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Homework Help: General solution of third order DE

  1. Aug 19, 2010 #1
    1. The problem statement, all variables and given/known data

    y'''-3y'+2y=0

    initial conditions y(0)=0, y'(0)=1,y''(0)=1

    2. Relevant equations

    Assume [itex]y=e^{rt}[/itex]


    3. The attempt at a solution

    By the substitution I'm left with

    [itex]r^3-3r+2=0[/itex]

    which gives me the roots of -2 and 1.

    my question is a lot of times with this type of question I can get three roots and get a genral solution of

    [itex]y=c_1e^{rx}+c_2e^{rx}+c_3e^{rx}[/itex]

    this time my solution is

    [itex]y=c_1e^{-2x}+c_2e^{t}[/itex]
    therefore
    [itex]y'=c_1-2e^{-2x}+c_2e^{t}[/itex]
    and
    [itex]y''=c_14e^{-2x}+c_2e^{t}[/itex]

    because I've only got the two constants I solved for [itex]c_1=\frac{-1}{3}[/itex] and [itex]c_2=\frac{1}{3}[/itex]


    The problem is when I substitue back into the solutions for y,y',y''

    I'm getting

    initial conditions y(0)=0, y'(0)=1 which is good but I'm getting y''(0)=-1.


    can some please let me know where I'm going wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 19, 2010 #2

    rock.freak667

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    r3-3r+2=0 should give you three solutions.
     
  4. Aug 19, 2010 #3

    vela

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    If you have a third-order differential equation, you need three linearly independent solutions. You only have two because one of the roots is repeated. What do you usually do when you have a repeated root? You do the same thing here.
     
  5. Aug 19, 2010 #4
    So I have the roots of -2 ,1 and the repeated root 1.
    forgive my ignorance but I don't know what to do with a repeated root
     
  6. Aug 19, 2010 #5

    vela

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    Did you check your textbook? I'm sure it discusses what to do in that case. Here's a quick summary to jog your memory.

    http://www.sosmath.com/diffeq/second/constantcof/constantcof.html

    That page talks about second-order equations, but it applies to higher-order equations as well.
     
  7. Aug 19, 2010 #6
    Hi Vela,

    I see that pesky extra x sneaking in there when r1=r2.

    So my general solution should be

    [itex]y=c_1e^{-2x}+c_2e^{x}+c_2xe^{x}[/itex]
     
    Last edited: Aug 19, 2010
  8. Aug 19, 2010 #7

    vela

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    Yes, and now with three linearly independent solutions, you can get a consistent solution for the constants that will satisfy the initial conditions.
     
  9. Aug 22, 2010 #8
    Hi Guys,
    using the general solution I found the following.

    [itex]y=c_1e^{-2x}+c_2e^{x}+c_3xe^{x}[/itex]
    [itex]y'=c_1-2e^{-2x}+c_2e^{x}+c_3e^{x}+xe^{x}[/itex]
    [itex]y''=c_14e^{-2x}+c_2e^{x}c_32e^{x}+xe^{x}[/itex]


    Using the coefficients and initial conditions in a matrix.

    [itex]\[ \left( \begin{array}{cccc}
    1 & 1 & 1 & 0 \\
    -2 & 1 & 1 & 1\\
    4 & 1 & 2 & 1\end{array} \right)\][/itex]

    I solved for the row reduced echelon matrix.

    [itex]\[ \left( \begin{array}{cccc}
    1 & 0 & 0 & -\frac{1}{3} \\
    0 & 1 & 0 & -\frac{5}{3}\\
    0 & 0 & 1 & 2\end{array} \right)\][/itex]


    which gives the solution of

    [itex]y=-\frac{1}{3}e^{-2x}+-\frac{5}{3}e^{x}+2xe^{x}[/itex]

    which satisfies

    [itex]y(0)=0[/itex]
    [itex]y'(0)=1[/itex]
    [itex]y''(0)=1[/itex]


    thanks for all your help guys.

    It was a pretty challenging problem!
     
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