How to solve the differential equation: yy''+2y'''=0 How do I solve yy''+2y'''=0 with B.Cs y(0)=y'(0)=0 & y'(∞)=1 ? Can somebody please hint at some substitution or refer any text related to these type of ode?
 A: NOT A FINAL ANSWER TO THE QUESTION
First, we use the substitution $v=y',u=y$. 
We have: $$\frac{\mathrm{d}v}{\mathrm{d}u}=\frac{\mathrm{d}y'}{\mathrm{d}y}=\frac{\frac{\mathrm{d}y'}{\mathrm{d}x}}{\frac{\mathrm{d}y}{\mathrm{d}x}}=\frac{y''}{y'}=\frac{y''}{v}$$ 
So $$y''=\frac{\mathrm{d}v}{\mathrm{d}u}v$$ 
And: $$\frac{\mathrm{d}^2v}{\mathrm{d}u^2}=\frac{\mathrm{d}\left(\frac{\mathrm{d}v}{\mathrm{d}u}\right)}{\mathrm{d}u}=\frac{\mathrm{d}\left(\frac{y''}{y'}\right)}{\mathrm{d}y}=\frac{\frac{\mathrm{d}\left(\frac{y''}{y'}\right)}{\mathrm{d}x}}{\frac{\mathrm{d}y}{\mathrm{d}x}}=\frac{\left(\frac{y'''y'-\left(y''\right)^2}{\left(y'\right)^2}\right)}{y'}=\frac{y'''v-\left(\frac{\mathrm{d}v}{\mathrm{d}u}v\right)^2}{v^3}$$ 
So $$y'''=\frac{\mathrm{d}^2v}{\mathrm{d}u^2}v^2+\left(\frac{\mathrm{d}v}{\mathrm{d}u}\right)^2v$$ 
Plugging in the above pair of expressions for $y''$ and $y'''$ along with $y=u$ (which holds by definition) into the original differential equation, we get: $$2\frac{\mathrm{d}^2v}{\mathrm{d}u^2}v^2+2\left(\frac{\mathrm{d}v}{\mathrm{d}u}\right)^2v+\frac{\mathrm{d}v}{\mathrm{d}u}uv=0$$ 
Dividing through by $v$ and denoting $\frac{\mathrm{d}v}{\mathrm{d}u}$ and $\frac{\mathrm{d}^2v}{\mathrm{d}u^2}$ by $v'$ and $v''$, respectively: $$2v''v+2(v')^2+v'u=0$$ 
And this way, we have reduced the order of the differential equation by $1$. We can go further. First, divide through the second-order equation by $v'\sqrt v$: $$2\frac{v''\sqrt v}{v'}+2\frac{v'}{\sqrt v}+\frac{u}{\sqrt v}=0$$ 
Now, we employ the substitution $w=2\mathrm{ln}(u)-\mathrm{ln}(v),z=2\mathrm{ln}(v')-\mathrm{ln}(v)$.  
It is immediate that $e^{\frac{w}{2}}=\frac{u}{\sqrt v}$ and $e^{\frac{z}{2}}=\frac{v'}{\sqrt v}$. Also: $$\frac{\mathrm{d}z}{\mathrm{d}w}=\frac{\mathrm{d}\left(2\mathrm{ln}(v')-\mathrm{ln}(v)\right)}{\mathrm{d}\left(2\mathrm{ln}(u)-\mathrm{ln}(v)\right)}=\frac{\frac{\mathrm{d}\left(2\mathrm{ln}(v')-\mathrm{ln}(v)\right)}{\mathrm{d}u}}{\frac{\mathrm{d}\left(2\mathrm{ln}(u)-\mathrm{ln}(v)\right)}{\mathrm{d}u}}=\frac{2\frac{v''}{v'}-\frac{v'}{v}}{2\frac{1}{u}-\frac{v'}{v}}$$$$=\frac{2\frac{v''\sqrt v}{v'}-\frac{v'}{\sqrt v}}{2\frac{\sqrt v}{u}-\frac{v'}{\sqrt v}}=\frac{2\frac{v''\sqrt v}{v'}-e^{\frac{z}{2}}}{2e^{-\frac{w}{2}}-e^{\frac{z}{2}}}$$ 
Rearranging and plugging everything back into the second-order equation, we get: $$\left(2e^{-\frac{w}{2}}-e^{\frac{z}{2}}\right)\frac{\mathrm{d}z}{\mathrm{d}w}+3e^{\frac{z}{2}}+e^{\frac{w}{2}}=0$$ Finally, a first-order equation. It is still an extremely unfriendly equation, though. We may rectify this with an additional substitution, $s=e^{\frac{z}{2}}, t=e^{-\frac{w}{2}}$. We then have: $$\frac{\mathrm{d}s}{\mathrm{d}t}=\frac{\mathrm{d}\left(e^{\frac{z}{2}}\right)}{\mathrm{d}\left(e^{-\frac{w}{2}}\right)}=\frac{\frac{\mathrm{d}\left(e^{\frac{z}{2}}\right)}{\mathrm{d}w}}{\frac{\mathrm{d}\left(e^{-\frac{w}{2}}\right)}{\mathrm{d}w}}=-\frac{\frac{\mathrm{d}z}{\mathrm{d}w}e^{\frac{z}{2}}}{e^{-\frac{w}{2}}}=-\frac{\frac{\mathrm{d}z}{\mathrm{d}w}s}{t}$$ Rearranging and substituting, once again, we get: $$(s-2t)\frac{\mathrm{d}s}{\mathrm{d}t}+3\frac{s^2}{t}+\frac{s}{t^2}=0$$ 
Finally, substitute $p=s-2t$ and leave $t$ as it is (I'm going to skip a bunch of working here): $$\frac{\mathrm{d}p}{\mathrm{d}t}+\frac{3}{t}p+\left(14+\frac{1}{t^2}\right)+\left(12t+\frac{2}{t}\right)\frac{1}{p}=0$$ 
This is a Chini equation. 
You can find some links in the accepted answer here. But this particular case of the Chini equation does not look like it'll have a clean solution. This exercise in reduction/ simplification of the differential equation serves a couple of purposes. 
First, you may be able to find references on solving the final equation (or, at least, bona fide approximations), as it is a previously studied type. Once you have something of a solution to the final equation, you should be able to reverse (at least, partially) the substitutions.
Secondly, even if you are not able to solve it in the general, these reduced differential still provide different ways of looking at the problem that can lend to better solution guessing. 
For instance, if you look at the second-order equation, $2v''v+2(v')^2+v'u=0$, you may easily guess that setting $v$ equal to a quadratic in $u$ may solve the equation (for some suitable parameters). You'd be right. And this guess will give you a solution to the original equation, $y'=\frac{6}{x+D}$. Making similar guesses based on observations about other versions of the equation may yield solutions that conform more to the requirements of your question. 
In any case, I hope this answer gives you an appreciation for how difficult finding closed-form solutions (if they even exist) to an 'average' differential equation can be. And this one was still relatively amenable to partial simplification. 
By the way, some of the substitutions used were pure ansatzes. But others used a working knowledge of the theory of Lie point symmetries. You can find many resources for the basics of how to use symmetry methods online. 
Note: if there any mistakes in the answer, feel free to edit in the corrections or comment them 
EDIT: A final substitution of $q=\frac{1}{p}$ turns the equation into an Abel equation which does seem to really have a closed-form solution, although, given its complexity and the amount of reverse-substitutions that would need to be performed, on top of having to solve for (or, more likely, approximate) the unknown constants to satisfy the boundary conditions, a numerical solution (like the one presented in Robert Israel's answer) would probably be more appropriate for this problem.
A: I don't know if there's a closed-form solution, and I don't know if this will help.  But it looks like  with initial conditions $y(0)=0, y'(0)=0, y''(0)=2c$ you have a series solution of the form 
$$ y(x) = \sum_{i=0}^\infty a_i c^{i+1} x^{2+3i}
$$
where the first few $a_i$ are 
$$ 1, -\frac1{60}, \frac{11}{20160}, -\frac{5}{266112}, \frac{9299}{14529715200}, -\frac{1272379}{59281238016000}$$
That is, $y(x) = cx^2 g(c x^3)$ for some $g$.  For this to be true, $g$ must satisfy 
$$ 54 t^2 g'''(t) + (9 t^2 g(t) + 216 t) g''(t) + (18 t g(t) + 120) g'(t) + 2 g(t)^2 =0$$
EDIT:
It appears (from numerical solution) that in the case $c=1$ we have $y(x) \sim 3.31 x$ as $x \to \infty$.  If that is true, then $g(t) \sim 3.31/t^{1/3}$ as $t \to \infty$, so to get $y(x) \sim x^2/2$ we'd want $c \approx  3.31^{-3/2} \approx 0.166$.
 Indeed, with $c = 0.166$, Maple's numerical solution gives us $y'(1000) \approx 0.999884746627945$.
