1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Undetermined coefficients

  1. Jun 22, 2005 #1
    Hello,
    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

    where
    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

    saltydog

    User Avatar
    Science Advisor
    Homework Helper

    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:

    [tex](D-1)(D-3)y=0[/tex]

    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:

    [tex](D-1)(D-3)(D^2+2D-3)y=0[/tex]

    Are you following this?

    So the solution to this is:

    [tex]y_c(x)=C_1e^x+Axe^x+C_3e^{-3x}+Be^{3x}[/tex]

    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

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

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

    Daniel.
     
  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.

    ARRRGGGH @ DE!

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Undetermined coefficients
  1. Coefficient of drag (Replies: 6)

Loading...