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Can someoen verify i did this differentiation right?

  1. Feb 18, 2006 #1
    Hello everyone, these problems take like 2 pages to do so i don't want to redo it all if i messed up on a derivative, can someone tell me if they get the same thing for y''?
    Thanks:
    [​IMG]

    I think I better do these problems in pencil...:bugeye:
     
  2. jcsd
  3. Feb 18, 2006 #2

    Fermat

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    y'' is the correct derivative from y'.
    But where did y' come from?
    What is y?
    Is y that first line on the page, with the number 2 written after it (in a circle).
    If so, then your y' should have used the product rule like y'' did.
     
  4. Feb 18, 2006 #3
    Thanks for the responce Fermat, here is what y is:
    [​IMG]
    [​IMG]

    So your thinking I did it wrong?
     
  5. Feb 18, 2006 #4

    Fermat

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    You got the y' wrong. I'm afraid.

    y = u1.f(t) + u2.g(t)

    you should have, (using the product rule)

    y' = u1'.f(t) + u1.f'(t)+ u2'.g(t) + u2.g'(t)

    where f(t) = e^(-7t)
    and g(t) = t.e^(-7t)

    to get g', you have to use the product rule on g(t).

    Also, I don't see hoe you got the line marked (2), from the line above it.

    HOLD ON. more to come.
     
  6. Feb 18, 2006 #5

    Fermat

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    I just noticed. You got a repeated root.

    With repeated roots, the solution is,

    y = (A + Bt)e^(rt)

    where r is the repeated root.

    Also, c1 and c1 are supposed to be constants of integration. Why did you replace them with functions ot time c1 = u1(t) ??
     
  7. Feb 18, 2006 #6
    Thanks for the responce fermat, i'm replacing the constants with the functions of t because i'm doing a varation of parameters technique to solve this, unless you know an easier way to solve it. Check my last thread: https://www.physicsforums.com/showthread.php?t=111140
    The problem takes like 2 pages to do!
     
  8. Feb 18, 2006 #7

    Fermat

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    I guess this must be just exercise in the use of variation of parameters and wronskians - yes ?
    It's a lot easier to do it the usual way.

    y1 = (A + Bt) e^(-7t), but you've got that haven't you.

    and y2 comes out quite simply at,

    y2 = 5e^t

    so,

    y = y1 + y2
    y = (A + Bt)e^(-7t) + 5e^t
    (takes about half a page)

    Mind you. I've not done variation of parameters or wronskians, so I hope I'm not misleading you.
     
  9. Feb 18, 2006 #8
    Actually i'm not forced to use variation of paramters, i thought that was the only way to solve this problem because it its not homogenous!
    I understand hwo You found y1, but how did u come out with y2 = 5e^t?
     
  10. Feb 18, 2006 #9
    WOw that was right and took like no work at all, if you can explain to me how you found y2 i will eat this cupcake infront of me.
    [​IMG]
     
  11. Feb 18, 2006 #10

    Fermat

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    finding y2 is just "standard procedure"

    If the "non-homogeneous" function is say,
    f(x) = a polynomial e.g f(x) = 2x² + 3x - 1, then try y2= Ax² + Bx + C
    if f(x) is a trig function e.g. f(x) = sin(3x), then try y2 = Asin(3x) + Bcos(3x)
    if f(x) is an exponential e.g. f(x) = 5e^(-3x), then try y2 = Ae^(-3x)

    that sort of thing.

    Whatever the function f(x) is, try a similar function for y2.

    p.s. this standard procedue doesn't always work. Try it on y'' + 25y = 50sin(5t).
     
    Last edited: Feb 18, 2006
  12. Feb 18, 2006 #11
    hm...I c, but on the right hand size of the equation it was 320e^t
    r^2+14r+49 = 320e^t,
    so how did u figure 5 as being the coefficent that will work?
    I would try y2 = Ae^(t) right? because its 320e^t?
     
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