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Homework Help: Variation of Parameters

  1. Apr 30, 2008 #1
    [SOLVED] Variation of Parameters

    1. The problem statement, all variables and given/known data

    y^(4)-6y^(3)=-5sinx


    3. The attempt at a solution
    I factored this at x^3(x-6)=0
    so my r values are 0,6
    also using for y(p) Dcosx + Esinx

    y=Ae^0 + Be^6x + Dcosx + Esinx ???
    y' =6Be^6x -Dsinx + Ecosx
    y'' =36Be^6x-Dcosx - Esinx
    y''' =216Be^6x + Dsinx - Ecosx
    y'''' =1296Be^6x +Dcosx + Esinx

    (1296Be^6x +Dcosx + Esinx) - 6(216Be^6x + Dsinx - Ecosx)=-5sinx

    is this the correct way to set up this problem?
     
  2. jcsd
  3. Apr 30, 2008 #2

    HallsofIvy

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    This is a fourth order equation. The solution space of the homogeneous equation is a four[\b] dimensional vector space. You need 4 independent solutions to it and you only have 2. What do you do when you have a double or triple root to your characteristic equation?

    Obviously the "1295Be6x" and -6(216Be6x)" terms will cancel out. When solving for a "particular solution", there is no need to carry the solutions to the homogeneous equation along. Just use y= A sin(x)+ B cos(x) and then, after you have found a particular solution, add it to the homogeneous solution.
     
  4. Apr 30, 2008 #3
    you put in a faxtor of x and x^2 ?
    y(c)= Ae^0 + Be^6x + Dxe^0 + Ex^2e^0
    I used 0 because that would be consitent with the x^3 term-

    y(c)=A + Be^6x + Dx + Ex^2

    y(p)=Asin(x) + Bcos(x)
    y'(p)=Acosx - Bsinx
    y"(p)=-Asinx - Bcosx
    y'''(p)=-Acosx + Bsinx
    y""(p)=Asinx + Bcosx

    (Asinx + Bcosx) - 6(-Acosx + Bsinx) = -5sinx

    how can you solve this??
    A(sinx + 6cosx) + B(cosx-6sinx)=-5sinx
     
  5. Apr 30, 2008 #4

    HallsofIvy

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    Yes, that is correct for the homogeous equation.

    Better is to "combine like functions": (A- 6B)sinx+ (6A+ B)cos x= -5sin x

    Now, since sin x and cos x are "independent" functions, The coefficients of like functions, on opposite sides of the equation, must be equal. That gives you two equations to solve for A and B.
     
  6. Apr 30, 2008 #5
    two equations would then be:
    1.(A-6B)sinx=-5sinx
    2.(6A+B)cosx=0

    1.(A-6B)sinx=sinx
    A-6B=-5
    -B/6 - 6B=-5
    B=-30/37

    6Acosx+Bcosx=0
    6Acosx=-Bcosx
    A=-B/6

    A=-(-37/30)/6= 5/37

    So general solution:
    y=A + Be^6x + Dx + Ex^2 + 5/37sinx -30/37cosx

    There seems like there is a different way to solve this??
     
  7. May 1, 2008 #6
    -5/37sinx +30/37cosx
     
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