# rearranging an equation

• November 27th 2006, 02:54 PM
mesterpa
rearranging an equation
hi
im trying to rearrange an equation to make y the subject to then use partial fraction expansion and laplace inverses to solve.
so far i have simplified to give

y(S^4 + 3S^2 -12) = 104/s+3 -24 +12S^3 - 3S^2 + 25S -15

i know i need to divide by the bracketed terms on the left but i dont know how to do this. i.e what is (104/s+3)/(S^4 + 3S^2 -12)

do i need a common denominator

any help would be appreciated
• August 14th 2013, 02:03 AM
brosnan123
Re: rearranging an equation
Solution :- Given y(S^4 + 3S^2 -12) = (104/s)+3 -24 +12S^3 - 3S^2 + 25S -15
s(S^4+3S^2-12)Y=104+s(12S^3-3S^2+25S-36)
because s(S^4+35^2-12)?0
Y=(12sS^3-3sS^2+25sS-36s+104)/s(S^4+3S^2-12)
• August 14th 2013, 10:42 AM
ebaines
Re: rearranging an equation
Please clarify the right hand side of yuor equation. What you wrote is this:

$\frac {104} {S} + 3 -24 +12S^3-3S^2 + 25S - 15$

But I suspect what you meant is this:

$\frac {104} {S + 3} -24 +12S^3-3S^2 + 25S - 15$

Why didn't you combine the -24 and -15 terms?

One option is to multiply both sides through by (S+3), then divide both sides by the Y coeffeicient:

$\small y(S^4 + 3S^2 -12)(S+3) = (\frac {104}{s+3} -24 +12S^3 - 3S^2 + 25S -15)(s+3)$

$\small y(S^5+3S^4+3S^3+9S^2-12S-36) = 104 -39(S+3)+12S^3(S+3)-3S^2(S+3)+25S(S+3)$

$\small y= \frac {(12S^4+33S^3+16S^2+10S+36S+179)}{(S^5+3S^4+3S^3+9 S^2-12S-36)}$