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Variation of parameters

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve for general solution with variation of parameter

    [tex]y'''(x) - y'(x) = x[/tex]

    3. The attempt at a solution

    I initially looked at [tex]y'''(x) - y'(x) = x[/tex] only and I foudn my answer to be

    [tex]y(x) = C_1e^{x} + C_2e^{-x} + 1 - x[/tex]

    Now i looked through my book and it says it works for ay'' + by' + c = f(t) only (second order).

    So I "integrated" [tex]y'''(x) - y'(x) = x[/tex]

    And I got [tex]y''(x) - y(x) = \frac{x^2}{2} + C[/tex], solving I got

    [tex]y(x) = C_1e^{x} + C_2e^{-x} + C_3 - \frac{x^2}{2}[/tex]

    Using the computer, it gave me

    http://www.wolframalpha.com/input/?i=Solve[y%27%27%27+-+y%27+%3D+x]

    Why does computer have negative sign??
     
  2. jcsd
  3. Aug 15, 2011 #2

    HallsofIvy

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    I have no idea what you mean when you say you "looked at y'''- y'= x only". That is the entire problem! Do you mean the associated homogeneous equation, y'''- y'= 0? That has characteristic equation [itex]r^3- r= r(r^2- 1)= r(r- 1)(r+ 1)= 0[/itex] so the general solution to the associated homogeneous equation is [itex]y_h(x)= C_1e^x+ C_2e^{-x}+ C_3[/itex]

    Now, you have title this "variation of parameters". To use that method for this differential equation ("undetermined coefficients" would be simpler but I presume this is for practice of variation of parameters specifically), look for a solution of the form
    [tex]y(x)= u(x)e^x+ v(x)e^{-x}+ w(x)[/tex]
    That is, we treat those constants (parameters) as variables- that is the reason for the name "variation of parameters". Differentiating, [itex]y'= u'e^x+ ue^x+ v'e^{-x}- ve^{-x}+ w'[/itex]. There will be many different possible functions that will work- we limit our search by requiring that [itex]u'e^x+ v'e^x+ w'= 0[/itex]

    That means that we have [itex]y'= ue^x- ve^{-x}[/itex]. Differentiating again, [itex]y''= u'e^x+ ue^x- v'e^{-x}+ ve^{-x}[/itex]. Once again, we "limit our search" (and simplify the equation) by requiring that [itex]u'e^x- v'e^{-x}= 0[/itex].

    That gives, now, [itex]y''= ue^{x}+ ve^{-x}[/itex] and, differentiating one more time, that [itex]y'''= u'e^{x}+ ue^{x}+ v'e^{-x}- ve^{-x}[/itex]. Putting that and the formula for y' into the equation,
    [tex]u'e^x+ ue^x+ v'e^{-x}- ve^{-x}- (ue^x- ve^{-x})= u'e^x+ v'e^{-x}= x[/tex]

    You see what has happened? We have no derivative of u, v, and w higher than first because of our "requirements" and we do not have u, v, and w themselves because they satify the homogenous equation and cancel. We have, instead the three equations
    [tex]u'e^x+ v'e^x+ w'= 0[/tex]
    [tex]u'e^x- v'e^{-x}= 0[/tex]
    and
    [tex]u'e^x+ v'e^{-x}= x[/tex]
    which we can treat as three linear equations in u', v', and w'. Solve for those and integrate to find u, v, and w.
     
  4. Aug 15, 2011 #3

    vela

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    If you're asking about the negative sign in front of c2e-x, it doesn't matter because the sign could be pulled into the arbitrary constant c2.
     
  5. Aug 16, 2011 #4
    HallsofIvy, my book did the same thing, they imposed on the requirement that

    [tex]u'e^x+ v'e^x+ w'= 0[/tex]

    Why? Was I right though initially when I 'integrated' [tex]y''' - y' = x[/tex]
     
  6. Aug 16, 2011 #5

    HallsofIvy

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    Yes, because there is no "y" in the original equation you can integrate [itex]y'''- y'= x[/itex] to get [itex]y''- y= (1/2)x^2+ C_1[/itex]. It will give you the same result.

    However, you titled this "variation of parameters" which will be of about the same difficulty for either equation.
     
  7. Aug 16, 2011 #6
    But why this condition

    [tex]u'e^x+ v'e^x+ w'= 0[/tex]

    My book threw this (similar, except they used y_1 and y_2 for general derivation) out of nowhere and I am left blank minded.

    How do they know it's going to work out? How did you know (don't say intuition...) the condition u'e^x+ v'e^x+ w'= 0 will enable to you solve the problem?
     
  8. Aug 16, 2011 #7
    Oh okay I see, because if "y" was there, then we can't assume what y really is? Is that why?

    Why does it make it equally correct that I "integrated" through [tex]y''' - y' = x[/tex] because y wasn't there? Why is it okay then that I don't need to know y', yet I would get the right answer?

    By the way, why do you use [/itex]? What does the i do?
     
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