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Undetermined coefficients

  1. Jun 22, 2005 #1
    I have this DE:

    y'' + 2y' - 3y = 8ex - 12e3x

    when I find homogeneous solution I get

    yh = c1ex + c2e-3x;

    so now to find the particular solution by method of undetermined coefficients, do I set y to smth like this:

    y = y1 + y2

    y1 = Axex,
    y2 = A3x ?

    since one of the solutions to auxiliary equation appears on the RHS of the DE and the other does not?
    I don't need the full solution, just confirmation/correction of this part.

    Thanks much!

    EDIT: if I take the fact that if y1 + y2 is a solution, then y1 is a solution and y2 is a solution. I guess I answered my own question. :frown:
    Last edited: Jun 22, 2005
  2. jcsd
  3. Jun 22, 2005 #2


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    Well EvLer, it's nice if you first recognize that the RHS is a particular solution to a homogeneous equation with roots 1, 3 so that equation would be:


    You familiar with those differential operations right?

    Thus, applying that operator to both sides of your equation will make the RHS 0 right (it's a solution to that homogeneous operator which is set to zero). So applying it we get:


    Are you following this?

    So the solution to this is:


    Now, take:

    [tex]y_p(x)=Axe^x+Be^{3x}[/tex] and back-substitute into your original equation, equate coefficients to find A and B.
  4. Jun 22, 2005 #3


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    Why don't you apply Lagrange's method of variation of constants...?

  5. Jun 22, 2005 #4
    Yeah, thanks, I got it 5 minutes after posting.

    @saltydog: it's the same technique, only we follow it more step-by-step, where we solve homogeneous equation first and then based on the solution of non-hom. and hom. eq. we pick appropriate form of particular solution, and then do differentiation and plug it all in.


    ps: don't know what "Lagrange's method of variation of constants" is but thanks, i'll look that up.
  6. Jun 22, 2005 #5
    Lagrange's method of variation of constants is also known as variation of parameters. It is based on the fact that if [itex]y_1[/itex] and [itex]y_2[/itex] are solutions to an homogeneous ODE, then so is [itex]c_1 y_1[/itex] and [itex]c_2 y_2[/itex] by principle of superposition. But we aim to find the particular solution of the form [itex]u(t) y_1(t)[/itex] and [itex]v(t) y_2(t)[/itex]

    The bottom line is that the particular solution is

    [tex]Y_p(t) = -y_1(t) \int \frac {y_2(t) g(t)}{W(y_1, y_2) (t)} dt ~+~ y_2(t) \int \frac {y_1(t) g(t)}{W(y_1, y_2)(t)}dt[/tex]
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