General solution for $\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = y$? Start with
$$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = y$$
then
$$\frac{1}{\mathrm{d} x} \, \mathrm{d} \left(\frac{\mathrm{d} y }{\mathrm{d} x}\right) = y$$
$$\frac{\mathrm{d} y}{\mathrm{d} x} \, \mathrm{d} \left(\frac{\mathrm{d} y }{\mathrm{d} x}\right) = y \, \mathrm{d} y$$
$$\frac{\mathrm{d} y}{\mathrm{d} x}  = \sqrt{y^2 + c}$$
$$\int \frac{1}{\sqrt{y^2+c}} \mathrm{d} y = x + c_1$$
$$ \ln\left(\sqrt{y^2 +c_0} + y\right) = x + c_1$$
$$ \sqrt{y^2 +c_0} + y = e^{x + c_1}$$
This does look trigonometric and exponential as it should but I don't know how to proceed to simplify it.
I actually know the actual solution but I want to prove it from first principles. This means I don't want to use guess and check methods because they can't handle solving problems in general and I don't want to assume the result in order to prove it.

For linear equations with constant coefficients, the "guess-and-check" method, which amounts to assuming $y=e^{\lambda x}$ and solving for $\lambda$, actually does generalize to all possibilities (provided that you can solve the necessary equation for $\lambda$, and with appropriate adjustment for duplicate roots). One can also prove using general linear ODE theory that it doesn't miss any solutions.
That said, if you want to do it without this, one way is operator factorization: $y''-y=0$ is the same as $(D^2-1)y=0$ which is the same as $(D+1)(D-1)y=0$. You can do this by finding the general solution $z$ to $(D+1)y=0$ (i.e. $y'+y=0$) and then solving $(D-1)y=z$. Each of these can be solved using the integrating factor technique. (Of course in a way the integrating factor technique is itself guess-and-check...)
