Find the explicit solution for the following differential equation. I have the following differential equation
$$\frac{dx}{dt} = x^2-4$$
Separating the variables, I get
$$\frac{dx}{x^2-4} = dt$$
Let us write it in partial form
$$\frac{dx}{(x-2)(x+2)} = dt$$
$$\frac{dx}{4(x-2)} - \frac{dx}{4(x+2)} = dt $$
$$ \frac{dx}{(x-2)} - \frac{dx}{(x+2)} = 4dt $$
$$ \ln{|x-2|} - \ln{|x+2|} + C_1 = 4t + C_2 $$
Let $C_2 - C_1 = C$
$$ \ln{|\frac{x-2}{x+2}|} = 4t + C $$
$$e^{\ln{|\frac{x-2}{x+2}|}} = e^{4t+C}$$
$$e^{\ln{|\frac{x-2}{x+2}|}} = e^{4t}e^C$$
Let $e^C = C$ Since it is a constant
$$\frac{x-2}{x+2} = Ce^{4t}$$
Let $x(0) = x_0$
$$\frac{x_0-2}{x_0+2} = C$$
Substituting for C
$$\frac{x-2}{x+2} = \frac{(x_0-2)e^{4t}}{x_0+2}$$
I am rather stuck in here. The solutions manual to this question gives:
$$x(t) =\frac{2[x_0 + 2 + (x_0 - 2)]e^{4t}}{x_0 + 2 - (x_0 - 2)}$$
The solutions manual does not elaborate on how it came to the solution above. How do I approach the problem? Any hints?
Source: Differential Equations and Boundary Value Problems: Computing and Modeling (5th Edition) 
by C. Henry Edwards (Author), David E. Penney (Author), David T. Calvis (Author)
Question 5 Chapter 2.2
NOTE: $x(0)$ is not given at all so this is not a mistake. Hence, we simply do $x(0) = x_0$. 
 A: All your steps are correct.
From your last equation,
$$
\frac{x-2}{x+2} = \frac{(x_0-2)e^{4t}}{x_0+2}
$$
clear denominators
$$
(x + 2 ) (x_0-2)e^{4t} = (x-2)(x_0+2)
$$
sort for $x$-terms
$$
x ((x_0+2)- (x_0-2)e^{4t}) = 2  (x_0-2)e^{4t} +2(x_0+2)
$$
and divide for the correct answer
$$
x(t) =\frac{2[x_0 + 2 + (x_0 - 2)e^{4t}]}{x_0 + 2 - (x_0 - 2)e^{4t}}
$$
You can verify it by plugging it into the DEQ  $\frac{dx}{dt} = x^2-4$ which perfectly holds true.
Hence, indeed the book has two printing errors: the $e^{4t}$ has to show up in the denominator,  and the  $e^{4t}$ in the numerator has to be inside the bracket.
Anyway, it's always a good idea to verify the solution (your own one, or the book's one) with a symbolic computer program like Mathematica, Matlab, Wolfram Alpha (free) or SAGE (free). In a publication (see here) on computational maths I was shocked to read the following:
"exhaustive testing eventually revealed that the problem actually lay with the textbooks. In some published tables the error rates exceeded 25%." 
So there is good reason for healthy scepticism even in textbooks.
A: If $$ \dfrac{a}{b}=\dfrac{c}{d}, $$
then 
$$ \dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}. $$ 
This is Componendo& Dividendo Rule of elementary algebra which if applied to your last but one equation gives the last equation of the manual.
