Are there solutions to $y''+a(y')^2=0$? I can't find any solved answers online. I don't know very much about solving differential equations, only first order ODEs...
 A: Hint:
Set $u(x)=y'(x)$ and solve for $u$ first.
A: \begin{align}
  y''+a(y')^2 &= 0 \\
  \frac{dy'}{dx}+a(y')^2 &= 0 \\
  \frac{dy'}{dy} \frac{dy}{dx}+a(y')^2 &= 0 \\
  y'\frac{dy'}{dy}+a(y')^2 &=0 \\
  y' \left( \frac{dy'}{dy}+ay' \right) &= 0
\end{align}


*

*Case I:  $y'=0$


$$\fbox{$y(x)=C$}$$


*

*Case II:  $y'\ne 0$


\begin{align}
  \frac{dy'}{dy}+ay' &= 0 \\
  \frac{dy'}{y'} &= -a\, dy \\
  \int   \frac{dy'}{y'} &= -a\int dy \\
  \ln y' &= -ay+c \\
  \frac{dy}{dx} &= e^{-ay+c}  \\
  \int e^{ay}\, dy &= \int e^{c}\, dx \\
  \frac{e^{ay}}{a} &= e^{c}x+k \\
  e^{ay} &= Ax+B \tag{$A=ae^{c}, B=ak$} \\
\end{align}
$$\fbox{$y=\frac{\ln (Ax+B)}{a}$}$$
In particular, case II will degenerate into case I when $A=0$ is included.
A: Preliminary remarks:  the equation needs initial conditions to completely specify a solution.  We thus assume, for some $x_0 \in \Bbb R$, we are given $y(x_0)$ and $y'(x_0)$; we also observe that if $a = 0$, we have $y''(x) = 0$ everywhere, and thus 
$y(x) = y'(x_0)(x - x_0) + y(x_0); \tag 0$
we therefore assume for the remainder of this answer that $a \ne 0$.  Furthermore, if $y'(x_0) = 0$, then we see that $y(x) = y(x_0)$ is the unique solution of $y''(x) + a (y'(x))^2 = 0$ which is valid for all $x$; thus we may assume $y'(x_0) \ne 0$.  
Near a point $x_0$ where 
$y'(x_0) \ne 0 \tag 1$
we may write
$(y'(x))^{-2}y''(x) = -a; \tag 2$
we note that
$((-y'(x))^{-1})' = (y'(x))^{-2}y''(x); \tag 3$
thus
$((-y'(x))^{-1})' = -a; \tag 4$
near $x_0$ such that $y'(x_0) \ne 0$ we may integrate 'twixt $x_0$ and $x$
$\displaystyle \int_{x_0}^x ((-y'(s))^{-1})' \; ds = -\int_{x_0}^x a \; ds; \tag 5$
since
$\displaystyle \int_{x_0}^x ((-y'(s))^{-1})' \; ds = -\int_{x_0}^x ((y'(s))^{-1})' \; ds, \tag 6$
(5) becomes
$\displaystyle \int_{x_0}^x ((y'(s))^{-1})' \; ds = \int_{x_0}^x a \; ds; \tag 7$
therefore
$(y'(x))^{-1} - (y'(x_0))^{-1} = a(x - x_0), \tag 8$
whence
$(y'(x))^{-1} = a(x - x_0) +  (y'(x_0))^{-1}; \tag 9$
thus,
$y'(x) = \dfrac{1}{a(x - x_0) + (y'(x_0))^{-1}} = \dfrac{y'(x_0)}{y'(x_0)a(x - x_0) + 1}, \tag{10}$
from which
$y(x) - y(x_0) = \displaystyle \int_{x_0}^x y'(s) \; ds = y'(x_0) \int_{x_0}^x \dfrac{ds}{y'(x_0)a(s - x_0) + 1}$
$ = \dfrac{1}{a}\ln(y'(x_0)a(x - x_0) + 1), \tag{11}$
or
$y(x) = \dfrac{1}{a}\ln(y'(x_0)a(x - x_0) + 1) + y(x_0). \tag{12}$
this is very easily checked:  clearly differentiating (12) yields (10), from which
$y''(x) = -y'(x_0)(y'(x_0) a)(y'(x_0)a(x - x_0) + 1)^{-2}$
$= -a \left (\dfrac{y'(x_0)}{y'(x_0)a(x - x_0) + 1} \right )^2, \tag{13}$
which is easily seen to satisfy the equation
$y''(x) + a (y'(x))^2 = 0. \tag{14}$
In closing, we observe that the solution (10) may be extended in $x$ as long as
$y'(x_0)a(x - x_0) + 1 \ne 0; \tag{15}$
for $ay'(x_0) > 0$, $y(x)$ may be continued for arbitrarily large $x > x_0$, but only as long as
$x > x_0 - \dfrac{1}{ay'(x_0)} \tag{16}$
in the negative direction.  The corresponding result, with the directions reversed, holds for $ay'(0) < 0$.  Lastly, it is easily seen from (10) that $y'(x)$ never changes sign over its domain of definition.
