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

Question

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?

Answer 1

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*}

These are separable which can be solved by integration.

Answer 2

$$y'+y^2=0$$ 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. $$y'' +y^3 = 0$$