Mathematica gives the following result: $6^{1/3} \times$ WeierstrassP[$\frac{x + C_1}{6^{1/3}}$, {$0, C_2$}], where $C_1 \& C_2$ are constants of integration.

It is possible to find the series expansion of the same to an arbitrary precision. But I am interested in the method through which this solution has been arrived at. Also, is the solution unique?

Thanks in advance.

  • 1
    $\begingroup$ did you try to multiply by y' and integrate $\endgroup$ – hamam_Abdallah Sep 27 '16 at 16:23
  • $\begingroup$ @AbdallahHammam Thanks. Apparently that should give the closed form solution. The detailed answer is given below. Do you happen to know, how this solution could be transformed to a Weierstrass P function? $\endgroup$ – Ayyappadas Sep 27 '16 at 16:37

$$y''(x)=y(x)^2\Longleftrightarrow\int y'(x)y''(x)\space\text{d}x=\int y'(x)y(x)^2\space\text{d}x$$


  1. Substitute $u=y'(x)$ and $\text{d}u=y''(x)\space\text{d}x$: $$\int y'(x)y''(x)\space\text{d}x=\int u\space\text{d}x=\frac{u^2}{2}+\text{C}=\frac{y'(x)^2}{2}+\text{C}$$
  2. Substitute $s=y(x)$ and $\text{d}s=y'(x)\space\text{d}x$: $$\int y'(x)y(x)^2\space\text{d}x=\int s^2\space\text{d}s=\frac{u^3}{3}+\text{C}=\frac{y(x)^3}{3}+\text{C}$$

So, we get:

$$\frac{y'(x)^2}{2}=\frac{y(x)^3}{3}+\text{C}\Longleftrightarrow\int\frac{y'(x)}{\sqrt{\text{C}+\frac{2y(x)^3}{3}}}\space\text{d}x=\pm\int1\space\text{d}x=\text{K}\pm x$$

Where $\text{C}$ and $\text{K}$ are arbitrary constants.

So, you need to solve, using a substitution $p=y(x)$ and $\text{d}p=y'(x)\space\text{d}x$:

$$\int\frac{y'(x)}{\sqrt{\text{C}+\frac{2y(x)^3}{3}}}\space\text{d}x=\text{K}\pm x\Longleftrightarrow\sqrt{3}\int\frac{1}{\sqrt{\text{C}+2p^3}}\space\text{d}p=\text{K}\pm x$$

For $\int\frac{1}{\sqrt{\text{C}+2p^3}}\space\text{d}p$, using the integration by parts:


That gives us:


  • $\begingroup$ Thanks for the solution. Is there a way to reduce this to a Weierstrass P function? $\endgroup$ – Ayyappadas Sep 27 '16 at 16:32
  • $\begingroup$ @Ayyappadas Look at WolframAlpha, because I don't know that. $\endgroup$ – Jan Sep 27 '16 at 16:39

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