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Well, I can't seem to find an error in my derivation and I can't decipher your writing, so I will assume the derivation I posted is correct. As someone already mentioned in this thread, it seems impossible that the equation can be solved in a closed form. So, I'm afraid you are left with solving it numerically.

Last time we ended at this step:

 Quote by Dickfore $$\left\{\begin{array}{rcl} \frac{d v}{d x} & = & p \\ \frac{d p}{d x} & = & \frac{v}{a} \, \left[p + \frac{n}{2 \, b} \, v^{2} \, (v^{2} - c) \right] \end{array}\right.$$ where we also made the transformation $p/2 \rightarrow p$ again. The boundary conditions that you gave translate to the following: $$\begin{array}{l} x \rightarrow -\infty \Rightarrow v = \sqrt{c} \\ x = 0 \Rightarrow v = \sqrt{b + c} \end{array}$$
The connection with the old variables is given by:

$$u(x) = [v(x)]^{2}$$

When working with numerics, it is best to get rid of as many parameters as possible. Let's scale everything:

$$\begin{array}{l} x = x_{0} \, \bar{x} \\ v = v_{0} \, \bar{v} \\ p = p_{0} \, \bar{p} \end{array}$$

Then the equations become:

$$\begin{array}{rcl} \frac{v_{0}}{x_{0}} \, \frac{d \bar{v}}{d \bar{x}} & = & p_{0} \, \bar{p} \\ \frac{p_{0}}{x_{0}} \, \frac{d \bar{p}}{d \bar{x}} & = & \frac{v_{0}}{a} \, \left[ p_{0} \, \bar{p} + \frac{n}{2 b} \, v^{2}_{0} \, \bar{v}^{2} \left(v^{2}_{0} \, \bar{v}^{2} - c \right) \right] \end{array}$$

I think it is most convenient to make this choice:

$$v^{2}_{0} = c, \; p_{0} = \frac{n}{2 b} \,v^{4}_{0}. \; \frac{p_{0}}{x_{0}} = \frac{v_{0} \, p_{0}}{a}$$

$$x_{0} = \frac{a}{\sqrt{c}}, p_{0} = \frac{n \, c^{2}}{2 b}, \; v_{0} = \sqrt{c} \Rightarrow \frac{x_{0} \, p_{0}}{v_{0}} = \frac{a n c}{2 b} \equiv k$$

Also, let's get rid of the bars above the symbols again:

$$\begin{array}{rcl} \frac{d v}{d x} & = & k \, p \\ \frac{d p}{d x} & = & v \, \left[ p + v^{2} \, (v^{2} - 1) \right] \end{array}$$

With the boundary conditions being:

$$\left\{\begin{array}{l} \sqrt{c} \, v = \sqrt{c}, \; x \rightarrow -\infty \\ \sqrt{c} \, v = \sqrt{b + c}, \; x = 0 \end{array}\right. \Rightarrow \left\{\begin{array}{l} v = 1, \; x \rightarrow -\infty \\ v = \sqrt{1 + \frac{b}{c}} = M, \; x = 0 \end{array}\right.$$

Instead of having this limit as $x \rightarrow -\infty$, let us make the simultaneous substitution $x \rightarrow -x, p \rightarrow -p$. The equations become:

$$\begin{array}{rcl} \frac{d v}{d x} & = & k \, p \\ \frac{d p}{d x} & = & v \, \left[ p + v^{2} \, (v^{2} - 1) \right] \end{array}$$

with the boundary conditions:

$$\left\{\begin{array}{l} v = \sqrt{1 + \frac{b}{c}} = M, \; x = 0 \\ v \rightarrow 1, \; x \rightarrow \infty \\ \end{array}\right.$$

Can you find the stationary points for this autonomous non-linear system. What is their type?