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Finding a particular solution of a differential equation

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the particular solution of this differential equation:
    y`` −3y` −10y=10t2+16t−19


    2. Relevant equations



    3. The attempt at a solution
    I'm not really sure what the roots look like for 10t^2 + 16t - 19. I thought t had roots (0,0). Does that mean t^2 has roots (0,0,0)? And -19 has no roots? So the roots of the entire right hand side is (0, 0, 0, 0, 0)?

    And yp = at + bt^2 + ct^3 + dt^4 + et^5?
     
  2. jcsd
  3. Oct 24, 2009 #2

    Mark44

    Staff: Mentor

    You don't need to factor the right-hand side or find roots. All you need is a particular solution that produces what's on the right side of your diff. equation. You do need to check the roots of the characteristic equation, though, because they can affect what you choose for your particular solution. The presence of the constant and two powers of t on the right side combined with one or more factors of r in the characteristic equation would change things.

    In your problem, though, the roots of the characteristic equation are 5 and -2, so the solutions to the homogeneous problem aren't going to affect the particular solution.

    For your problem, try yp = At2 + Bt + C. This will work.

     
  4. Oct 24, 2009 #3
    Hmm... but how do you know At^2 + Bt + C will work? I can't just look at it and see that that's going to give me something that matches the left hand side always, can I? There's no method you can follow to find yp?

    Or what about if you had something like y`` + 16y = 8sin(4t)? Then the roots overlap, so how can you find something that works?
     
    Last edited: Oct 24, 2009
  5. Oct 24, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You should be able to! Yes, there is a "method you can follow". I'm surprised you haven't learned it.

    First, the "method of undetermined coefficients" only work if the right hand side consists of the kind of functions that normally satisfy homogeneous linear equations with constant coefficients. Those are
    a) exponentials
    b) sine and cosine
    c) polynomials and
    d) sums and products of such things.

    There is one caveat: if such a function already satisfies the homogeneous equation you must multiply by the independent variable.

    In this problem the right side is a polynomial. Since neither of the roots of the characteristic equation (r2- 3r- 10= 0) is 0, a polynomial cannot satisfy the homogenous equation so you try yp= At^2+ Bt+ C.

    (I have no idea what you mean by "I thought t had roots (0,0). Does that mean t^2 has roots (0,0,0)? And -19 has no roots? So the roots of the entire right hand side is (0, 0, 0, 0, 0)?" t is a variable, not an equation, and does not have roots. If you mean roots of t2+ 16t- 19= 0, that is a quadratic equation and has two roots which have nothing to do with "t2= 0", "t= 0", and "-19= 0"!

    For y"+ 16y= 8 sin(4t), the roots of the characteristic equation are 4i and -4i which means that sin(4t) and cos(4t) are solutions to the homogeneous equation. As I said above, you try multiplying by t: yp= At sin(4t)+ Bt cos(4t). (In general, if you have "sin(at)" or "cos(at)" you have to try both. In this case, since there are only"even" derivatives, you will find that A= 0 and you really only need Bt cos(4t).
     
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