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Can someone solve the following differential equation for me please? The second derivative of a function equals the function squared.

Find $y(x)$ if $$ \frac{d^2 y}{dx^2} = y^2 $$

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

2 Answers 2

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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.

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

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  • $\begingroup$ 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. $\endgroup$
    – amsmath
    Aug 29, 2019 at 15:57

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