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Math Help - differential equation

  1. #1
    Junior Member
    Joined
    Oct 2009
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    differential equation

    I have a 2nd order differential equation:

    (d^2*y)/(d*x^2) = cosh(x)sinh(x)

    How do you find a general solution?

    Also how would you find a particular solution satisfying the conditions y(0) =1 and y'(0)= 2?

    I really am having trouble understanding differential equations so help would be really appreciated.
    Last edited by JQ2009; February 9th 2010 at 08:11 AM.
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  2. #2
    Junior Member
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    Recall that the general solution of a linear differential equation is the sum of a particular solution with the general solution of the corresponding homogeneous equation. The homogeneous equation has a trivial solution in this case:

    \frac{d^2y}{dx^2}=0 \Leftrightarrow y(x)=ax+b (with a,b\in \mathbb R)

    As for the particular solution, notice that \sinh and \cosh are derivatives of one another. Thus, one could expect that a particular solution is given by \alpha \sinh (x)\cosh (x) for some \alpha \in \mathbb R. Now simply observe that

    (\sinh(x)\cosh(x))''=(\sinh^2(x)+\cosh^2(x))'
    =2\sinh(x)\cosh(x)+2\sinh(x)\cosh(x)=4\sinh(x)\cos  h(x)

    so \alpha =\frac{1}{4}. The general solution is thus

    y(x)=\frac{1}{4}\sinh(x)\cosh(x)+ax+b (with a,b\in \mathbb R)
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