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Show that the origin is a critical point (linear differential equation)

  1. Mar 7, 2009 #1

    JJBladester

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    1. The problem statement, all variables and given/known data

    Show that the origin (0,0) is a critical point. Write the linear differential equation in operator format and solve.

    2. Relevant equations

    x'' + 10x' + 25x = 0

    3. The attempt at a solution

    I am not sure how to show that the origin is a critical point (without using a graph).

    As for solving the 2nd-order linear differential equation, here's what I did:

    x'' + 10x' + 25x = 0

    Auxillary equation: m2 + 10m + 25 = 0
    Roots: m1 = m2 = -5

    General solution: x = c1e-5x + c2xe-5x

    I believe my general solution is correct, but am not sure if I solved it by "operator method".
     
    Last edited: Mar 7, 2009
  2. jcsd
  3. Mar 7, 2009 #2

    tiny-tim

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    Hi JJBladester! :wink:
    Yes, that's the correct general solution :smile:

    I think by operator format they just mean D2 + 10D + 25 = 0 :wink:

    But I don't quite understand what is meant by (0,0) :confused:

    does that mean that there is an initial condition of x' = 0 at x = 0?

    if so, (0,0) is a critical point just by looking at x'' + 10x' + 25x = 0 … you don't need to solve it. :wink:
     
  4. Mar 7, 2009 #3

    JJBladester

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    My professor said I should:

    "Change the 2nd order DEQ into a system of first order DEQ's.
    Basically...Let y=dx/dt and then substitute.... When you get these two equations, set both derivatives equal to zero and solve for the x and y value that solves the system."

    I'm still a little confused what that means.

    Here is my attempt:

    Let y = x', y'=x''

    So x'' + 10x' + 25x = 0 becomes y' + 10y + 25x = 0.

    -10y - 25x = 0
    y = 0
    -25x = 0
    x = 0
     
    Last edited: Mar 7, 2009
  5. Mar 7, 2009 #4

    tiny-tim

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    hmm … I've no idea what he means :redface:

    unless maybe he means put y = x' + 5x …

    then y' + 5y = 0 …

    dunno … as I said in my last post, I don't really understand the question :confused:
     
  6. Mar 8, 2009 #5

    HallsofIvy

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    Let y= x'. Then x'' + 10x' + 25x = 0 becomes y'+ 10y+ 25x= 0 and we have the system of equations x'= y, y'= -10x-25y. A "critical point" is (x,y) such that x'= 0 and y'= 0.
     
  7. Mar 8, 2009 #6

    tiny-tim

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    why can't we just put x'' = x' = 0 in the original equation, then, and forget about y? :confused:
     
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