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Using variation of parameters to derive a general solution?

  1. Feb 24, 2017 #1
    1. The problem statement, all variables and given/known data
    "By choosing the lower limit of integration in Eq. (28) in the text as the initial point ##t_0##, show that ##Y(t)## becomes

    ##Y(t)=\int_{t_0}^t(\frac{y_1(s)y_2(t)-y_t(t)y_2(s)}{y_1(s)y_2'(s)-y_1'(s)y_2(s)})g(s)ds##

    Show that ##Y(t)## is a solution of the initial value problem:

    ##L[y]=g(t)##
    ##y(t_0)=0##
    ##y'(t_0)=0##"

    2. Relevant equations
    Equation 28: ##Y(t)=-y_1(t)\int_{t_0}^t\frac{y_2(s)g(s)}{W(y_1,y_2)(s)}ds+y_2(t)\int_{t_0}^t\frac{y_1(s)g(s)}{W(y_1,y_2)(s)}ds##

    3. The attempt at a solution
    My main method of trying to solve this would be to differentiate ##Y(t)## twice and substituting in the derivatives into the general equation ##y''+p(t)y'+q(t)y=g(t)##. So...

    ##Y(t)=\int_{t_0}^t(\frac{y_1(s)y_2(t)-y_t(t)y_2(s)}{y_1(s)y_2'(s)-y_1'(s)y_2(s)})g(s)ds##
    ##\frac{d}{dt}Y(t)=(-y_1'(t)\int_{t_0}^t\frac{y_2(s)g(s)}{W(y_1,y_2)(s)}ds-y_1(t)\frac{y_2(t)g(t)}{W(y_1,y_2)(t)})+(y_2'(t)\int_{t_0}^t\frac{y_1(s)g(s)}{W(y_1,y_2)(t)}ds+y_2(t)\frac{y_1(t)g(t)}{W(y_1,y_2)(t)})##

    I feel like there should be an easier way to do this, because I don't really wish to calculate the second derivative of ##Y(t)##. Perhaps I should do something with the ##L[y]## operator? I don't exactly know how it works.
     
    Last edited: Feb 24, 2017
  2. jcsd
  3. Feb 24, 2017 #2

    LCKurtz

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    Think about Leibnitz rule:$$
    \frac d {dt}\int_{t_0}^t f(s,t)~ds = f(t,t) + \int_{t_0}^t f_t(s,t)ds$$Try applying that to$$
    Y(t)=\int_{t_0}^t(\frac{y_1(s)y_2(t)-y_1(t)y_2(s)}{y_1(s)y_2'(s)-y_1'(s)y_2(s)})g(s)ds$$Try applying that directly to your integral without taking anything out of the integral. I think you will see it works easier than what you did and you will notice immediately that part of your answer cancels out. Not that your work is incorrect but I think you will find it seems less messy and won't be so terrified about taking another derivative.
     
  4. Feb 24, 2017 #3
    Do you mean I take the partial derivative with respect to ##t##? Anyway, thanks. This simplified the problem very much.
     
    Last edited: Feb 25, 2017
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