# What is the solution to this non-linear second order differential equation?

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 $$u(x)$$?

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

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.

• Thanks a lot. I was expecting some trigonometric functions, or the Elliptic function at worst. This solution is sadly awefull!
– Cham
Jul 4 at 2:25
• 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 $\operatorname{sn}$. Jul 4 at 3:59

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 $$y(x) = y_0 \pm x$$.