Variation of parameters

  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. HallsofIvy

    HallsofIvy 40,932
    Staff Emeritus
    Science Advisor

    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. vela

    vela 12,779
    Staff Emeritus
    Science Advisor
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    Education Advisor

    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. 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. HallsofIvy

    HallsofIvy 40,932
    Staff Emeritus
    Science Advisor

    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. 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. 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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