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D'Alembert's Reduction of Order Method

  1. Oct 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Use the reduction of order metho to find a second linearly independent solution. What is the general solution of the differential equation?
    [itex] y'' - y = 0 [/itex]
    [itex] y_1(x) = e^x [/itex]

    2. Relevant equations
    Reduction of order formula

    3. The attempt at a solution
    First, I set:
    [itex] y = ve^x [/itex]
    [itex] y' = ve^x + e^{x}v' [/itex]
    [itex] y'' = ve^x + 2e^{x}v' + e^{x}v'' [/itex]
    [itex] (ve^x + 2e^{x}v' + e^{x}v'') - (ve^x) = 0 [/itex]
    [itex] 2e^{x}v' + e^{x}v'' = 0 [/itex]
    [itex] e^{x}v'' + 2e^{x}v' = 0 [/itex]
    [itex] v'' + e^{x}v' = 0 [/itex]
    Then I made a substitution:
    [itex] w = v' [/itex]
    So the equation becomes:
    [itex] w' + e^{x}w = 0 [/itex]
    At this point, I tried to find an integration factor. However, the integrating factor I obtained is a bit unusual, which leads me to believe that I have made a mistake somewhere. This is the integrating factor I obtained:
    [itex] p(x) = e^x [/itex]
    [itex] u(x) = e^{\int e^x} = e^{e^x} [/itex]
    At this point, due to the odd integrating factor, I am not sure what I have done wrong or how to continue the problem.
     
  2. jcsd
  3. Oct 13, 2014 #2

    BvU

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    ##e^{x}v'' + 2e^{x}v' = 0## does not lead to ##v'' + e^{x}v' = 0## !
     
  4. Oct 13, 2014 #3

    vela

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    I hope this mistake was a total brainfart. If not, you really need to get a handle on basic algebra. It'll make learning this stuff a lot easier when you're not wasting time tracking down avoidable errors like this one.
     
  5. Oct 13, 2014 #4
    It totally was, I promise my algebra is not that bad :p ...anyway thanks you both for the help
     
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