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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 |
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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 |
| Nov2-09, 09:31 PM | #2 |
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bump...pleaseee help
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| 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
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| Nov2-09, 09:38 PM | #4 |
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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.
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| Nov2-09, 09:47 PM | #6 |
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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
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| 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?
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| Nov2-09, 10:00 PM | #8 |
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yes.
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| Nov2-09, 10:03 PM | #9 |
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| Nov2-09, 10:04 PM | #10 |
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ok i think i get it... the y values from the initial conditions are the ones put in the vector c matrix
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| Nov2-09, 10:05 PM | #11 |
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got it thanks!
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