This is from a text book by Tsai Tai-Peng "Lectures on Navier-Stokes Equations", page 148 if anyone wants to read. But I am only asking about how to solve this ODE,
$$(1-t^2) L'' +2L+LL' = 0, \quad t\in(-1,1)\\L(-1)=L(1)=0.$$ The author begins by noticing that $(1-t^2)$ is a solution to the linear problem (just deleting the $LL'$ term) $$ (1-t^2) L'' +2L = 0.$$ "Therefore", as a kind of variation of parameters method, he decides to set $$ L(t) \overset{\Delta}=u(t)(1-t^2).$$ Apparently, it follows that $u$ solves $$ u' + \frac{u^2}2 = 0?$$
I really can't prove this, and must be going crazy. If you want some scratch work to see:
\begin{align} L &= (1-t^2)u\\ L’ &= (1-t^2)u’ -2tu\\ L’’ &= (1-t^2)u’’ - 2tu’ - 2tu’ - 2u\\ &= (1-t^2)u’’ - 4tu’ - 2u \end{align} divide L equation by $1-t^2$, \begin{align} (1-t^2)u’’ - 4tu’ - 2u + 2u + u L’ &= 0\\ (1-t^2)u’’ - 4tu’ + u[ (1-t^2)u’ -2tu ]&=0 \end{align}
Any pointers? This is apparently from the thesis of the author Tsai Tai-Peng but I don't have access to it at the moment.
$$\begin{align} (1-t^2)u’’ - 4tu’ + u[ (1-t^2)u’ -2tu ]&=0 \end{align}$$ You are almost there, Calvin, because the author split the original differential equation into two differential equations that are equivalents for some value of the constant of integration: $$(1-t^2)(u''+uu')=0 \tag 1$$ And also : $$-4tu'-2tu^2=0 \implies u'+\dfrac {u^2}2=0 \tag 2$$
These two equations are equivalents I guess if you equate the costant to zero since first equation gives after integration: $$(1-t^2)(u''+uu')=0 \tag 1$$ $$u''+uu'=0 $$ $$u'+\dfrac {u^2}2=C$$ Set $c=0$. Then the author deduces that: $$u(t)=\dfrac 2 {t-a}$$ where $a$ is a constant.
Another method:
You can also use substitution method : $$\begin{align} (1-t^2)u’’ - 4tu’ + u[ (1-t^2)u’ -2tu ]&=0 \end{align}$$ $$\begin{align} (1-t^2)(u’’+uu') - 4t(u’+\dfrac {u^2}2) &=0 \end{align}$$ $$\begin{align} (1-t^2)z' - 4tz &=0 \end{align}$$ Where $z=u'+\dfrac {u^2}{2}$
You can choose $z=0 \implies u'+\dfrac {u^2}2=0$.