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Math Help - Differential equation

  1. #1
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    Differential equation

    For what nonzero values of k does the function y=sinhkt satisfy the DE  y''-25y=0?

    I got k=5 and k=-5 which I think is correct, but I'm struggling with the other two parts of the question:

    For those values of k, show that every member of the family of functions Asinhkt+Bcoshkt (where  A and B are constants) is also a solution of the DE.

    Come up with a second order DE for which y=sinkt is a solution for the same values of k obtained before.
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  2. #2
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    Quote Originally Posted by Konidias View Post
    For what nonzero values of k does the function y=sinhkt satisfy the DE  y''-25y=0?

    I got k=5 and k=-5 which I think is correct
    Yes, it is. If y= sinh(kt) then y'= k cosh(kt) and y"= k^2 sinh(kt). Putting that into the differential equation, y"- 25y= k^2 sinh(kt)- 25 sinh(kt)= (k^2- 25)sinh(kt)= 0 for all t. Since sinh(kt) is not alway7s 0 itself, we must have k^2- 25= 0 so k= 5 or k= -5.

    but I'm struggling with the other two parts of the question:

    For those values of k, show that every member of the family of functions Asinhkt+Bcoshkt (where  A and B are constants) is also a solution of the DE.
    Just DO it! With k= 5, those functions are A sinh(5t)+ B cosh(5t). Find the second derivative, plug it into the equation and see what happens! With k= -5, you haved A sinh(-5t)+ B cosh(-5t). Since sinh is an odd function and cosh is an even function, that is just -A sinh(5t)+ B cosh(5t) which is really the same thing since A itself can be positive or negative.

    [/quote]Come up with a second order DE for which y=sinkt is a solution for the same values of k obtained before.[/QUOTE]
    So now youy are working with y= sin(5t). Take the second derivative, stare at it and think!
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