Third order nonlinear ODE I am looking to solve the following third order nonlinear ODE:
$$\frac{\textrm{d}^{3}y}{\textrm{d}x^{3}}+\biggl(\frac{\textrm{d}y}{\textrm{d}x}\biggr)^{2}-y\frac{\textrm{d}^{2}y}{\textrm{d}x^{2}}=0,$$
subject to 
$$y(x=0)=0,\qquad\frac{\textrm{d}y}{\textrm{d}x}(x=0)=-1,\qquad\frac{\textrm{d}y}{\textrm{d}x}(x\to\infty)\to0.$$
From inspection I can see that the solution is $y(x)=e^{-x}-1$. However, I would like to be able to derive this solution for myself. I've made a couple of attempts that so far have proved unsuccessful. For example, if I set $z=\textrm{d}y/\textrm{d}x$ then
\begin{align*}
\frac{\textrm{d}^{2}y}{\textrm{d}x^{2}}&=z\frac{\textrm{d}z}{\textrm{d}y},
\\
\frac{\textrm{d}^{3}y}{\textrm{d}x^{3}}&=z\biggl(\frac{\textrm{d}z}{\textrm{d}y}\biggr)^{2}+z^{2}\frac{\textrm{d}^{2}z}{\textrm{d}y^{2}}.
\end{align*}
So that
$$z\biggl(\frac{\textrm{d}z}{\textrm{d}y}\biggr)^{2}+z^{2}\frac{\textrm{d}^{2}z}{\textrm{d}y^{2}}+z^{2}-yz\frac{\textrm{d}z}{\textrm{d}y}=0.$$
The above could be rewritten like so
$$\frac{\textrm{d}}{\textrm{d}y}\biggl(z\frac{\textrm{d}z}{\textrm{d}y}\biggr)+z-y\frac{\textrm{d}z}{\textrm{d}y}=0,$$
which is equivalent to 
$$\frac{\textrm{d}}{\textrm{d}y}\biggl(z\frac{\textrm{d}z}{\textrm{d}y}\biggr)+z^{2}\frac{\textrm{d}}{\textrm{d}y}\biggl(\frac{y}{z}\biggr)=0.$$
Any suggestions as to where to go from here or am I barking up the wrong tree?
Thanks
 A: THIS NOTE MAY HELP
We want to solve
$$
y'''+(y')^2-yy''=0\tag 1
$$
Instead of (1) I will solve
$$
y'''-(y')^2-yy''=0\tag 2
$$
We have
$$
y'''-(y')^2-yy''=0\Leftrightarrow y'''-(yy')'=0\Leftrightarrow y''-yy'=-C_1\Leftrightarrow
$$
$$
(y'-\frac{y^2}{2})'=(-C_1x)'\Leftrightarrow y'-y^2/2=-C_1x-C_2
$$
If we set $y=-2u'/u$ we arrive to
$$
u''=\frac{1}{2}(C_1x+C_2)u
$$ 
The last equation is solvable with Airy $\textrm{Ai}(x)$,$\textrm{Bi}(x)$ functions see Wikipedia. 
$$
y(x)=-2^{2/3}C_1^{1/3}\frac{\textrm{Bi}'\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)+\textrm{Ai}'\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)C_3}{\textrm{Bi}\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)+\textrm{Ai}\left(\frac{C_1x+C_2}{2^{1/3}C_1^{2/3}}\right)C_3}
$$
For the conditions $y(0)=0$, $y'(0)=-1$, $y'(\infty)=0$, we easily get  $C_1=1/2$,$C_2=1$,$C_3=-\textrm{Bi}'(2^{1/3})/\textrm{Ai}'(2^{1/3})$.
A: I found a solution to the ODE, but it does not fit all initial conditions.
Ignoring initial conditions, a solution to this ODE is $$y=\frac{-6c}{cx+d}. $$
How I found this solution:
We have $y'''=yy''-(y')^2$. By induction, we can prove
$$y^{(2n)}=\sum_{r=0}^{n}b_r^{2n}y^{(r)}y^{(2n-1-r)}$$
and $$y^{(2n+1)}=\sum_{r=0}^{n}b_r^{2n+1}y^{(r)}y^{(2n-r)}.$$
Now let $n=1$, (I know that it is not true, but it helps to find a solution.) we have
$$y''=b_0^{2}yy'+b_1^{2}yy'=(b_0^{2}+b_1^{2})yy'=\frac{b_0^{2}+b_1^{2}}{2}(y^2)'$$
which has the solution of the form
$$y=\frac{a}{cx+d}. $$
Plugging this $y$ in the ODE, shows that $a=-6c$.
A: Here is also my note (unfortunately, probably also not super helpful).
$$y'''+(y')^2-yy''=0\quad\quad(1)$$
Can be rewritten as:
$$y'''=y^2\left(\frac{y'}{y}\right)'$$.
Integrating both sides (using integration by parts), this leads to:
$$y''=yy'-2\int(y')^2dx + C.$$
Thus, $$2\int(y')^2dx=yy'-y''+C=\left(\frac{y^2}{2}\right)'-y''+C=\left(\frac{y^2}{2}-y'\right)'+C.\quad\quad(2)$$
Thus the differential equation is also equal to:
$$\int \left(y'\right)^2dx = \left(\left(\frac{y}{2}\right)^2-\frac{y'}{2}\right)'+C$$
From this point on, the derivation gets somehow fishy, but if we assume that the integral $\int (y')^2dx$ is zero (or a constant) (but of course, I don't know why this should be the case), we arrive to a differential equation
$$\left(\frac{y^2}{2}-y'\right)'=-C,$$
which has a solution
$y(x)=ae^x+be^{-x}-C$. Plugging in the initial conditions, we get $a=0, b=-1, C = 1$ as we guessed at the first place.
But I hope someone will come up with a better solution:)
A: THIS NOTE MAY (NOT) HELP 
$y'''-y'=0\rightarrow y=Ae^x+Be^{-x}+Ce^{0x}\rightarrow_{B.C} A=0,B=1,C=-1$ so this equation determines uniquely that $y=e^x-1$ but it also solves $(y^{(6)}-y^{(4)})+(y'''+(y')^2-yy'')=0$ for example so it lives in the space of solution of this last equations but it is not clear it belong to the basis of that space.
Instead of solving $y'''-y'=0$ directly one could use it to reduce the other equation, 
$$0=(y^{(6)}-y^{(4)})+(y'''+(y')^2-yy'')$$
$$\rightarrow_{[y'''=y'\Rightarrow y^{(6)}=y^{(4)}]} 0=y'''+(y')^2-yy'' \quad\text{(our equation)}$$
$$\rightarrow_{\frac{d}{dx}}0=y''''+y'y''-yy'''$$
$$\rightarrow_{[y'''=y'\Rightarrow y''''=y'' \text{ & } y''=y+D]}0=(y+D)+y'(y+D)-y(y')=Dy'+y+D=De^{-\frac{x}{D}}(e^{\frac{x}{D}}y)'+D$$
$$\rightarrow y=Fe^{-\frac{x}{D}}-D$$
$$\rightarrow_{BC\Rightarrow F=1,D=1} y=e^{-x}-1$$
And we get the same solution again anyway, so $y'''-y'=0$ helps obtain a solution but it doesn't really helps us solve the original equation $0=y'''+(y')^2-yy''$ (or $0=(y^{(6)}-y^{(4)})+(y'''+(y')^2-yy'')$)
A: Thanks all for the suggestions. Some interesting ideas.
I noticed that making the substitution $z=(\textrm{d}y/\textrm{d}x)^{2}$, gives the following
$$\frac{\textrm{d}^{2}z}{\textrm{d}y^{2}}+4z\frac{\textrm{d}}{\textrm{d}y}\biggl(\frac{y}{2\sqrt{z}}\biggr)=0.$$
Not really sure that helps though...
A: Here would be my solution. We start with the equation $$y'''+(y')^2-yy''=0.$$
Then, if we differentiate it twice we get $$y^{(5)}+(y'')^2-yy^{(4)}=0$$
which is intriguingly similar.
Now, one of the solution (unique?) would be if the set of equations simultaneously hold, i.e.,
$$y^{(5)}=y''' \tag{1}$$
$$y'' = y'\quad\text{or}\quad y''=-y'\tag{2}$$
$$y^{(4)}=y''\tag{3}$$
General solutions of the differential equations:
$$(1):\quad y(x)=a_1x^2+b_1x+c_1e^x+d_1e^{-x}+f_1$$
$$(2):\quad y(x)=a_2e^x+b_2\quad\text{or}\quad y(x)=a_2e^{-x}+b_b$$
$$(3):\quad y(x)=a_3x+b_3e^x+c_3e^{-x}+d_3$$
Plugging the initial conditions into $(1)$:
$$y(0)=0\ \to \ c_1+d_1+f_1=0\\
y'(0)=-1\ \to \ b_1+c_1-d_1=-1\\
y'(\infty)=0\ \to \ a_1=b_1=c_1=0,\ \text{and}\ d_1\ \text{is any number},$$
and thus $a_1=0, b_1=0, c_1=0,d_1=1,f_1=-1, $ leading to a unique solution $$y(x)=e^{-x}-1.$$
We now have to check if that solution holds also for equations $(2)$ and $(3)$. But that holds obviously for the right constants.
