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Solve for the solution of the differential equation

  1. Oct 21, 2016 #1
    1. The problem statement, all variables and given/known data
    Solve for the solution of the differential equation and use the method of variation of parameters.
    x`` - x = (e^t) + t

    2. Relevant equations

    W= (y2`y1)-(y2y1`)
    v1 = integral of ( g(t) (y1) ) / W
    v2 = integral of ( g(t) (y2) ) / W

    3. The attempt at a solution

    yc= c1 e^t + c2 e^-t
    yp = v1 e^t + v2 e^-t

    W= -2
    v1 = integral of ((e^-t)(e^t + t)) / -2 = (1/2) (t - te^-t - e^-t)
    v2 = integral of ((e^2t)(e^t + t)) / -2 = (1/2) (1/2 e^2t + te^t - e^t)

    v1y1 = 1/2 te^t - 1/2 t - 1/2
    v2y2 = 1/4 e^t - 1/2t + 1/2

    adding these, yp = -t + 1/2 te^t + 1/4 e^t
    yc + yp = (c1 e^t + c2 e^-t ) + (-t + 1/2 te^t + 1/4 e^t)
    but the answer is (c1 e^t + c2 e^-t ) + (-t + 1/2 te^t)

    I got an extra term. Where did I go wrong?
    Thank you!
     
  2. jcsd
  3. Oct 21, 2016 #2

    Simon Bridge

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    I do not see your reasoning - it looks like you have attempted to apply a method of solution by rote instead of understanding it.

    Guessing - yc is supposed to be the general solution to the homogeneous part of the DE. In which case, yp, as you have written it, is a specific solution to the homogeneous part - but it is supposed to be a particular solution to the entire DE.
    However, yc is not a solution to the homogeneous part.
    On top of all that... the DE does not contain the variable "y" that you keep using.

    But I am only guessing about your reasoning here.
     
    Last edited: Oct 21, 2016
  4. Oct 21, 2016 #3
    I got the formula eq0027MP.gif from http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx to solve for the particular solution of the DE.
    In this problem, g(t) = e^t + t, y1 = e^t and y2 = e^-t .
    The Wronskian of these two functions:
    | e^t e^-t |
    | e^t -e^-t |

    = (e^t)(-e^-t) - (e^t)(e^-t) = -1-1 = -2
    Then plugging in the values to the equation, I got 1/2 (t - te^-t - e^-t) for v1 (or u1 in the formula) and -1/2 (1/2 e^2t + te^t - e^t) for v2 (u2 in the formula).
    Multiplying these by y1 and y2 respectively, I got these:
    v1y1 = 1/2 te^t - 1/2 t - 1/2
    v2y2 = 1/4 e^t - 1/2t + 1/2

    So from what you have said, I have solved for the specific solution to the homogeneous part, not the DE.
    And you have also said that yc is not the solution for the homogeneous part. I do not understand.

    m^2 - 1 = 0 (is this not to solve the roots?)
    m= 1, - 1
    leading to yc = c1 e^1t + c2 e^-1t

    So what should I start with?
    Thank you for your reply.
     
  5. Oct 21, 2016 #4

    Simon Bridge

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    Well, first of all, the DE is ##\ddot x + x = t+e^t## ... there is no ##y## in that equation.
    But you mean: ##x_c = Ae^t + Be^{-t}## ... where A and B are arbitrary constants
    ... is the general solution to: ##\ddot x + x = 0## ... that about right?
    OK then: prove it.

    [edit] excuse me that's a bit flip.
    If you want to understand how x_c is not a solution to the homogeneous part, just substitute it into the homogeneous part and see what happens.
     
  6. Oct 21, 2016 #5
    trial solution: x= e^mt
    x` = me^mt
    x`` = m^2 (e^mt)

    x`` - x = 0
    m^2 (e^mt) - e^mt = 0 (characteristic equation)
    dividing both sides by e^mt,
    m^2 = 1
    m = 1, -1

    for distinct real roots,
    xc = A (e^ (m1)t + B (e^ (m2)t) = A e^t + B e^-t

    ?
     
  7. Oct 21, 2016 #6

    Simon Bridge

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    Oh I misread - I thought it said ##\ddot x + x = 0## but that should be a minus sign.
    So the second derivative should be the same as the function. Well done.

    ##x_c = Ae^t + Be^{-t}## ... you proved your case: well done :)
    Now you need a particular solution to ##\ddot x - x = t + e^t## ... any solution will do...

    Notice that if you now write ##x_p = Ce^t + De^{-t}## you are just rewriting the original ##x_c##, just using different labels for the "arbitrary constants"?

    Since any solution will do, don't sweat the whole wronskian thing - just guess:
    You need x_p to end up with a second derivative that has a "t+"something in it. t^3+something will do that...
    You could try ##x_p=at^3+be^t## ... is that close to being a solution to ##\ddot x - x = t+e^t##?
    How does it need to change to make it a solution?
     
  8. Oct 21, 2016 #7
    Any solution will do?
    ooohhh
    I have tried substituting what I got from my solution into the DE.

    xp = 1/2 te^t - t - 1/4 e^t
    xp` = 1/2 (te^t + e^t) - 1 - 1/4 e^t
    xp`` = 1/2 (te^t + 2e^t) - 1/4 e^t

    1/2 (te^t + 2e^t) - 1/4 e^t - ( 1/2 te^t - t - 1/4 e^t) = (1/2 te^t + 3/4 e^t) - (1/2 te^t - t - 1/4 e^t) = t + (4/4) e^t = t + e^t which is equal to the right hand side of the given DE.

    Did I understand it right?
     
    Last edited: Oct 21, 2016
  9. Oct 21, 2016 #8

    Simon Bridge

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    That checks out. Note: you originally (post #1) had:
    Now you have
    ... actually, since any particular solution will work, there is no special reason that your answer will agree in detail to theirs.

    You should probably check that ##-t+te^t## (theirs) also works as a particular solution ...
    BTW: also see: https://www.physicsforums.com/help/latexhelp/ for help writing equations.
     
  10. Oct 22, 2016 #9
    Thank you very much!
     
  11. Oct 23, 2016 #10

    Simon Bridge

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    No worries :)
     
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