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Math Help - Differential Equations Help

  1. #1
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    Differential Equations Help

    Stuck on the following problem, unsure of how to go around it

    1a.

    For what none values of k does the function y=sinh (kt) satisfy the differential equation

    y'' - 25y = 0

    1b.

    for those values of k, show that every member of the family of functions Asinh(kt) + Bcosh(kt) (where A and B are constants) is also a solution to the differential equation.

    1c.

    come up with a second order differntial equation for which y=sinh (kt)is a solution for the same values of k obtained in part (a)

    im usually ok with differentail equations but unsure when it comes to the worded versions

    for 1a i tryed the following but werent sure where else to go

    y=sinh (kt)
    y'' = k^2sinh(kt)
    25y = k^2sinh(kt)
    25sinh(kt) = k^2sinh(kt)
    sinh(kt)(25-k^2) = 0

    so one possible solution is k = +/- 5 but im not sure if that is the right thing to do

    Any help is most appreciated
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  2. #2
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    Quote Originally Posted by jordanrs View Post
    Stuck on the following problem, unsure of how to go around it

    1a.

    For what none values of k does the function y=sinh (kt) satisfy the differential equation

    y'' - 25y = 0

    1b.

    for those values of k, show that every member of the family of functions Asinh(kt) + Bcosh(kt) (where A and B are constants) is also a solution to the differential equation.

    1c.

    come up with a second order differntial equation for which y=sinh (kt)is a solution for the same values of k obtained in part (a)

    im usually ok with differentail equations but unsure when it comes to the worded versions

    for 1a i tryed the following but werent sure where else to go

    y=sinh (kt)
    y'' = k^2sinh(kt)
    25y = k^2sinh(kt)
    25sinh(kt) = k^2sinh(kt)
    sinh(kt)(25-k^2) = 0

    so one possible solution is k = +/- 5 but im not sure if that is the right thing to do

    Any help is most appreciated
    We are told that y=\sinh(kx) solves y'' - 25y=0. Now, y'' = k^2\sinh(kx), therefore we want k^2\sinh(x) - 25\sinh(x) = 0 \implies \sinh(x) (k^2 - 25) = 0. Thus, k=\pm 5 and so \sinh(5x) and -\sinh(5x) solve this differencial equation.

    In general we have shown that \sinh(kx) solves the differencial equation y'' -k^2y=0.
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  3. #3
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    Crucial point: you must have sinh(kt)(25- k^2)= 0 for all t. Since there exist values of t such that sinh(kt) is not 0, we must have 25- k^2= 0 or k^2= 25.
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