Solving $x'-2xt-t=0$. Find the general and particular solution of:
$$x'-2xt-t=0\\ x(0)=1$$
Separating the variable I got $\frac {dx}{2x+1}=tdt \implies \ln|2x+1|=t^2+c$.
After this I got confused, after exponentiating everything, how do I get rid of the absolute values?
Also, what does "the maximal interval of existence" mean? Aren't all solutions to an ODE defined in the same domain?

Using separation of variables
$$ \frac{1}{2x+1}dx = t \ dt $$
You can then integrate both sides
And then:
$$ \frac{\ln |2x+1|}{2} = \frac{t^2}{2} + c$$
$$ 2x+1 = e^{t^2+2c} = Ae^{t^2}$$
where $A = e^{2c}$
You can use your initial condition to obtain a value for $c$
$$ c = \frac{\ln 3}{2}$$
$$ 2x+1 = e^{t^2}\cdot e^{\ln 3}= 3 e^{t^2}$$
$$ x= \frac{3}{2}e^{t^2}-\frac{1}{2}$$
