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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?

1 Answers 1

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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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    Does $\cosh\,mx=\frac12(e^{mx}+e^{-mx})$ and similarly for $\sinh\,mx$ too?2011-05-06
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    Yes.$\qquad\quad$2011-05-06
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    @J.M:I am trying but not getting there,I never really did anything over hyperbolic may be cold feet because of that,could you please give me some more hints how to convert between the forms quickly as this is a objective question and I am suppose to solve it fast.2011-05-06
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    Let $c_1=\frac{c_3+c_4}{2}$ and $c_2=\frac{c_3-c_4}{2}$...2011-05-06
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    sorry,I could not understand :(2011-05-06
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    Make that substitution, expand accordingly ($(\frac{c_3}{2}e^{mx}+\cdots)$), bring together the stuff with $c_3$ and the stuff with $c_4$, and use the relations between exponential and hyperbolics.2011-05-06
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    I did it with someway different,using the two results I notice that $(\cosh mx + \sinh mx) = e^{mx}$ and $ (\cosh mx - \sinh mx) = e^{-mx}$ after that it is fairly simple :) Thanks for your time :) +$1$ and accepted.2011-05-06
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    Yes, that's another way of doing it. :)2011-05-06