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Math Help - Differential Equation help

  1. #1
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    Differential Equation help

    Hi,

    For what nonzero values of k does the function  y = sinh (kt) satisfy the diff erential equation y'' - 25y = 0?


    I don't really understand what the general method of solving this differential equation is. Do I differentiate w.r.t to k??
    The only thing I understand regarding DE's is separating the variables but I dont think it's used here..is it? :S.

    What I've tried:

    <br />
y' = k cosh (kt)

    y'' = k^2 sinh (kt) + cosh (kt)
    Since
     y'' = 25y

     k^2 sinh (kt) + cosh (kt) = 25 sinh (kt)

     (k^2 - 25) sinh (kt) + cosh (kt) = 0

    --
    But I don't know where to go from there...or if I've gone in a completely wrong direction. Explanation and/or answer would be greatly appreciated ASAP. Thanks!
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  2. #2
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    Quote Originally Posted by Solo View Post
    Hi,

    For what nonzero values of k does the function  y = sinh (kt) satisfy the differential equation y'' - 25y = 0?


    I don't really understand what the general method of solving this differential equation is. Do I differentiate w.r.t to k??
    The only thing I understand regarding DE's is separating the variables but I dont think it's used here..is it? :S.

    What I've tried:

    <br />
y' = k cosh (kt)

    \color{red}y'' = k^2 sinh (kt) + cosh (kt)
    where'd you get that from?


    your approach is good, but you messed up where i indicated
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  3. #3
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    Ah I think I've differentiated first WRT t, and then WRT k, which is worng

    Should it just be:

    y'' = k^2 sinh (kt)

    then:

    k^2 sinh (kt) = 25 sinh (kt)
    k^2 = 25

    k = 5

    Is that right? Surely it can't be that easy?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Solo View Post
    Ah I think I've differentiated first WRT t, and then WRT k, which is worng

    Should it just be:

    y'' = k^2 sinh (kt)

    then:

    k^2 sinh (kt) = 25 sinh (kt)
    k^2 = 25

    k = 5

    Is that right? Surely it can't be that easy?
    yup, that's it

    (note that zero also works, but they asked for nonzero solutions, so we deliberately forget about it)
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  5. #5
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    Cheers

    Now I'm stuck on the next part:

    Come up with a second order di fferential equation for which y = sin (kt) is a solution for the same values of k obtained in part (a).


    For k = 5

    y = sin (5t)
    y' = 5 cos (5t)
    y'' = -25 sin (5t)
    Equation: y'' + 25y = 0

    For k = -5

    y = sin (-5t)
    y' = -5 cos (-5t)
    y'' = 25 sin (-5t)
    Equation: y'' - 25y = 0


    The question only asks for one differential equation...What have I done wrong?
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  6. #6
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    y=sin(-5t)=-sin(5t) since sin(x) is an odd function i.e.  sin(-x)=-sin(x)

    y=-sin(5t)
    y'=-5cos(5t)
    y''=25sin(5t)

    25sin(5t)-25sin(5t)=0
    Last edited by Konidias; February 8th 2009 at 12:04 PM.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Solo View Post
    Come up with a second order di fferential equation for which y = sin (kt) is a solution for the same values of k obtained in part (a).
    .
    .
    .
    Equation: y'' + 25y = 0
    this equation works for both the values of k.
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