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Math Help - transform out 1st order term from 2nd order linear ode

  1. #1
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    transform out 1st order term from 2nd order linear ode

    Hello,
    I am a PHD student in engineering with a strong background in ODEs and PDEs.
    I am working through a text called "Combustion Physics" by C. K. Law. On p.327, Law says:

    "It is, however, well known that in a second order linear ordinary differential equation, the first order differential can be transformed away."

    Such a technique is NOT well known to me and I cannot find anything like it in my undergrad ODEs text (Boyce and DiPrima). Unfortunately, Law doesn't provide a reference, nor does the example he provides explain the method. Can someone refer me to a book where I can learn more about this?
    Thanks!
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  2. #2
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    Re: transform out 1st order term from 2nd order linear ode

    Consider

    y'' +  p(x) y' + q(x) y = f(x)

    Let  y = a(x) u so

    y' = a u ' + a' u,
    y'' = au'' + 2 a' u' + a''u

    Substitute. This gives

    au'' + 2 a' u' + a''u + p\left(a u ' + a' u \right) + q a u = f

    Re-grouping gives

    au'' + \left(2a' + p a \right)u' + \left(a'' + pa' + qa \right)u = f

    Now pick a such that 2a' + pa = 0. This gives an ODE absent of the term u'.
    Last edited by Jester; May 13th 2012 at 07:37 AM.
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  3. #3
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    Re: transform out 1st order term from 2nd order linear ode

    Danny,

    Ah, a clever solution! What if, following your nomenclature, q(x) = 0 and in the end, I wish to have only a second order term? In other words, is there a method by which I can eliminate the first order differential from

    y" + p(x) y' = f(x)

    leaving only a second order differential and no u term?

    Thanks,
    T
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  4. #4
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    Re: transform out 1st order term from 2nd order linear ode

    Only for certain p's because what you want is

    au'' + \left(2a' + p a \right)u' + \left(a'' + pa' \right)u = f

    to become

    au''  = f

    meaning that both 2a' + pa = 0 and a'' + pa' = 0 must be satisfied. However, why go to all that trouble. If you have

    y'' + p(x) y' = f(x)

    then letting u = y' gives a linear ODE in u
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