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Finding A Second Solution For A DE Via Reduction Of Order

  1. Oct 20, 2011 #1
    I'm given
    [tex]y''-4y'+4y=0[/tex]
    and there is a solution [tex]y_{1}=e^{2x}[/tex]
    Using this I need to find a second solution.

    Starting with the assumption:
    [tex]y_{2}=u(x)*y_{1}[/tex]
    Then:
    [tex]y=ue^{2x}[/tex]
    [tex]y'=2ue^{2x}+u'e^{2x}[/tex]
    [tex]y''=4ue^{2x}+2u'e^{2x}+u''e^{2x}+2u'e^{2x}[/tex]

    When I substitute back into the original equation, after doing the cancellation I wind up with:
    [tex]u''e^{2x}=0[/tex]
    Then using the following order reduction:
    [tex]w=u'[/tex]
    I get:
    [tex]w'e^{2x}=0[/tex]
    I'm not sure what to do next here. The example in my textbook had two terms, and they used an integrating factor. How can I go about solving this problem? The answer given in the book is:
    [tex]y_{2}=e^{2x}[/tex]
     
  2. jcsd
  3. Oct 20, 2011 #2

    lurflurf

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    so
    w' e2x=0
    reduces to
    w'=0
    which is easy to solve
    the answer given is wrong it should be
    y2=(a x+b)e2x
    where a and b are any numbers except a is not 0
     
  4. Oct 20, 2011 #3
    I actually found a silly algebra error, so there should be two terms. However the answer is even stranger. I'll talk to my prof about it, thanks for looking!
     
  5. Oct 21, 2011 #4

    lurflurf

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    You were right the first time
    w' e2x=0
    reduces to
    w'=0
    notice when y is a solution so is y'-2y or conversely when
    y'-2y=e2x
    y is a solution of
    y''-4y'+4=0
     
    Last edited: Oct 21, 2011
  6. Oct 21, 2011 #5
    I guess that does work. Thanks!

    EDIT: I'm not sure how though. You have that w'=0, but how do you then solve for y being e^2x?
     
    Last edited: Oct 21, 2011
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