I think you can use the substitution: .
How do i show this mathematically, preferably without using auxillary equations, that the solution to second order DE (d2y/dx2)-a2y=0 is in form y=Asinh(ax)+Bcosh(ax)
where to from there? I end up with:
I'm not sure if this is right. There are no general constants !!!
One of the most 'attracting' properties of the hyperboloc function is ...
(1)
(2)
... and from (1) and (2)...
(3)
(4)
... so that both and satisfy the ODE...
(5)
But the (5) is a linear second order ODE and any linear combination of two linearly independent solutions is also a solution...
Kind regards
If the problem was, as initially stated, NOT to solve but simply to show that y(x)= Asinh(ax)+ Bcosh(ax) is a solution, then you simply need to note that , as chisigma said, and that is so that .
By the way, just as sin(ax) and cos(ax) are solutions to the initial value problems , y(0)= 0, y'(0)= 1 and , y(0)= 1, y'(0)= 0, respectively, so sinh(ax) and cosh(ax) are solutions to the initial value problems , y(0)= 0, y'(0)= 1 and , y(0)= 1, y'(0)= 0, respectively.