second derivative of a function equals the function squared Can someone solve the following differential equation for me please?
The second derivative of a function equals the function squared.
Find $y(x)$ if 
$$
\frac{d^2 y}{dx^2} = y^2
$$
 A: How familiar are up with Weierstrass elliptic functions?
Let $\mathcal{P}(x;a,b)$ be that value of $z$ which makes 
$$
\int_{-\infty}^z \frac{1}{\sqrt{4t^3 - at - b}}dt = x
$$ 
This is the Weierstrass $\mathcal{P}$ function.
The general solution to 
$$
\frac{d^2 y}{dx^2} = y^2
$$
is 
$$
y = \sqrt[3]{6} \,\mathcal{P} \left( \frac{x+c_1}{\sqrt[3]{6}}; 0, c_2\right)
$$
where $c_1$ and $c_2$ are constants determined by the initial conditions.
The Weierstrass $\mathcal{P}$ function looks kind of like your top row of front teeth, as seen by a nearsighted dentist.
A: This answer is just a partial solution because it consists of complicated integral that you and me might not know. I will try my best to do it.
Firstly, multiply both side by $\dfrac{dy}{dx}$ and we get $$\dfrac{d^2y}{dx^2}\dfrac{dy}{dx}=y^2\dfrac{dy}{dx}$$
Observe the antiderivative on both side, $\dfrac{d}{dx}\left[\dfrac{1}{2}\left(\dfrac{dy}{dx}\right)^2\right]=\dfrac{d^2y}{dx^2}\dfrac{dy}{dx}\text{ and }\dfrac{d}{dx}\left(\dfrac{y^3}{3}\right)=y^2\dfrac{dy}{dx}$. Therefore, $$\dfrac{1}{2}\left(\dfrac{dy}{dx}\right)^2=\dfrac{y^3}{3}+C \text{ (}C\text{ is a constant)}$$
After that, we start to rearrange it. We multiply by $2$ and take a square root both side, we get $$\dfrac{dy}{dx}=\sqrt{\dfrac{2y^3}{3}+2C}$$
and we can arrange into a better way for integral which is $$\dfrac{dy}{\sqrt{\dfrac{2y^3}{3}+2C}}=dx \rightarrow \int\dfrac{dy}{\sqrt{\dfrac{2y^3}{3}+2C}}=\int dx=x+c_1\text{ that }c_1\text{ is a constant}$$
Then, I stopped here because the left hand side is really hard to solve. There is a link which you may take a look. If I can, I may try to finish the integral. Thank you.
