# Reduction of Order to Solve $y'' =-y^{3}$

I want to solve $$y'' +y^3 = 0$$ with the boundary conditions $$y(0) = a$$ and $$y(k) = b$$. My goal is to reduce this problem to $$y' +y^2 = 0$$ while solving but I'm not sure it can be done.

I tried reduction of order substitutions (ie. taking $$y' = w$$ and $$y'' = \frac{dw}{dy}y'$$) but that did not work. Then I tried to solve in the following way

$$y'' y' = -y^3 y'$$

$$\frac{1}{2}[(y')^2]' = -[\frac{1}{4} y^4]'$$

$$\frac{1}{2}(y')^2 = -\frac{1}{4} y^4+C$$

$$(y')^2 = -\frac{1}{2} y^4+C$$

$$y' = \pm \sqrt{-\frac{1}{2} y^4+C}$$

It seems to me if I take my original problem to be $$y'' - 2y^3 = 0$$ instead, I get $$y' = \pm \sqrt{y^4+C}$$. If $$C=0$$, this would reduce to $$y' - y^2 = 0$$, which is close enough to what I want for my purposes. But I'm not sure how to get $$C=0$$ without a condition on the derivative, so maybe this was the wrong way to go.

1. Can I reduce my original problem, $$y'' +y^3 = 0$$, to $$y' +y^2 = 0$$?
2. Where do my boundary conditions come into play?
• $y'' = y^{-3}$ is not equivalent to $y'' + y^3 = 0$.
– Dan
Commented Nov 9, 2022 at 20:02
• @Dan sorry that was a silly typo. Fixed it :) Commented Nov 9, 2022 at 20:03
• Also, $y’+y^2=0$ has a simple solution, but the solution to $y’’=cy^a$ is here and involves elliptic functions Commented Nov 9, 2022 at 20:43

Solve $$\begin{gather*} \boxed{y^{\prime \prime}+y^{3}=0} \end{gather*}$$ Multiplying the ode by $$y^{\prime}$$ gives $$y^{\prime} y^{\prime \prime}+y^{3} y^{\prime} = 0$$ Integrating the above w.r.t $$x$$ gives \begin{align*} \int \left(y^{\prime} y^{\prime \prime}+y^{3} y^{\prime}\right)d x &= 0 \\ \frac{\left(y^{\prime}\right)^{2}}{2}+\frac{y^{4}}{4} = c_2 \end{align*} Which is now solved for $$y$$. Solving for $$y^{\prime}$$ gives \begin{align*} y^{\prime}&=\frac{\sqrt{-2 y^{4}+8 c_{2}}}{2}\tag{1} \\ y^{\prime}&=-\frac{\sqrt{-2 y^{4}+8 c_{2}}}{2}\tag{2} \end{align*}
• Thank you for your response! This matches exactly what I was getting. Is there instead a way to get the ODE $y' + y^2 = 0$ from the original problem? Commented Nov 9, 2022 at 20:48
• @k12345 I am little confused. The ode $y''+y^3=0$ does not have the same solution as $y'+y^2=0$. So how could the first lead to the second then? If there was a way, then both should have same solution, right? Which tells me it is not possible convert the first to the second. Commented Nov 9, 2022 at 20:51
$$y'+y^2=0$$ Differentiate: $$y''+2yy'=0$$ $$y''+2y(-y^2)=0$$ $$y''-2y^3=0$$ So you will never end with the second order DE you posted. $$y'' +y^3 = 0$$