derive $\delta _{tt} y=y$ without guessing Say we have a second order differential equation $\delta _{tt} y=y$ 
We can on the basis of our intuition say that the functions $e^t$ and $e^{-t}$ satisfy this second order ODE. We can then reason that the superposition must also satisfy the ODE.
But is there a way to derive directly from the ODE, without make the abovementioned educated guess, the general solution of that ODE?
(I don't have much experience with second order ODE's).
 A: You are trying to solve:
$$\frac{d^2 y}{dt^2}=y$$
Here is how to do it without using the characteristic polynomial (Using the ansatz $y=e^{\lambda t}$). Note that the characteristic polynomial method is a lot easier but this is a method without any guesses.

If you multiply both sides by $\frac{dy}{dt}$, you should obtain: $$\frac{dy}{dt}\cdot \frac{d^2 y}{dt^2}=y\cdot \frac{dy}{dt}$$ Integrating both sides with respect to $t$ gives: $$\int \frac{dy}{dt}\cdot \frac{d^2 y}{dt^2}~dt=\int y\cdot \frac{dy}{dt}~dt$$
One can integrate the LHS by substituting:
$$u=\frac{dy}{dt} \iff du=\frac{d^2 y}{dt^2}~dt$$
As a result, we obtain:
$$\frac{1}{2}\left(\frac{dy}{dt}\right)^2=\frac{y^2}{2}+c_1 \tag{1}$$

It remains to solve the first order ODE. Be careful to not forget the case when $\frac{dy}{dt}$ is negative!
$$\frac{dy}{dt}=\pm \sqrt{y^2+2c_1}$$
It is a separable ODE:
$$\int \frac{1}{\sqrt{y^2+2c_1}}~dy=\int \pm dt$$
Integrating both sides gives:
$$\ln\left(\sqrt{y^2+2c_1}+y\right)=c_2\pm t$$
Now, one can rearrange this to obtain two explicit solutions for $y(t)$:
$$y(t)=\frac{e^{t+c_2}-2c_1e^{-t-c_2}}{2} \tag{2.1}$$
$$y(t)=\frac{e^{-t+c_2}-2c_1e^{t-c_2}}{2} \tag{2.2}$$
Redefining the arbitrary constants for both cases $(2.1)$ and $(2.2)$ gives the same general solution!
$$\bbox[5px,border:2px solid #C0A000]{y(t)=Ae^t+Be^{-t}}$$
A: One way to motivate the introduction of the characteristic equations associated to the ODEs with constant coefficients is by noticing that such differential equations can be factorized exactly as the polynomials. For example, in the given example,
$$0 = (\partial^2_t - 1)y(t) = (\partial_t - 1)(\partial_t + 1)y(t) \ ,$$
we can "split" the original 2nd order ODE into two simpler 1st order ODEs, $y'(t) - y(t) = 0$ and $y'(t) + y(t) = 0$. Each of them is easy to integrate and the solutions are, respectfully, $y_1(t) = A e^t$ and $y_2(t) = Be^{-t}$, with some constants $A$ and $B$. Finally, both $y_1(t)$ and $y_2(t)$ are solutions of the original ODE since each of them is "killed" by one of the operators, $\partial_t - 1$ or $\partial_t + 1$.
