Question

How do you solve for A, B, C, D? As^3+2As^2+3As+Bs^2+2Bs+3B+Cs^3+Cs+Ds^2+D=1

Answer 1

nope, don't go on that way!!!!! there is an outlet but it is not the best way to solve
However, if you want, you can. combine like term, first s^3 (A +C) , the right handside doesn't have s^3 therefore, A+C =0
do the same with s^2 , s and constant. you will have another system to solve. My way is the easiest way to solve partial fraction decompose.

Answer 2

So there's another easier way to do this? How?

Answer 3

give me the original one or the partial decompose form . I mean A /...+ B/... +C/...=1

Answer 4

So A+C=0, 2A+B+D=0, 3A+2B+C=0, and 3B+D=1, right?

Answer 5

yes

Answer 6

Okay, please don't leave, wait a minute.

Answer 7

I told you, don't go on that way

Answer 8

Why?

Answer 9

other way is wayaaaayy easier. Ok, you can experiment. do it, when get stuck or miserable with the matrix, I will show you mine . Stubborn girl, (like me!! hehehe)

Answer 10

Can you show me your way?

Answer 11

You mean woman...

Answer 12

Okay, so I got A=-1/4, C=1/4, B=1/4, D=1/4. Now what?

Answer 13

plug into the decompose form, for example \[\frac{A}{s^2+1}\text{replace A =-1/4 to get }~~\frac{-1}{4(s^2+1)}\]

Answer 14

It's (As+B)/(s^2+1)+(Cs+D)/(s^2+2s+3).

Answer 15

same,

Answer 16

So it's -1/4(s-1)/(s^2+1)+1/4(s+1)/(s^2+2s+3), right?

Answer 17

hey, stubborn girl!!! Anyway, you don't have laplace inverse of As +B , have to break it down to As/(s^2+1) + B/(s^2+1) right? so, plug A and B into them.

Answer 18

Let me look at the table first. And I'm proud to be stubborn.

Answer 19

hahahaa, meeeee too.!! I am proud to be meeeeaaan!! hehehehe

Answer 20

You're not mean, you're evil. And I wanna be like you.

Answer 21

Do you know how to find the inverse L.T. of e^(-3pi*s)/(s^2+2s+3)?