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I have to find the general solution for the differential equation $y''(x) - m^2y(x) = 0$I tried to solve this one by deriving the auxiliary equation as $M^2-m^2=0$ which gives $M = \pm m$ hence the general solutions is $c_1e^{mx} + c_2e^{-mx}$ but in the paper it's given the answer should be in the form of summation of hyperbolic functions,which is $c_1 \sinh mx + c_2 \cosh mx$

I haven't done anything much on hyperbolic,could anybody help me in this regard?

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$c_1e^{mx} + c_2e^{-mx}$

and

$c_3\cosh\,mx + c_4\sinh\,mx$

are two different ways of writing the same thing, since $\cosh\,x=\frac12(e^x+e^{-x})$ and $\sinh\,x=\frac12(e^x-e^{-x})$. You should be able to figure out how to express $c_1$ and $c_2$ in terms of $c_3$ and $c_4$, and vice-versa, to go back and forth between these two forms.

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    Yes, that's another way of doing it. :)2011-05-06