Hi,

For what $\displaystyle nonzero$ values of $\displaystyle k$ does the function $\displaystyle y = sinh (kt)$ satisfy the differential equation $\displaystyle y'' - 25y = 0$?

I don't really understand what the general method of solving this differential equation is. Do I differentiate w.r.t to $\displaystyle k$??

The only thing I understand regarding DE's is separating the variables but I dont think it's used here..is it? :S.

What I've tried:

$\displaystyle

y' = k cosh (kt)$

$\displaystyle y'' = k^2 sinh (kt) + cosh (kt)$

Since

$\displaystyle y'' = 25y$

$\displaystyle k^2 sinh (kt) + cosh (kt) = 25 sinh (kt)$

$\displaystyle (k^2 - 25) sinh (kt) + cosh (kt) = 0$

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But I don't know where to go from there...or if I've gone in a completely wrong direction. Explanation and/or answer would be greatly appreciated ASAP. Thanks!