<s>[INST] $y''+y'^{2}+y=0$ equation solution How would you solve this differential equation $y''+y'^{2}+y=0$? I can't apply the ansatz method (or more formally apply the characteristic polynomial method). Thanks
 [/INST] Consider $y=ax^2+bx+c,$ then we have $y'=2ax+b, (y')^2=4a^2x^2+4abx+b^2, y''=2a$ which leads to
$$c=-2a-b^2\\
-ax^2-bx=4a^2x^2+4abx\\
-ax-b=4a^2x+4ab\\
-ax-b=4a(ax+b)\\
a=-\frac 14\text{ or }ax+b=0$$
$b$ is a free variable when $a=-\frac 14$, leading to a set of solutions $y=-\frac 14x^2+bx+\frac 12-b^2$.  When $ax+b=0$ we get the "set" of solutions $a=b=c=0$.  Double checking with the original for $a=-\frac 14$, we get
$$y''+(y')^2+y=0\\
=-\frac 12+\frac 14x^2-bx+b^2-\frac 14x^2+bx+\frac 12-b^2$$
which reduces to $0$ as required.
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