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.
 A: You are over complicating. 
Re-write it as $y'' - y = 0$. Then it has auxiliary equation $\lambda^2 - 1 = 0$. So the general solution is $y = \alpha e^{x} + \beta e^{-x}$. 
A: It does look like trigonometric and exponential function indeed:
The characteristic equation is $\lambda^2-1=0$, with roots $\pm1$, hence a basis of the space of solutions is $\mathrm e^x$ and  $\mathrm e^{-x}$, or if one prefers
$$\frac{\mathrm e^x+\mathrm e^{-x}}2=\cosh x,\qquad \frac{\mathrm e^x-\mathrm e^{-x}}2=\sinh x.$$
For the analogy to  be complete, the harmonic oscillator equation $\; y''=-y\;$ has characteristic equation $\lambda^2+1=0$, with roots $\pm\mathrm i$, hence a basis of the space of solutions is $\mathrm e^{\mathrm i x}$ and  $\mathrm e^{-\mathrm i x}$, or
$$\frac{\mathrm e^{\mathrm i x}+\mathrm e^{-\mathrm i x}}2=\cos  x,\qquad \frac{\mathrm e^{\mathrm i x}-\mathrm e^{-\mathrm i x}}{2\mathrm i}=\sin  x.$$
A: 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...)
A: Your approach is a slightly garbled version of a perfectly fine heuristic (which then is easily checked to have produced genuine solutions):
From $y''=y$, get ${d\over dx}(y')=y$ as you do, then, one way or another, multiply through by $dy$ as you do, to get $y'\,dy'=y\,dy$. It is at this point that you made some manipulation errors, because one would (at least heuristically) get ${1\over 2}(y')^2={1\over 2}y^2+C$. Taking $C=0$, we can easily take the square root(s): $y'=\pm y$, then $dy/y=\pm d$, or $\log y=\pm x+C$. Exponentiating, $y=e^{\pm x}$, which gives the two solutions we knew all along.
This kind of manipulation does appear in many older differential equations sources, and does give a way to anticipate the solutions (without having to know the constant-coefficient equation riff, as standard and basic as the latter is).
EDIT: after @Ian's and Willie Wong's comments... Indeed, taking $C=1$, for example, instead of $0$, gives $\int \sqrt{1+y^2}\;dy=x$, which is easily solved by letting $y=(t+t^{-1})/2$ (although knowing such a trick is perhaps already contrary to the spirit?)...
2nd EDIT: and, also, "discovering" the solutions via a non-linear differential equation (as a result of the manipulation) does create hazards that linear equations don't share, e.g., that solutions don't necessarily add... :)
A: 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
A: Here is another way to solve $$y''=y.$$
Rewrite the equation as 
$$y''+y'-y' -y=0.$$
This is just 
$$y''+y'-1(y'+y)=0.$$
Let $z=y'+y.$
The equation now becomes
$$z'-z=0$$
This has solution $z=Ce^x$ with $C$ being a constant. Verify this using separation of variables.
Now we solve $$Ce^x=y'+y$$
Using the integrating factor method,
$$y=\frac{\int \mu(x)Ce^x dx+ D}{\mu(x)}$$, where $D$ is a constant and
$\mu(x)=e^{\int 1 dx}=e^x$
This tells us that:
$$y=\frac{\frac{Ce^{2x}}{2}+D}{e^x}=Ee^x+De^{-x}$$ where $E$ is a constant.
