Help needed to solve third order differential equation I am newbie to calculus and having hard time to figure out
$y^{\prime\prime\prime} + \frac{1}{2}y\cdot y^{\prime\prime} = 0 $
One way to think about this was
$D^3 + \frac{1}{2}\cdot y\cdot D^2$
Where $D=y^\prime$
$D^2(D + \frac{1}{2}\cdot y) = 0 $
$D^2 = 0 $, 
$D = \frac{1}{2}\cdot y$
Am I doing it right?

 A: $$\frac{d^3y}{dx^3}+\frac12 y\frac{d^2y}{dx^2}=0$$
This ODE is of the autonomous kind. The usual change of function is 
$$\frac{dy}{dx}=u(y)\qquad\text{to not be confused with } u(x)$$
$$\frac{d^2y}{dx^2}=\frac{du}{dy}\frac{dy}{dx}=u\frac{du}{dy}$$
$$\frac{d^3y}{dx^3}=\frac{d(u\frac{du}{dy})}{dy}\frac{dy}{dx}=\left(u\frac{d^2u}{dy^2}+(\frac{du}{dy})^2\right)u$$
$$\left(uu''+(u')^2\right)u+\frac12 yuu'=0$$
There is a trivial solution $u(y)=0$ which leads to $y(x)=$constant, rejected because it doesn't satisfy the boundary conditions.
$$uu''+(u')^2+\frac12 yu'=0$$
Again the trivial $u'(y)=0$  leading to $y(x)=c_1x+c_2$ is rejected for the same reason.
$$\left(uu'\right)'+\frac12 yu'=0$$
Unfortunately, this second order non-linear ODE seems not solvable for a general solution in terms of standard functions. The most likely numerical calculus is required.
Of course, it is possible to find some particular solutions such as $u(y)=-\frac16 y^2$ which leads to some particular solutions of $y'''+\frac12 yy''=0$ such as $y(x)=\frac{6}{x+c}$ not compatible to the boundary conditions.
Nevertheless, if the whole problem is an academic problem, this draw to think that something might be wrong in the wording. May be a typo in the ODE or in the boundary conditions $y(0)=y'(0)=0$ and $y(\infty)=1.$
A: Hint:
With reference to http://eqworld.ipmnet.ru/en/solutions/ode/ode0503.pdf,
Let $u=\left(\dfrac{dy}{dx}\right)^2$ ,
Then $\dfrac{du}{dx}=2\dfrac{dy}{dx}\dfrac{d^2y}{dx^2}$
$\dfrac{du}{dy}\dfrac{dy}{dx}=2\dfrac{dy}{dx}\dfrac{d^2y}{dx^2}$
$2\dfrac{d^2y}{dx^2}=\dfrac{du}{dy}$
$2\dfrac{d^3y}{dx^3}=\dfrac{d}{dx}\left(\dfrac{du}{dy}\right)=\dfrac{d}{dy}\left(\dfrac{du}{dy}\right)\dfrac{dy}{dx}=\pm\sqrt u\dfrac{d^2u}{dy^2}$
$\therefore\pm2\sqrt u\dfrac{d^2u}{dy^2}+y\dfrac{du}{dy}=0$
Which reduces to a generalized Emden-Fowler equation
