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I'm trying to solve the following non-linear second order differential equation: $$\tag{1} \frac{d\, }{dx} \Bigl( \frac{1}{y^2} \, \frac{dy}{dx} \Bigr) = -\, \frac{2}{y^3}, $$ where $y(x)$ is an unknown function on the real axis. I already know the "trivial" solution $y(x) = y_0 \pm x$. The solution I'm looking may involve trigonometric functions, but I'm not sure. Take note that we may pose $$\tag{2} u = \frac{1}{y}, $$ so that (1) takes another form: $$\tag{3} \frac{d^2 u}{dx^2} - 2 \, u^3 = 0. $$ So what is the non-linear solution $y(x)$? Or $u(x)$?

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    $\begingroup$ Try multiplying by $u’$ and integrating! $\endgroup$ Jul 4 at 2:05
  • $\begingroup$ @gt6989b, I don't understand... your last equation doesn't make any sense. $\endgroup$
    – Cham
    Jul 4 at 2:18

2 Answers 2

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Solve
\begin{align*} u^{\prime \prime}&=2 u^{3} \end{align*} Multiplying both sides by $u^{\prime}$ gives \begin{align*} u^{\prime} u^{\prime \prime}&=2 u^{3} u^{\prime} \end{align*} Integrating both sides w.r.t. $x$ gives \begin{align*} \int{u^{\prime} u^{\prime \prime}\, \mathrm{d}x} &=\int{2 u^{3} u^{\prime}\, \mathrm{d}x}\\ \int{u^{\prime} u^{\prime \prime}\, \mathrm{d}x} &=\int{2 u^{3}\, \mathrm{d}u} \tag{1} \end{align*} But $$ \int{u^{\prime} u^{\prime \prime}\, \mathrm{d}x} = \frac{1}{2} \left(u^{\prime}\right)^2 $$ Hence equation (1) becomes \begin{align*} \frac{1}{2} \left(u^{\prime}\right)^2 &=\int{2 u^{3}\, \mathrm{d}u} \tag{2} \end{align*} But $$ \int{2 u^{3}\, \mathrm{d}u} = \frac{u^{4}}{2} $$ Therefore equation (2) becomes \begin{align*} \frac{1}{2} \left(u^{\prime}\right)^2 &=\frac{u^{4}}{2} + c_2 \end{align*} Where $c_2$ is an arbitrary constant of integration.

This is first order ODE which is now solved for $u$.

Solving for $u^{\prime}$ gives \begin{align*} u^{\prime}&=\sqrt{u^{4}+2 c_{2}}\tag{1} \\ u^{\prime}&=-\sqrt{u^{4}+2 c_{2}}\tag{2} \end{align*}

Let just solve (1) as (2) is similar.

\begin{align*} \frac{1}{\sqrt{u^{4}+2 c_{2}}}\mathop{\mathrm{d}u} &= \mathop{\mathrm{d}x}\\ \int \frac{1}{\sqrt{u^{4}+2 c_{2}}}\mathop{\mathrm{d}u} &= \int \mathop{\mathrm{d}x}\\ \int \frac{1}{\sqrt{u^{4}+c_{3}}}\mathop{\mathrm{d}u} &= x +c_{1} \end{align*} Using the computer, integrating the left side above gives the solution

\begin{align*} \frac{\sqrt{1-\frac{i u^{2}}{\sqrt{c_{3}}}}\, \sqrt{1+\frac{i u^{2}}{\sqrt{c_{3}}}}\, \operatorname{EllipticF}\left(u \sqrt{\frac{i}{\sqrt{c_{3}}}}, i\right)}{\sqrt{\frac{i}{\sqrt{c_{3}}}}\, \sqrt{u^{4}+c_{3}}} = x +c_{1} \end{align*}

Similar solution for the second ode.

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  • $\begingroup$ Thanks a lot. I was expecting some trigonometric functions, or the Elliptic function at worst. This solution is sadly awefull! $\endgroup$
    – Cham
    Jul 4 at 2:25
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    $\begingroup$ The solution can be written more simply: When $c_3 < 0$, for example, we can write $c_3 = -c^4$ for some c, in which case the solution is $x = \frac{1}{c} F\left(\frac{i}{c} u, i\right) + d$, where $F$ is the incomplete elliptic integral of the first kind, more or less immediately from the definition of that function. Inverting then gives $u(x)$ (and thus $y(x)$) explicitly in terms of the Jacobi elliptic function $\operatorname{sn}$. $\endgroup$ Jul 4 at 3:59
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Multiplying by $u'$ and regrouping gives an equality of derivatives $$\frac{1}{2} ((u')^2)' = \frac{1}{2} (u^4)',$$ and integrating gives $$(u')^2 = u^4 + c .$$ This equation is separable, and rearranging and integrating gives $$\pm \int \frac{du}{\sqrt{u^4 + c}} = x + d .$$

If $c < 0$, we can write $c = -a^4$ for some $a > 0$, and in the variable $v = \frac{u}{a i}$ the left-hand side is $$\pm \frac{i}{a} \int \frac{dv}{\sqrt{v^4 - 1}} = \pm \frac{i}{a} F(iv, i) = \pm \frac{i}{a} F\left(\frac{u}{a}, i\right) ,$$ where $F$ is the incomplete elliptic integral of the first kind and where we've absorbed the constant of integration into $d$. Then, we can write $u$ in terms of the Jacobi elliptic function $\operatorname{sn}$: $$u(x) = \mp a i \operatorname{sn} (a(x + d), i) .$$ The reciprocal of $\operatorname{sn}$ is denoted $\operatorname{ns}$, and so we may write $$y(x) = \pm \frac{i}{a} \operatorname{ns} (a(x + d), i) .$$

The case $c > 0$ can be handled similarly, and the case $c = 0$ yields the linear solutions $y(x) = y_0 \pm x$.

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