Show that
$$
h_{L,M} = L\cosh\left(k\left[x-a\right]\right) + M\sinh\left(k\left[x-a\right]\right)
$$
is a general solution for the differential equation $y'' – k^{2}y = 0$.
Express the function h using an exponential function.
Okay so my way of going about this would be simply to start off by finding the second derivative of $L\cosh\left(k\left[x-a\right]\right) + M\sinh\left(k\left[x-a\right]\right)$ and then somehow compare that to $y'' – k^{2}y = 0$.
That seems very tedious though… unless there is some sort of a rule for finding the second derivative for something so complex. I'm wondering if there's a "better" way of approaching this.
Best Answer
You're right, just take derivatives and plug in.
$$ \begin{align} h&=L\cosh{k(x-a)}+M\sinh{k(x-a)}\\ h'&=Lk\sinh{k(x-a)}+Mk\cosh{k(x-a)}\\ h''&=Lk^2\cosh{k(x-a)}+Mk^2\sinh{k(x-a)} \end{align} $$
so then notice that, indeed, $$h''-k^2h=0$$ so your proposed $h$ is a solution to this differential equation.
I can help withthe second part, expressing $h$ as an exponential function, but I want to see what you've tried so far, so we can help the most with what you don't understand.