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Y′′=−20⋅4x^3, Second order linear ordinary DE

  1. Mar 31, 2015 #1
    1. The problem statement, all variables and given/known data
    y′′=−20⋅4x^3

    2. Relevant equations
    Undetermined coefficients method

    3. The attempt at a solution
    so at first, solving the associated homogeneous equation I find the fundamental set of solutions to be: y1=1 and y2=x.

    I know that these are correct. Now for the part that confuses me.

    I'm trying to find Yp, the particular solution.

    g(x) = -80x^3 so Yp is of the form Ax^3 + Bx^2 + Cx + D

    but when I solve this I get Yp=0, that is not correct.

    It seems to me that Ax^3+Bx^2+Cx+D is Linearly independent from y1 and y2 so why doesn't this work? The book lists the solution as Yp=-4x^5.
     
  2. jcsd
  3. Mar 31, 2015 #2

    Mark44

    Staff: Mentor

    This is much simpler than you're making it out to be.
    Just integrate both sides twice.
     
  4. Mar 31, 2015 #3
    Ok fair enough, it could be done like that. But the point of this chapter is to learn this method. The question specifically asks for the fundamental set of solutions, Yp and the general solution. At any rate, I'm stuck on another similar problem which I can't integrate. I'm sure I can solve it if I can figure out this simpler case first.
     
  5. Mar 31, 2015 #4

    Mark44

    Staff: Mentor

    Since y'' = -80x3, a reasonable choice for yp would be ##y_p = Ax^5 + Bx^4 + Cx^3 + Dx^2 + c_1x + c_2##. Differentiate twice and compare coefficients to what you have in the differential equation. That seems like a pointless technique, though, IMO, at least for this problem.

    For the nonhomogeneous equation, y'' = 0, the fundamental set of solutions is as you say, {1, x}. Note that in my particular solution above, the complementary solutions are included with coefficients of c1 and c2.

    BTW, is there some reason you wrote -20*4x3 instead of -80x3?
     
  6. Mar 31, 2015 #5
    Ok that works but how did you pick your Yp? I know Yp has to be linearly independent from the fundamental set of solutions but it seems to me like Ax^3+Bx^2+Cx+D already is, so why doesn't that satisfy the conditions? I feel like I'm missing something very fundamental here.
     
  7. Mar 31, 2015 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    If yp = Ax^3+Bx^2+Cx+D, what happens after you differentiate yp twice w.r.t x? Do you get a cubic in x?
     
  8. Mar 31, 2015 #7
    I get 6Ax + 2B
     
  9. Mar 31, 2015 #8
    So essentially what you're telling me that if g(x) is a polynomial of degree n, then for y'' the 2nd derivative of yp has the be a polynomial of degree n?
     
  10. Mar 31, 2015 #9

    Mark44

    Staff: Mentor

    Something like that.

    There is the Method of Annihilators, which is applicable in this problem.
    Homogeneous equation: y'' = 0
    The characteristic equation is r2 = 0, which has r = 0 as a repeated root. The complementary function (solution of the homogeneous problem) is yc = c1 + c2x.

    The nonhomogeneous problem (your problem) is y'' = -80x3. This could also be written as D2y = -80x3, where Dy means dy/dx or y', and D2y means ##\frac{d^2y}{dx^2}## or y''.

    We note that the D4 operator "annihilates" the right side, as D4(x3) = 0. IOW, the fourth derivative of x3 = 0. Because of this fact, we apply the D4 operator to both sides of the nonhomogeneous problem to get
    D4(D2y) = D4(-80x3) = 0
    Or, D6y = 0, which is a homogeneous equation.

    The characteristic equation is r6 = 0, so that r = 0 is a repeated root. The general solution of this equation is ##y = c_1 + c_2x + Ax^2 + Bx^3 + Cx^4 + Dx^5##. I wrote the first two differently, as they were the solution of y'' = 0, and I want to keep them separate. The particular solution is the portion with the coefficients A, B, C, and D.
     
  11. Mar 31, 2015 #10
    Perfect, that was exactly the kind of explanation that I was looking for. Thanks!
     
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