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Obtaining particular solution of second order linear DE from first

  1. Feb 18, 2012 #1

    K29

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    In cullen-zill chapter 6 equation 23 it says that

    [itex]y_{2}(x)=y_{1}(x)\int\frac{e^{-\int P(x)dx}}{y_{1}^{2}(x)}dx[/itex]
    is a solution of
    [itex]y''+P(x)y'+Q(x)y=0[/itex]
    whenever [itex]y_{1}(x)[/itex] is a known solution

    Where does this come from? I would like to be able to prove this or find a proof somewhere.

    My first thought is that since the general solution solution is [itex]y_{h}=C_{1}y_{1}+C_{2}y_{2}[/itex] then
    [itex]y_{2}=v(x)y_{1}[/itex] where v(x) is just a function of x.
    Thought maybe I could substitute that into the general form of the DE, but it doesn't seem to help much
     
  2. jcsd
  3. Feb 18, 2012 #2

    HallsofIvy

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    Staff Emeritus
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    Yes, that method, "variation of parameters", will work. With [itex]y_2= v(x)y_1[/itex], [itex]y_2'= v'(x)y_1+ v(x)y_1'[/itex] and [itex]y_2''= v''(x)y_1+ 2v'(x)y_1'+ v(x)y_1''[/itex]. Putting those into the equation,
    [tex]v''y_1+ 2v'y'+ vy_1''+ P(x)v'y_1+ P(x)vy_1'+ Q(x)vy_1= y_1v''+ 2v'y_1'+ P(x)v'y_1+ v(y_1''+ P(x)y_1'+ Q(x)y_1)= 0[/tex]

    Of course, that last term in parentheses is 0 so we have [itex]y_1v''+ (2y_1+ P(x)y_1)v'= 0[/itex]. Notice that there is no "v(x)" (undifferentiated) so if we let u= v', we have [itex]y_1u'+ (2y_1+ P(x)y_1)u= 0[/itex], a separable first order equation.
     
    Last edited: Feb 19, 2012
  4. Feb 19, 2012 #3

    K29

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    Solved. Thank you.
     
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