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ODE with variable coefficient

  1. Nov 22, 2011 #1
    i want to solve the following differential equation:

    y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = M*exp(-N*x)
    A,B,C,M,N are constants.

    -is there any solution of the above equation (except series solution)?
    -is there any proper substitution that can turn the variable coefficient into constant?
    -can i use method of undetermined coefficient to obtain particular integral?
  2. jcsd
  3. Nov 22, 2011 #2
    I'd try using laplace transforms to see if it has any analytical non-series solution. No idea if you can use undetermined coefficients, but I can't imagine why you'd want to. A transform looks much simpler, especially with the e term in there.
  4. Nov 23, 2011 #3
    thanks for the reply.......

    for the homogeneous equation:
    y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = 0

    if the substitution z =exp(-C*A*x) is used,it becomes something like the following one( if i am not wrong):

    z*y''(z) + (1 + 1/C)*y'(z) - B/(C*A)^2*y(z) = 0

    which has a solution containing bessel function. But how can i obtain the particular integral (without containing any integral form) for :

    y''(x) - A*y'(x) - B*exp(-C*A*x)*y(x) = M*exp(-N*x)

    i can use method of variation of parameter for particular integral, but the wronskian becomes complicated (i have to leave the solution in integral form).
    therefore i was thinking about method of undetermined coefficient to obtain the particular integral.

    i need the general solution without keeping any integral form in the solution.
    please help me...............
    many many thanks in advance...................................
    Last edited: Nov 23, 2011
  5. Nov 23, 2011 #4


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    I don't think that will work well. I don't remember Laplace transforms being very good for ODE's with variable coefficients. In particular, the [itex]\exp(-c x)y(x)[/itex] term will give a [itex]Y(s+c)[/itex] term, which means the Laplace-domain equation for the Laplace transform variable [itex]Y(s)[/itex] is a functional equation instead of an algebraic equation.

    You may have to leave the solution in integral form. There's not necessarily a nice, closed form solution. Do you know what a Green's function is? Consider a modified form of your equation:

    [tex]\mathcal L G(x;\xi) = \delta(x-\xi),[/tex]

    where [itex]\delta(x-\xi)[/itex] is the dirac delta function and [itex]\mathcal L[/itex] is the differential operator,

    [tex]\mathcal L = \frac{d^2}{dx^2} -A \frac{d}{dx} - B\exp(-ACx)[/tex]

    If you solve that equation for [itex]G(x;\xi)[/itex], then to the solution for your equation with the [itex]M\exp(-Nx)[/itex] is just

    [tex]y(x) = M\int d\xi G(x;\xi)\exp(-N\xi),[/tex]

    where the integral is taken over your range of x. In a one-dimensional problem like yours, this is very similar to the variations of parameters method, I think, but this one generalizes to higher dimensions. Your solution may still be in terms of an integral, and there may be no way around that.
    Last edited: Nov 23, 2011
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