1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook