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.

Your first manipulation is incorrect. Something like this is true:
$\frac{d}{dx} \frac{dy}{dx} = y$
But this doesn't help you. 
A common method for solving this is to look at it as two applications of the derivative-with-respect-to-x operator:
$\frac{d}{dx} \frac{d}{dx} y = y$
Let $D = \frac{d}{dx}$, then the above is equivalent to $(D^2 - 1)y = 0$ (we moved the sole $y$ to the left hand side). Factoring the expression gives us:
$(D-1)(D+1)y = 0$. The general solution is a linear combination of the solutions of $(D+1)y = 0$ or $(D-1)y = 0$. Well the first equation is just $y'=-y$ and its solution is $e^-x$, the second option is $y' = y$ and hence has the solution $e^x$. The general solution is thus $y(x) = c_1 e^{-x} + c_2 e^x$ for some $c_1,c_2 \in \mathbb R$.
http://www.wolframalpha.com/input/?i=y%27%27+%3D+y
