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1st order, nonhomogeneous, linear DE - particular solution

  1. Oct 11, 2013 #1
    I know nothing about DEs, so this may be a silly question.

    I'm given some time varying [itex](x_t)_t[/itex] and a constant [itex]r[/itex], and I want to solve the equation [itex]u_t = rx_t + \dot u_t[/itex] for [itex]u[/itex].

    What I know so far is that (solving the homogeneous equation) if [itex]\bar u[/itex] is some particular solution, then any [itex]u[/itex] is a solution iff it takes the form [itex]u_t=\bar u_t + \alpha e^{rt}[/itex] for some constant [itex]\alpha[/itex].

    I'm wondering whether there's a brute force way of finding some [itex]\bar u[/itex]. Anywhere I've looked suggests the "method of undetermined coefficients", but I know it's not useful in my setting. Is there a formula I can blindly apply to get a particular solution? I'm happy to assume [itex]x[/itex] as well behaved as needed, and I'm happy to have my formula be some horrible definite integral I can't compute.
     
  2. jcsd
  3. Oct 11, 2013 #2

    fzero

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    The general method for finding particular solutions is called variation of parameters, though in this case, we can use an integrating factor to save quite a bit of work. Note that

    $$ e^t \frac{d}{dt} \left( e^{-t} u\right) = \dot{u} - u,$$

    so that the integrating factor ##e^{-t}## will lead to a particular solution in the form of an indefinite integral involving ##x_t##.
     
  4. Oct 12, 2013 #3

    HallsofIvy

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    So the subscript t just means that u and x are functions of t?

    No. That would be the case if the "r" were multiplying [itex]u_t[/itex], not x. The general solution to the equation [itex]u'(t)+ rx(t)= u(t)[/itex] is [itex]\bar u(t)+ \alpha e^t[/itex].

    I'm not sure you would call it "brute force" but I would use "variation of parameters". Knowing that [itex]u(t)= Ce^t[/itex], for C any constant, satisfies [itex]u'- u= 0[/itex], to solve the problem [itex]u'- u= -rx(t)[/itex], look for a solution of the form [itex]u(t)= v(t)e^t[/itex]. Then [itex]u'(t)= v'(t)e^t+ ve^t[/itex]. so [itex]u'- u= v'e^t+ ve^t- ve^t= v'e^t= -rx[/itex] then [itex]v'= -re^{-t}x(t)[/itex]. Then [itex]v(t)= -r\int_0^t e^{-\tau}x(\tau)d\tau+ C[/itex] so that
    [tex]u(t)= -re^t\int_0^t e^{-\tau}x(\tau)d\tau+ Ce^t[/tex]
     
  5. Oct 13, 2013 #4
    Whoops, I'd meant to write [itex]ru_t = rx_t + \dot u_t[/itex], but anyway...

    Thank you very much, folks! You have been extremely helpful. :)
     
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