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Suppose i have ODE:

$y''-4y=0$

We know that the solution is

$y=c_1 e^{2x} + c_2 e^{-2x}$

And I realize that this also has the following solution

$y=c_1 \cosh(2x) + c_2 \sinh(2x)$

are the two solutions equivalent?

So, if i have this ODE

$ay''-by=0\qquad a>0, b>0$

May I write the general equation as follows?

$y=c_1 \cosh(\sqrt{\frac{b}{a}}x) + c_2 \sinh(\sqrt{\frac{b}{a}}x)$

Thanks.

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    $\begingroup$ Yes, the two solutions are equivalent (with different values of $c_1$ and $c_2$). If you have the first solution, you can always re-write it in the form of the second solution, and vice versa. Yes, your last expression is the general solution. $\endgroup$ – John Barber May 11 at 22:08
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    $\begingroup$ It's simply a changing of the basis. Instead of $(e^{2x},e^{-2x})$ you take $(\cosh(2x),\sinh(2x))$. Of course coordinates, $c_1,c_2$, will change. $\endgroup$ – Wang May 11 at 22:11
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Yes the two forms of solutions are equivalent.

Any linear combination of $e^{\lambda_1}$ and $e^{\lambda _2}$ is a solution

Hyperbolic functions are linear combination of exponentials.

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