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Help with variation of constants

  1. May 29, 2012 #1
    1. The problem statement, all variables and given/known data
    solve the following differential equation:
    t4x'' - 4t3t' + 6t2x = - 12t - 20

    2. Relevant equations
    substitution x(t) = tn
    3. The attempt at a solution
    this is a Euler equation with the following general solution: x(t) = c1t2 + c2t3 worked out using the above substitution.

    The particular solution should be obtainable through variation of constants but I just get a nonsense result:

    The wronksian = W = 3c1c2t4 - 2c1c2t4 = c1c2t4

    therefore:

    [itex]x(t) = - x_{1} \int \frac{x_{2}b(t)}{W} dt + x_{2} \int \frac{x_{1}b(t)}{W}dt = x_{1} \int \frac{c_{2}t^{3}(12t + 20)}{c_{1}c_{2}t^{4}}dt - x_{2} \int \frac{c_{1}t^{2}(12t + 20)}{c_{1}c_{2}t^{4}} dt [/itex]
    [itex]x(t) = \frac{x_{1}}{c_{1}} \int (12 + \frac{20}{t})dt - \frac{x_{2}}{c_{2}} \int (\frac{12}{t} + \frac{20}{t^{2}}) dt[/itex]

    the integration is trivial but definitely isn't a particular solution!
     
    Last edited: May 29, 2012
  2. jcsd
  3. May 29, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Using "variation of parameters" (I would not call it "variation of constants") we do NOT include the constants- they become the "parameters". Knowing that [itex]t^2[/itex] and [itex]t^3[/itex] are solutions to the associated homogeneous equation, we look for solutions for the entire equation of the form [itex]x(t)= t^2u(t)+ t^3v(t)[/itex] where we have replaced the constants with the "parameters" u and v. There are many such solutions- given any solution, we could find u and v to work.

    Differentiating, we have [itex]x'= 2tu+ t^2u'+ 3t^2v+ t^3v'[/itex]. Because there are many such solutions we "narrow the search" and simplify the calculations, by requiring that [itex]t^2u'+ t^3v'= 0[/itex]. That leaves [itex]x'= 2tu+ 3t^2v[/itex]. Differentiating again, [itex]x''= 2u+ 2tu'+ 6tv+ 3t^2v'[/itex].

    Putting those into the original equation,
    [tex]t^4x''- 4t^3x'+ 6t^2x= 2t^4u+ 2t^5u'+ 6t^5v+ 3t^6v'- 8t^4u- 12t^5v+ 6t^4u+ 6t^5v= -12t- 20[/tex]
    [itex]2t^4u- 9t^4u+ 6t^4u= 0[/itex] and [itex]6t^5v- 12t^5v+ 5t^6v= 0[/itex] so the equation reduces to
    [tex]2t^5u'+ 3t^6v'= -12t- 20[/tex]
    That, together with [itex]t^2u'+ t^3v'= 0[/itex] gives us two linear equations to solve for u' and v'. (That's where the Wronskian comes in.)

    If we multiply the second equation by [itex]2t^3[/itex], and subtract from the first equation, we get [itex]t^6v'= -12t- 20[/itex] or [itex]v'= -12t^{-5}- 20t^{-6}[/itex]. If, instead, we multiply the second equation by [itex]3t^3[/itex], and subtract from the first equation, we get [itex]-t^5u'= -12t- 20[/itex] or [itex]u'= 12t^{-4}+ 20t^{-5}[/itex].

    Integrating those will give u and v to put into the original form.
     
  4. May 29, 2012 #3
    Thanks this is quite different from how it was explained to us. BTW I study in German and here it is called "Variation der Konstanten" but I see your point
     
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