In this MSE question we are presented with an orthogonal basis for the space $H^1(-\pi, \pi)=W^{1,2}[-\pi, \pi]$, the Hilbert space of absolutely continuous functions on the interval $[-\pi, \pi]$ that have their first derivative in $L^2[-\pi, \pi]$ (Lebesgue space). The inner product is defined as $$\langle f, g \rangle = \langle f, g\rangle_{L^2[-\pi,\pi]} + \langle f', g'\rangle_{L^2[-\pi,\pi]}.$$
Now to my question. It is known that $\{e^{i n x}\}$ defines an orthogonal basis in $L^2[-\pi, \pi]$ (e.g. see another MSE question), but it turns out it's not enough in $W^{1,2}[-\pi, \pi]$. I can verify, with the explicit computations of inner products, that $\sinh(x)$ is in the orthogonal complement of exponents, making the system incomplete. But I wonder how we could come up with this $\sinh(x)$ in the first place. I'm sure it's not just some educated guess. I suspect that we can somehow make a differential equation out of the orthogonality condition $$\langle e^{i n x}, g\rangle_{L^2[-\pi,\pi]} + \langle i n e^{i n x}, g'\rangle_{L^2[-\pi,\pi]}=0.$$ I know that $\sinh(x)$ is the solution to the equation $f''(x)=f(x)$ with some conditions for $f(0), f'(1)$. I know next to nothing about DEs, so I don't see how this inner product can be possibly related to them.
The idea about deriving a differential equation is correct, this is somewhat similar to deriving the Euler-Lagrange equation in the calculus of variations. Integrating by parts, we have $$\langle f, g \rangle =\langle f, g\rangle_{L^2}+\langle f', g'\rangle_{L^2}= \langle f, g\rangle_{L^2} - \langle f'', g\rangle_{L^2}+f'g\mid_{-\pi}^{\pi}=\langle f-f'', g\rangle_{L^2}+f'g\mid_{-\pi}^{\pi}.$$
For $g(x)=e^{inx}$ due to $2\pi$ periodicity $g(\pi)=g(-\pi)$, and there are sufficiently many $2\pi$-periodic functions that vanish at the ends to show that each term must vanish separately. So for $f$ to be orthogonal to all imaginary exponents in $W^{1,2}$ we need $f''=f$ and $f'(\pi)=f'(-\pi)$. The general solution to the equation is $f(x)=A\sinh x+B\cosh x$, and the boundary condition then implies that $f(x)=A\sinh x$. Thus, supplementing imaginary exponents with $\sinh x$ we get an orthogonal basis of $W^{1,2}$.