# 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 $$\frac{d^2 y}{dx^2} = y^2$$

• Where the equation? Aug 28, 2019 at 20:39
• Here the equation. Aug 28, 2019 at 20:42
• Not the general solution - but $y = \frac{6}{(x+C)^2}$ forms a family of simple solutions. Aug 29, 2019 at 3:27

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.

• How do you integrate $1/|t|$ at $-\infty$? Aug 28, 2019 at 21:09
• Sorry, you caught a typo -- the first term under the radical was supposed to be $4t^3$, not $4t^2$. Corrected now. Aug 29, 2019 at 15:22
• Oh, I see. Thanks for clearing this up. Aug 29, 2019 at 15:41

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.

• Well, you can express it with the Weierstraß function posted by Mark Fishler in his answer. You just showed a nice way how to get there. Aug 29, 2019 at 15:57