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Math Help - why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

  1. #1
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    Question why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

    How do i show this mathematically, preferably without using auxillary equations, that the solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)
    Last edited by mr fantastic; July 31st 2011 at 03:50 AM. Reason: Copied title into main body of post.
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    MHF Contributor Siron's Avatar
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    Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

    I think you can use the substitution: y=e^{tx}.
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    Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

    where to from there? I end up with:


    y\left(\left(\frac{\ln{y}}{x}\right)^2-9\right)=0\\\\y=e^{\pm3x}

    I'm not sure if this is right. There are no general constants !!!
    Attached Thumbnails Attached Thumbnails why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)-capture-1.jpg   why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)-capture-2.png  
    Last edited by Istafa; July 30th 2011 at 02:51 AM. Reason: Latex???
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    MHF Contributor Siron's Avatar
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    Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

    I would solve it like this:
    Let y=e^{tx}
    Substituting in the equation gives:
    t^2\cdot e^{tx}-a^2\cdot e^{tx}=0
    t^2-a^2=0 \Leftrightarrow (t-a)(t+a)=0
    t=a \ \mbox{or} \ t=-a
    So the general solution is:
    y=c_1\cdot e^{ax}+c_2\cdot e^{-ax}
    Where c_1,c_2 are constants (which can be chosen).
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    Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

    Quote Originally Posted by Siron View Post
    I would solve it like this:
    Let y=e^{tx}
    Substituting in the equation gives:
    t^2\cdot e^{tx}-a^2\cdot e^{tx}=0
    t^2-a^2=0 \Leftrightarrow (t-a)(t+a)=0
    t=a \ \mbox{or} \ t=-a
    So the general solution is:
    y=c_1\cdot e^{ax}+c_2\cdot e^{-ax}
    Where c_1,c_2 are constants (which can be chosen).
    This is using the fact that any de with n solutions will also have solutions that are a linear combination of one or more of those n solutions, right? Would anyone know where there is a proof for this fact?
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    MHF Contributor chisigma's Avatar
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    Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

    Quote Originally Posted by Istafa View Post
    How do i show this mathematically, preferably without using auxillary equations
    One of the most 'attracting' properties of the hyperboloc function is ...

    \frac{d}{dx} \sinh a x = a\ \cosh a x (1)

    \frac{d}{dx} \cosh a x = a\ \sinh a x (2)

    ... and from (1) and (2)...

    \frac{d^{2}}{d x^{2}} \sinh a x = a^{2}\ \sinh a x (3)

    \frac{d^{2}}{d x^{2}} \cosh a x = a^{2}\ \cosh a x (4)

    ... so that both \sinh a x and \cosh a x satisfy the ODE...

    \frac{d^{2} y}{d x^{2}} = a^{2}\ y (5)

    But the (5) is a linear second order ODE and any linear combination of two linearly independent solutions is also a solution...

    Kind regards

    \chi \sigma
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    Re: why? solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)

    If the problem was, as initially stated, NOT to solve y''- a^2y= 0 but simply to show that y(x)= Asinh(ax)+ Bcosh(ax) is a solution, then you simply need to note that (Acosh(ax)+ Bsinh(ax))''= Aa^2cosh(ax)+ Ba^2sinh(ax), as chisigma said, and that is a^2(Acosh(ax)+ Bsinh(ax))= y so that y''- a^2y= a^2y- a^2y= 0.

    By the way, just as sin(ax) and cos(ax) are solutions to the initial value problems y''+ a^2y= 0, y(0)= 0, y'(0)= 1 and y''+ a^2y= 0, y(0)= 1, y'(0)= 0, respectively, so sinh(ax) and cosh(ax) are solutions to the initial value problems y''- a^2y= 0, y(0)= 0, y'(0)= 1 and y''- a^2y= 0, y(0)= 1, y'(0)= 0, respectively.
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