Question

need help with solving system of equations...

Answer 1

\[s^2x(s)+Y(s)[3s+3]=2\]
\[s^2x(s)+3Y(s)=2+\frac{ 1 }{ (s+1)^2 }\]

Answer 2

@zepdrix @dan815 @amistre64

Answer 3

i am using the determinant approach.

Answer 4

Wronskian if you will.

Answer 5

D(determinant) = \[-3s^3\]

Answer 6

Hmm I don't think I'm familiar with this type of problem.
So this system includes x ( a function of s) and y (also a function of s).

What are we trying to solve for? :U

Answer 7

Yes, we are trying to solve for the variables X(s) and Y(s).

Answer 8

or rather the functions.

Answer 9

this looks like a Laplace transform?

Answer 10

Dx=\[-6s-\frac{ 3s+3 }{ (s+1)^2 }\]

Answer 11

yes, but i only need help with solving the system of equations.

Answer 12

it seems pretty hairy and want somebody to check my work.

Answer 13

are u up for that @amistre64 ?

Answer 14

the system is just a "cramer rule" from algebra i believe

Answer 15

Dy=\[\frac{ s^2 }{ (s+1)^2 }\]

Answer 16

yes thats right.

Answer 17

can u check my work?

Answer 18

\[s^2~X(s)+(3s+3)Y(s)=2\]\[s^2~X(s)+~~~~~3~~~~~Y(s)=2+\frac{ 1 }{ (s+1)^2 }\]
\[X(s)=s^2(2+\frac{1}{(s+1)^2})-2s^2\]
\[Y(s)=6-(3s+3)(2+\frac{1}{(s+1)^2})\]

Answer 19

i got those mixed around

Answer 20

got that.

Answer 21

and i divided that by the wronskian to get teh X(s) and Y(s)

Answer 22

ok i think i might need some help with laplace stuff, too. what should i do get the inverse laplace? partial fractions for last part of x and y, right?

Answer 23

how would u break:\[\frac{ 1 }{ s^3(s+1) }\]

Answer 24

@hartnn

Answer 25

also, how would u break down:\[\frac{ -1 }{ 3s(s+1)^2 }\]

Answer 26

i setup my matrix like this to that i dont have to worry about negating the middle
\[\begin{vmatrix}Dx&D&Dy\\
s^2&2&(3s+3)\\
s^2&(2+\frac{1}{(s+1)^2})&3\\
\end{vmatrix}\]
\[Dx=6-(3s+3)(2+\frac{1}{(s+1)^2})\]\[Dx=6-6(s+1)+6\frac{1}{(s+1)}\]
\[Dy=s^2(2+\frac{1}{(s+1)^2})-2s^2\]\[Dy=\frac{s^2}{(s+1)^2}\]
\[D=-3s^3\]
-------------------
\[Dx/D=\frac2{s^3}-\frac{2(s+1)}{s^3}+\frac{2}{s^3(s+1)}\]
\[Dy/D=\frac{1}{3s(s+1)^2}\]

Answer 27

expand it all out, and complete a square on it i beleive

Answer 28

or, split it into a decomp

Answer 29

\[1=\frac{A}{3s}+\frac{B}{s+1}+\frac{C}{(s+1)^2}\]

Answer 30

can u just tell me what i should have for teh denominators because I cant seem to find great examples of the rules. I know the basics, but i think this requires a lot more specific rules.

Answer 31

what about the \[\frac{ 1 }{ s^3(s+1) }\]

Answer 32

s, s^2, s^3, s+1

Answer 33

last term should be Cx+D

Answer 34

a multiple of a linear combination has to account for all possible constructs of it

Answer 35

@zepdrix can u help with partial fractions?

Answer 36

these are all linear so the tops are constants

Answer 37

but are u sure because wouldn't teh term repeat?

Answer 38

like would A/s be the same if we broke down the the last term, D/s

Answer 39

\[\frac14=\frac{A}{2}+\frac{B}{2^2}\]

Answer 40

some of the tops might zero out, but you have to account for all the possibilities

Answer 41

yes they are linear, but some of them are quadratic.

Answer 42

so it should be Bs+C over s^2

Answer 43

none of them are quadratic .... they all reduce to a linear form; just multiples of the same linear form

Answer 44

r u sure?!

Answer 45

http://tutorial.math.lamar.edu/Classes/Alg/PartialFractions.aspx

Answer 46

i dont want to have to do the whole thing over...did that many times before.

Answer 47

thanks! ( i have been reading that page wrong haha)

Answer 48

just to clear the clutter from the page ....