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Math Help - Differential Equations and Variation of Parameters

  1. #1
    Senior Member tukeywilliams's Avatar
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    Differential Equations and Variation of Parameters

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    Grand Panjandrum
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    Quote Originally Posted by tukeywilliams View Post
    y'' - 4y' + 3y = e^-t + t.

    The general solution to this is the sum of the general solution to

    y'' - 4y' + 3y = 0,

    and a particular integral of the original equation.

    To find a particular integral look at a trial solution:

    y(t) = A e^{-t} + ax^2 + bx +c.

    Determine the coefficients by differentiating twice and substituting into
    the original equation.

    You will now have the general solution and you need only fit the initial conditions to find the solution you require.

    RonL
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  3. #3
    Senior Member tukeywilliams's Avatar
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    how would I do the second one? just guess and check?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by tukeywilliams View Post
    how would I do the second one? just guess and check?
    No, use exactly the same method as before, find the general solution of

    y^(4) = 0

    (which is the general cubic), and a particular solution of

    y^(4) = e^t,

    which we might as well take to be y=e^t. So the general solution is:

    y(t) = e^t + ax^3 + bx^2 + cx +d

    Now fit the initial conditions.

    RonL
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  5. #5
    Senior Member tukeywilliams's Avatar
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    Is it possible to solve the last one by hand?
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  6. #6
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    Hello, tukeywilliams!

    Here's the second one . . .


    y'''' = e^t, . y(0) = 1, .y'(0) = y''(0) = y'''(0) = 0.

    We have: .y'''' .= .e^
    t

    . . Integrate: .y''' .= .e^
    t + C1

    . . Since y'''(0) = 0, we have: .e^
    0 + C1 .= .0 . . C1 = -1

    . . Hence: .y'''(t) .= .e^
    t - 1


    Integrate: .y''(t) .= .e^
    t - t + C2

    . . Since y''(0) = 0, we have: .e^
    0 - 0 + C2 .= .0 . . C2 = -1

    . . Hence: .y''(t) .= . e^
    t - t - 1


    Integrate: .y'(t) .= . e^
    t - t/2 - t + C3

    . . Since y'(0) = 0, we have: .e^
    0 - 0/2 - 0 + C3 .= .0 . . C3 = -1

    . . Hence: .y'(t) .= .e^
    t - t/2 - t - 1


    Integrate: .y(t) .= .e^
    t - t/6 - t/2 - t + C4

    . . Since y(0) =1, we have: .e^
    0 - 0/6 - 0/2 - 0 + C4 .= .1 . . C4 = 0


    Therefore: .y(t) .= .e^
    t - t/6 - t/2 - t

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  7. #7
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    Quote Originally Posted by CaptainBlank View Post

    y(t) = A e^{-t} + ax^2 + bx +c.
    I think you made a mistake.

    It should be,

    y = Ate^{-t} + at^2+bt+c
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  8. #8
    Senior Member tukeywilliams's Avatar
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    no I think he is correct. This is not the serpent in paradise.

    you use y_p = Ate^2t only if we get L(y_p) = 0.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker View Post
    I think you made a mistake.

    It should be,

    y = Ate^{-t} + at^2+bt+c
    Well I did actualy do a bit more than I presented in my post and am
    fairly comfortable with the idea that my trial solution will give a particular
    integal (that is I did the work and have it here - of course there could be
    some algebraic infelicities which would lead to it being wrong, but I don't
    see any).

    [-1 is not a root of the indical equation]

    RonL
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  10. #10
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    What textbook you use?
    Attached Thumbnails Attached Thumbnails Differential Equations and Variation of Parameters-picture18.gif  
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Well I did actualy do a bit more than I presented in my post and am
    fairly comfortable with the idea that my trial solution will give a particular
    integal (that is I did the work and have it here - of course there could be
    some algebraic infelicities which would lead to it being wrong, but I don't
    see any).

    [-1 is not a root of the indical equation]

    RonL
    "infelicities"

    what impressive vocabulary you have Captain!

    (or is that a really common word that only I have never seen before?)
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  12. #12
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    For the last solution,

    y'''+y'=sec t

    You can let z=y'

    To get,

    z''+z=sec t

    The general solution to homogenous equation is:
    z=C_1sin(t)+C_2cos(t)

    Then you can find particular via Lagrange's Variation of Parameters techiqnue.

    Then you need to integrate the entire solution for z tto get y.

    Well I did actualy do a bit more than I presented in my post and am
    fairly comfortable with the idea that my trial solution will give a particular
    integal (that is I did the work and have it here - of course there could be
    some algebraic infelicities which would lead to it being wrong, but I don't
    see any).
    Yes, I did make a mistake. My first one in about 7 months.
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  13. #13
    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    "infelicities"

    what impressive vocabulary you have Captain!

    (or is that a really common word that only I have never seen before?)
    When you have been in this business as long as I have you build up an
    impressive list of synonyms for mistake or error.

    RonL
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    When you have been in this business as long as I have you build up an
    impressive list of synonyms for mistake or error.

    RonL
    indeed!
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  15. #15
    Senior Member tukeywilliams's Avatar
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    Differential Equations & Linear Algebra by Farlow Hall Mcdill West
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