I have recently thought of this problem and I am wondering if the technique used is at all valid for solving the differential equation, since this equation involves infinitely many derivatives of $y$. I'm not sure if the mathematics used to solve it is valid, or if the equation itself makes sense. If the method does not make sense please explain why, if it does, how can I check that the answer is right?
Consider the differential equation
$$0 = \frac{dy}{dx} -\frac{1}{3!}\frac{d^3y}{dx^3}+\frac{1}{5!}\frac{d^5y}{dx^5}-\frac{1}{7!}\frac{d^7y}{dx^7}+ ... + \frac{(-1)^{n}}{(2n+1)!}\frac{d^{2n+1}y}{dx^{2n+1}}+...$$
Similarly to how a second (or third, or some finite) order homogenous differential equation may be solved, guess that the general solution may be in the form $y=e^{\lambda x}$. This means the $n$th derivative of y, $\frac{d^ny}{dx^n}$, is $\lambda ^ne^{\lambda x}$.
So, using the above equation with $y=e^{\lambda x}$ gives the result $$0 = \lambda e^{\lambda x} - \frac{\lambda ^3}{3!} e^{\lambda x} + \frac{\lambda ^5}{5!} e^{\lambda x} - \frac{\lambda ^7}{7!} e^{\lambda x} + ... + \frac{(-1)^{n} \lambda^{2n+1}}{(2n+1)!}e^{\lambda x}+...$$ As $e^{\lambda x}$ is never $0$,
$$0 = \lambda - \frac{\lambda ^3}{3!} + \frac{\lambda ^5}{5!} - \frac{\lambda ^7}{7!} + ... + \frac{(-1)^{n} \lambda^{2n+1}}{(2n+1)!}+...$$
Which is the Maclaurin series for $\sin(\lambda)$, converging for all $\lambda$. Therefore $0 = \sin(\lambda)$, so $\lambda = \pi n$, where $n \in \mathbb{Z} $.
So the general solution is
$$y=...+C_{-2}e^{-2 \pi x}+C_{-1}e^{- \pi x}+C_{0}+C_{1}e^{ \pi x}+C_{2}e^{2 \pi x}+...$$
(or, alternatively written)
$$y = \sum_{r=-\infty}^{\infty}C_re^{\pi rx}.$$
I'm not sure if any problem arises at any point with trying to extend the method for solving a finite differential equation is applied to this infinitely long one.
The Taylor expansion is, in formal power series, $y(x+h)=(e^{h\frac{d}{dx}}y)(x)$. As the sine can be expressed in exponentials, you get $$ 0=[\sin(\tfrac{d}{dx})y](x)=\left[\frac{e^{i\tfrac{d}{dx}}-e^{-i\tfrac{d}{dx}}}{2i}y\right](x) =\frac{y(x+i)-y(x-i)}{2i} $$ assuming complex arguments make sense for $y$. This means that $y(z)$ has a period $2i$. Examples are $y(z)=e^{k\pi z}$, $k\in \Bbb Z$ and linear combinations, as the original equation is linear.