Finding general solution to DE subject to initial condition How do we solve the following Differential Equation?
$$2 x''' + xx'' =0$$
Subject to conditions:
$$ x(0)=0$$
$$ x'(0)=0$$
$$ x'(\infty)=1$$
Is there any numerical method to solve it or some general method??
 A: Numerically there does not seem to be any problem, $T=20$ as approximation of $T=\infty$ appears large enough, the solution can be computed via boundary value solver, here python's scipy.integrate.solve_bvp:
T = 20
def x_ode(t,x): return [x[1], x[2], -0.5*x[0]*x[2]] 
def x_bc(x0, xT): return [x0[0], x0[1], xT[1]-1]

s = np.linspace(0,1,11);
t_init = T*s
x_init = [ T*s , 1+0*s, 0+0*s ]

res = solve_bvp(x_ode, x_bc, t_init, x_init, tol=1e-5, max_nodes=80000)

print res.message

with the result 
The algorithm converged to the desired accuracy.

The plot of function and derivative

is then produced via
if res.success:
    plt.figure(figsize=(9,6))
    plt.subplot(211); plt.plot(res.x, res.y[0], '-'); plt.grid();
    plt.subplot(212); plt.plot(res.x, res.y[1], '-'); plt.grid(); 
    plt.show()


second approach
To transform the problem to a finite interval the boundary conditions suggest to use $s=x'$, $s\in[0,1]$, as new independent parameter, thus using the inverse function $u$ to $x'$, $t=u(x')$, $x=v_0(x')$, $v_1(s)=x'(u(s))=s$, $v_2(s)=x''(u(s))$ which leads to the derivatives
\begin{align}
1=\frac{dt}{dt}&=u'(x'(t))x''(t)\\
\frac{dt}{ds}&=u'(s)=\frac1{v_2(s)}\\
\frac{dv_0(s)}{ds}&=x'(u(s))u'(s)=\frac{s}{v_2(s)}\\
\frac{dv_2(s)}{ds}&=x'''(u(s))u'(s)=-\frac{v_0(s)}{2}
\end{align}
To get $v_0(s)\to\infty$ for $s\to 1$ we need $v_2(1)=0$ along with $u(0)=0$ and $v_0(0)=0$. To desingularize $1/v_2$ the approach using $v_2/(\epsilon^2+v_2^2)$ works best.
eps = 1e-8
def uv_ode(s,y): 
    u, v0, v2 = y; 
    v2inv = v2/(eps**2+v2**2); 
    return [v2inv, s*v2inv, -0.5*v0]

def uv_bc(u0, u1): 
    return [ u0[0], u0[1], u1[2]]

s = np.linspace(0,1,11);
y_init = [ 16*s**2 , 14*s**2, (1-s)**2/4 ]

res = solve_bvp(uv_ode, uv_bc, s, y_init, tol=1e-5, max_nodes=80000)
print res.message

if True or res.success:
    plt.figure(figsize=(9,9))
    plt.subplot(311); plt.plot(res.x, res.y[0], '-o',ms=1); plt.ylabel('$u=t$'); plt.grid();
    plt.subplot(312); plt.plot(res.x, res.y[1], '-o',ms=1); plt.ylabel('$v_0=x$'); plt.grid();
    plt.subplot(313); plt.plot(res.x, res.y[2], '-o',ms=1); plt.ylabel('$v_2=x\'\'$'); plt.grid();  plt.xlabel('$s=x\'$');
    plt.show()


A: The first step in a "general method" to solve this problem is to recognize that the given third-order ordinary differential equation (ODE) is autonomous.
We'll use subscripts for derivatives in the following, because we are going to use derivatives with respect to the independent variable $t$ but also with respect to the dependent variable $x$.
Starting with $2 x_{ttt} + x x_{tt} = 0$ we write $x_t = w(x)$ with some unknown function $w$, to obtain a boundary-value problem (BVP) with a second-order ODE for $w$:
\begin{equation}
w w_{xx} + w_x \left( w_x + \frac{x}{2} \right) = 0, \quad w(0) = 0, \quad w(\infty) = 1
\end{equation}
(note that the derivatives of $w$ are taken with respect to $x$!).
Once the function $w$ is found we may solve the following initial-value problem (IVP) with a separable first-order ODE for $x$:
\begin{equation}
x_t = w(x), \quad x(0) = 0.
\end{equation}
Thus we have split the initial-boundary-value problem (IBVP) with a third-order autonomous ODE into a BVP with a second-order ODE and an IVP with a separable first-order ODE.
Going for the inverse function $t(x)$ as suggested by JJacquelin we may write
\begin{equation}
t(x) = \int \limits_0^x \frac{1}{w(\xi)} \, \mathrm{d}\xi,
\end{equation}
with the function $w$ from above. Thus it remains to solve the BVP for $w$ and then to compute the integral.
Both tasks may be difficult, and I believe that there is a mistake in the solution provided by JJacquelin (in the calculation of the third derivative of the inverse function). I would normally write a comment on this but I am not yet allowed.
A: Follows a MATHEMATICA script showing a numerical solution. Here $t = 200 \approx\infty$

tmax = 200;
solx = NDSolve[{x1'[t] == x2[t], 
                x2'[t] == x3[t], 
2 x3'[t] + x1[t] x3[t] == 0, x1[0] == 0, x2[0] == 0, x2[200] == 1}, 
{x1, x2, x3}, {t, 0, tmax}][[1]]
Plot[Evaluate[{x1[t], x2[t], x3[t]} /. solx], {t, 0, 20}, 
 PlotRange -> {0, 2}, 
 PlotStyle -> {{Black, Thick}, {Blue, Thick}, {Red, Thick}}, 
 PlotLegends -> {Subscript[x, 1][t], Subscript[x, 2][t], 
   Subscript[x, 3][t]}]


$$
x_1(t) \to \mbox{black}\\
x_2(t) \to \mbox{blue}\\
x_3(t) \to \mbox{red}
$$
