Solving the ODE $\frac{\mathrm{d}x}{\mathrm{d}t} = x^{-n} -2x$ I'm preparing for my mock exams, and one of the past questions was to 

Solve $$\frac{\mathrm{d}x}{\mathrm{d}t}  = x^{-n} -2x$$ for $0<x<\infty$ where $n>0$, subject to $x(1)=1$.

My work so far is as follows: 
Multiplying both sides of the ODE by $x^{n}$ gives $$
 x^n\frac{\mathrm{d}x}{\mathrm{d}t}  = 1 -2x^{n+1}.
 $$
Put $p = x^{n+1}$. Then $$
\frac{\mathrm{d}p}{\mathrm{d}t} = (n+1)x^{n}\frac{\mathrm{d}x}{\mathrm{d}t},
$$
or $$
x^{n}\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{n+1}\frac{\mathrm{d}p}{\mathrm{d}t}.
$$
Therefore, the ODE in question may be rewritten in the form of $$
 \frac{1}{n+1}\frac{\mathrm{d}p}{\mathrm{d}t}  = 1 -2p.
$$
Assume, for the time being, that $1-2p\neq 0$. Then separating variables leads to $$
\frac{1}{n+1}\frac{\mathrm{d}p}{1 -2p} = \mathrm{d}t,
$$
and integrating both sides gives  $$
-\frac{1}{2}\frac{1}{n+1}\ln{|1-2p|} = t+C
$$
or $$
-\frac{1}{2}\frac{1}{n+1}\ln{|1-2x^{n+1}|} = t+C.
$$
The condition $x(1)=1$ leads to $$
-\frac{1}{2}\frac{1}{n+1}\ln{|1-2\cdot 1|}=-\frac{1}{2}\frac{1}{n+1}\ln{1}= 0  = 1+C,
$$
so that $C = -1$ and $$
-\frac{1}{2}\frac{1}{n+1}\ln{|1-2x^{n+1}|} = t - 1,
$$
and from here we get $$
|1-2x^{n+1}| = e^{2(n+1)(1-t)}.
$$
Now, I wanted to get $x(t)$ explicitly. I obtained that $1-2x^{n+1}\ge 0$ for $x\in\left(0, \frac{1}{2^{n+1}}\right]$ and $1-2x^{n+1}\le 0$ for $x\in\left[\frac{1}{2^{n+1}}, \infty \right)$. This leads to "solutions"(??)
$$ \begin{align}
x(t) = \begin{cases}
&\displaystyle\sqrt[n+1]{\displaystyle\frac{1 - e^{2(n+1)(1-t)}}{2}}\quad\text{ for }x\in\left(0, \displaystyle\frac{1}{2^{n+1}}\right]\\\
&\displaystyle \sqrt[n+1]{\displaystyle\frac{1 + e^{2(n+1)(1-t)}}{2}}\quad\text{ for }x\in\left[\displaystyle\frac{1}{2^{n+1}}, \infty \right).
\end{cases}
\end{align}$$
but this kinda doesn't make sense, because we're giving a formula for $x$ as a function of $t$ and at the same time imposing restrictions as to which interval has $x$ to be to be given by a particular formula...
My question(a) is(are): 

  
*
  
*Is there a simpler way to get $x(t)$?
  
*How can we reconcile the restrictions on $x$ (so that $1-2x^{n+1}$ is $><0$) with the formulas for $x$ as a function of $t$?
  

And frankly, I'm not sure how to word my question(s); tl;dr would be how to get explicit formulae for $x(t)$ from the ODE - and not implicit relation between $|1-2x^{n+1}|$ and some $f(t)$)
 A: $p=\frac12$, that is $x(t)=x^*=\sqrt[n+1]{1/2}$, is a constant solution. All other solutions $x$ will not cross this constant solution, that is, stay on one side of $x^*$ by the uniqueness theorem. Thus one concludes that as $2p-1=2x^{n+1}-1>0$ at the initial condition, so it stays for the full solution. So that you directly solve the absolute value in
$$
|1-2x^{n+1}|=e^{(n+1)(1-t)}
$$
to
$$
2x^{n+1}-1=e^{(n+1)(1-t)}\implies x(t)=\sqrt[n+1]{\frac{1+e^{(n+1)(1-t)}}2}.
$$

In general when solving linear first order ODE using the separation of variables method, it helps to translate the integration constant directly into a factor after the exponentiation,  the constant in $\ln|w|=u(t)+c$ gets replaced by $C=signum(w(t_0))e^c$ in $w=Ce^{u(t)}$. Here that leads to
$$
1-2x^{n+1}=Ce^{(n+1)(1-t)}\implies C=1-2x_1^{n+1}
$$
for general initial value $x(1)=x_1>0$, and thus to
$$
x(t)=\sqrt[n+1]{\frac{1+(2x_1^{n+1}-1)e^{(n+1)(1-t)}}2}.
$$
Note that this formula keeps the solution automatically on the right side of $\sqrt[n+1]{1/2}$.
A: This answer concerns only your question 1 :
$$\text{Is there a simpler way to get } x(t) ?$$ 
$$\frac{dx}{dt}=x^{-n}-2x$$
$$dt=\frac{1}{x^{-n}-2x}dx$$
$$t=\int \frac{dx}{x^{-n}-2x}$$
$$t=-\frac{1}{2(n+1)}\ln|1-2x^{n+1}|+C$$
$x(1)=1\quad\implies\quad 1=-\frac{1}{2(n+1)}\ln|1-2|+C\quad\implies\quad C=1$
$$t=-\frac{1}{2(n+1)}\ln|1-2x^{n+1}|+1$$
This is the solution on the form $t(x)$. The explicit solution is the inverse function $x(t)$.
$$|1-2x^{n+1}|=e^{-2(n+1)(t-1)}$$
If $2x^{n+1}<1$ then $x=\left(\frac{1-e^{-2(n+1)(t-1)}}{2} \right)^{\frac{1}{n+1}}$
If $2x^{n+1}>1$ then $x=\left(\frac{1+e^{-2(n+1)(t-1)}}{2} \right)^{\frac{1}{n+1}}$
At this stage we come to your question 2. This was already discussed by LutzL. No need for me to go further.
