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Let $y''+\lambda y=0$, and $\lambda<0$. Then

$ y(x)=c_1e^{-\sqrt{-\lambda}x}+c_2e^{\sqrt{-\lambda}x}\stackrel{?}{=}c_3\cosh\sqrt{-\lambda}x+c_4\sinh\sqrt{-\lambda}x. $

I do not understand how the RHS of this obtained. I know that use of the equalities

$ \cosh x=\frac{e^x+e^{-x}}{2}\qquad\text{and}\qquad\sinh x=\frac{e^x-e^{-x}}{2} $

is necessary, but the constants $c_i$ make this difficult.

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    Hint: $c_2 = (c_3+c_4)/2$.2012-11-14

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There are two ways to see this. The first is to note that $\{\cosh (\sqrt{- \lambda} x), \sinh(\sqrt{-\lambda} x)\}$ is a set of linearly independent solutions to a second order linear, homogeneous ODE (for confirmation, check the Wronskian). Then you apply the standard result that the set of solutions in such a problem forms a two-dimensional vector space.

That's the "abstract" way of looking at it. The not-so-abstract way is to verify algebraically what $c_3, c_4$ must be in terms of $c_1, c_2$, in which case you have $ c_1 = \frac{1}{2}(c_3 + c_4) \\ c_2 = \frac{1}{2}(c_3 - c_4) $ Viewing this as a linear system of real numbers, you get forms for $c_3, c_4$ in terms of $c_1, c_2$ (this amounts to inverting a matrix).