How were the solutions to these differential equations found? These two very strange differential equations came up yesterday while I was doing a physics problem that I made up:
EQ 1) $y'^{2} = k \sin(y)$ 
EQ 2) $y'' = k\cos(y)$
where $y'$ means $\frac{dy}{dx}$
With initial conditions:
$y(0) = y'(0) = 0$
Here is the thread which shows my derivation of these equations and the original problem
I then plugged these equations into Wolfram Alpha, and they gave me these results:
Wolfram: EQ 1
solution1 http://www4b.wolframalpha.com/Calculate/MSP/MSP13591hg9040bceih89ag00005eef63f376f0b02b?MSPStoreType=image/gif&s=55&w=257.&h=36.
solution2 http://www4b.wolframalpha.com/Calculate/MSP/MSP13621hg9040bceih89ag000056714a816hhac700?MSPStoreType=image/gif&s=55&w=257.&h=36.
Wolfram: EQ 2
solution1 http://www4a.wolframalpha.com/Calculate/MSP/MSP34961h7f9890752347ad0000552fdd38bccgc0bb?MSPStoreType=image/gif&s=34&w=357.&h=40.
solution2 http://www4a.wolframalpha.com/Calculate/MSP/MSP34991h7f9890752347ad000059ead9f1ca0157cb?MSPStoreType=image/gif&s=34&w=357.&h=40.
Where am is the Jacbi amplitude function
How do you arrive at the two above links as the solutions?
 A: Ansatz
This seems to be a case for separation of variables, as I don't see the variable which you differentiate to on the right hand side of the differential equations:
$$
y'(x) = f(y) \iff 
dy = f(y)\,dx \iff
\int \!\!\frac{dy}{f(y)} = \int\!\! dx = x + C
$$
for some integration constant $C$. 
Equation 1
This equation features the RHS $f(y) = k \sin y$. Using separation of variables we get
$$
G(y) := \int\limits_{y_0}^y \!\!\frac{dY}{\sqrt{k \sin Y}} = x - x_0
$$
For the initial conditions $(x_0, y_0) = (0, 0)$ this formula reduces to
$$
G(y) := \int\limits_{0}^y \!\!\frac{dY}{\sqrt{k \sin Y}} = x
$$
This will probably need numerical integration to yield a graph / table $x = G(y)$ and numerical inversion to yield $y = G^{-1}(x)$.
Connection to the Wolfram Alpha solution
$$
y(x) = \frac{1}{2} \left(\pi-4\, \mathrm{am}\left(\frac{1}{2} \left(\sqrt{k} \, x + C\right) \large | 2\right)\right)
$$
Asking Wolfram Alpha for an indefinite integral using 
this query yields:
$$
\int \!\!\frac{dt}{\sqrt{\sin t}} = 
- 2\, F\left(\frac{1}{4}(\pi-2 t) \,\large|\, 2\right) + C \qquad(*)
$$
for some constant $C$ and $F(z\,|\,m)$ is the elliptic integral of the first kind
$$
F(z \,|\, m) = \int\limits_0^z\!\!\frac{dt}{\sqrt{1 - m \sin^2 t}}
$$
Like $G$ above this $F$ is not an elementary function, but defined by an integral function too.
It is only easier in the sense that this is a well studied function where you can get numerical implementations for, instead of integrating it numerically yourself. 
Its inverse is the Jacobi amplitude $\mathrm{am}(w \,|\, m)$ which fulfills
$$
w = F(z \,|\, m) \iff \mathrm{am}(w \,|\, m) = z
$$
Justification for equation (*)
$$
F\left(\frac{1}{4}(\pi-2 t) \,\large|\, 2\right) = 
\int\limits_0^{\frac{1}{4}(\pi-2 t)}\!\!\frac{du}{\sqrt{1 - 2 \sin^2 u}} =
\int\limits_0^{\frac{1}{4}(\pi-2 t)}\!\!\frac{du}{\sqrt{\cos 2u}}
$$
because $1 - 2 \sin^2 u = \cos 2u$. With $v = 2u$ this yields
$$
F\left(\frac{1}{4}(\pi-2 t) \,\large|\, 2\right) = 
\frac{1}{2}\int\limits_0^{\frac{\pi}{2} - t}\!\!\frac{dv}{\sqrt{\cos v}}
$$
with $w = \frac{\pi}{2} - v$ we get
$$
F\left(\frac{1}{4}(\pi-2 t) \,\large|\, 2\right) = 
-\frac{1}{2}\int\limits_{\frac{\pi}{2}}^t\!\!\frac{dw}{\sqrt{\cos(\frac{\pi}{2} - w)}} =
-\frac{1}{2}\int\limits_{\frac{\pi}{2}}^t\!\!\frac{dw}{\sqrt{\sin w}}
$$
because $\cos(\frac{\pi}{2} - w) = \sin w$.
Equation 2
For the other equation we need to deal with the second derivative $y''$, the idea here is to treat the first derivative $y'$ formally as another independent variable together with $y$ and have one side of the equation with only $y'$ and the other with only $y$:
$$
y'' = \frac{dy'}{dx} = \frac{dy'}{dy}\frac{dy}{dx} = \frac{dy'}{dy} y' = k \cos y \iff
y' \, dy' = k \cos y \, dy
$$
Thus
$$
\int\limits_{y'_0}^{y'} \! Y' \, dY' = \int\limits_{y_0}^y \! k \cos Y \, dY \iff
$$
$$
(y')^2 - (y'_0)^2 = \int\limits_{y_0}^y \! 2 k \cos Y \, dY = 2 k \sin y - 2 k \sin y_0
$$
With $(y_0, y'_0) = (0, 0)$ this reduces to
$$
(y')^2 = 2 k \sin y
$$
and yields
$$
G(y) := \int\limits_0^y \!\! \frac{dY}{\sqrt{2 k \sin Y}} = x
$$
