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 !!!
... and from (1) and (2)...
... so that both and satisfy the ODE...
But the (5) is a linear second order ODE and any linear combination of two linearly independent solutions is also a solution...
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.