B7%29^4%29

Answer 12

Thank everything that is holy for WolframAlpha! Totally never knew you could do partial fractions on there! Cheers :D

Answer 13

you have to multiply out, equate like coefficients etc etc a real real pain
is this the end of some laplace question?

Answer 14

hey at least i got the -2 right!!

Answer 15

yeah its using inverse laplace transformations to solve ODEs :) using what wolfram gave me, I'm still getting the (x+7) and (x+4 coefficients wrong.... can't seem to find an easy way of figuring out whats wrong!

Answer 16

\[
\frac{6}{(s+4)
(s+7)^4}=\frac{A}{(s+7)^4}+
\frac{B}{(s+7)^3}+\frac{C}{
(s+7)^2}+\frac{D}{s+7}+\frac
{F}{s+4}
\]
Multiply the two sides by 4+s and make s =-4, you get
\[F = \frac 2 {27}
\]
Multiply both sides by (7+s)^4 and make s=-7, you get
A=-2
Multiply the two sides by (7+ s) let s go to Infinity
\[
0= F+ D=\frac 2 {27}\\
D=-\frac 2 {27}\\
\]
We still have B and C to determine.
Replace s by -6 and A, D and F by their values we get
\[-3=B+C-\frac{19}{9}\\
B+c = -\frac 89
\]
Replace s by -5 and A, D and F by their values we get

\[
-\frac{3}{8}=\frac{B}{8}+\frac
{C}{4}-\frac{17}{72}
\\
B+2 C =-\frac {10}9
\]

Solving the last equation for B and C , we get
\[
\left\{B=-\frac{2}{3},C=-\frac
{2}{9}\right\}

\]
Finally
\[
\frac{6}{(s+4)
(s+7)^4}=-\frac{2}{27
(s+7)}-\frac{2}{9
(s+7)^2}-\frac{2}{3
(s+7)^3}-\frac{2}{(s+7)^4}+
\frac{2}{27 (s+4)}
\]

Answer 17

ok that is much snappier. i will have to see if i can remember this, especially the "let x go to infinity" part

Answer 18

This is called the cover up method.