$y''+y'^{2}+y=0$ 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
 A: $$
y'' + y'^2 + y = 0
$$
we can rewrite as this (prove it)
$$
\frac{1}{2}\frac{d}{dy}p^2 + p^2 + y = \frac{d}{dy}p^2 + 2p^2 + 2y = 0
$$
where $p = y'$.
$$
p^2 = \mathrm{e}^{-2y}\left(-\int 2y\mathrm{e}^{2y}dy + C \right)
$$
thus solution  in terms of $p$
$$
p = \sqrt{\frac{1}{2} -y + C\mathrm{e}^{-2y}} = y'
$$
or
$$
\int \frac{dy}{\sqrt{\frac{1}{2} -y + C\mathrm{e}^{-2y}}} = \int dx = x + \lambda
$$
now the lhs probably has no closed form but if choose the special case $C=0$
$$
\int \frac{dy}{\sqrt{\frac{1}{2}-y}} = -2\left(\frac{1}{2}-y\right)^{1/2} = x + \lambda
$$
thus
$$
\frac{1}{2} - y = \left(-\frac{x}{2} + \lambda_1\right)^2 = \frac{x^2}{4} -\lambda_1 x + \lambda_1^2 
$$
or
$$
y = -\frac{x^2}{4} +\lambda_1 x - \lambda_1^2 +\frac{1}{2}
$$
A: 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.
A: Let us introduce $y$ as new independent variable. Then we have  $y'^2=u(y)$ with a certain unknown function $u$. It follows that $2y'y''=\dot u(y)\>y'$, or $y''={1\over2}\dot u$, where the $\cdot$ denotes differentiation with respect to $y$. Inserting this into the given ODE we obtain the first order equation
$${1\over2}\dot u+u=-y$$
with the general solution
$$ u(y)=Ce^{-2y}-y+{1\over2}\ .$$
As $y'=\sqrt{u(y)}$ (up to sign) we can separate variables:
$$dx={dy\over\sqrt{Ce^{-2y}-y+{1\over2}}}\ .$$
When $C\ne0$ the resulting integral will be nonelementary. The special case $C=0$ leads to
$$x-x_0=\int_{y_0}^y{dt\over\sqrt{{1\over2}-t}}=-2\sqrt{{1\over2}-t}\Biggr|_{y_0}^y\ ,$$
a family of parabolas.
