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converting an nth order equation to a system of first order equations

 
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Nov2-09, 09:00 PM   #1
 

converting an nth order equation to a system of first order equations


1. The problem statement, all variables and given/known data
convert y'' +x^2y'+12y=0 to a system of first order equations with initial conditions y(0)=0 y'(0)=7.


2. Relevant equations



3. The attempt at a solution
first i isolate highest derivative y'' = -x^2y'-12y
then i let u_1=y u_2=y'

then (u_1)' = u_2 and (u_2)= y''

then (u_2)' = (-12u_1)-(x^2u_2)

i then write these as

(u_1)' = 0*u_1 + 1u_2
(u_2)' = (-12u_1) + (-x^2u_2)


so then in matrix form i have

matrix [u_1 u_2] = [top -0 1 bottom -12 -x^2] *[u_1 u_2] + [0 0]

i think im close but i don't know how to get vector c but putting vec u(0) ....pls help and sorry for the poor notation...i can rewrite but i can't find link to use the symbols and such
 
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Nov2-09, 09:31 PM   #2
 
bump...pleaseee help
 
Nov2-09, 09:33 PM   #3

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I think you've basically got it. [u1,u2]'=[[0,1],[-12,-x^2]]*[u1,u2]. You can add [0,0] to that but there's no need to. The initial condition is then [u1(0),u2(0)]=[0,7], right? If you want some texing clues check out http://www.physicsforums.com/showthread.php?t=8997
 
Nov2-09, 09:38 PM   #4
 

converting an nth order equation to a system of first order equations


im sorry but if u can, please help me with that matrix algebra...my prof just copied chicken scratch notes and didn't explain anything.

edit :this is what i get

[u1,u2]'= [top 0 u1 bottom -12u2 -x^2u2]

how to i get my final answer
 
Nov2-09, 09:44 PM   #5

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There's not much to explain. The problem was to convert the problem to a system of first order ode's. You have already done that, I think. It didn't say you should solve it, right? It just said convert.
 
Nov2-09, 09:47 PM   #6
 
your right, it says to convert, but my prof also wants us to find vector c with the initial conditions (forgot to mention). thats how im not sure you get [0,7] which is the correct answer
 
Nov2-09, 09:51 PM   #7

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[u1(0),u2(0)]=[y(0),y'(0)], that was your definition of u1 and u2, right?
 
Nov2-09, 10:00 PM   #8
 
yes.
 
Nov2-09, 10:03 PM   #9

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Quote by hachi_roku View Post
yes.
Ok, so I'm guessing you also see why that's [0,7].
 
Nov2-09, 10:04 PM   #10
 
ok i think i get it... the y values from the initial conditions are the ones put in the vector c matrix
 
Nov2-09, 10:05 PM   #11
 
got it thanks!
 
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