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?
 A: 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 $c$
$$ c = \frac{\ln 3}{2}$$
$$ 2x+1 = e^{t^2}\cdot e^{\ln 3}= 3 e^{t^2}$$
$$ x= \frac{3}{2}e^{t^2}-\frac{1}{2}$$
A: You want $\int \frac{dx}{2x+1}=\int t\,dt$, and the next step is $\frac{1}{2}\ln(|2x+1|)=\frac{t^2}{2}+C$. You can simplify to $\ln(|2x+1|)=t^2+D$. Now evaluate $D$, and continue.
Added: Without the initial condition, we get $|2x+1|=Ke^{t^2}$. So $2x+1=\pm Ke^{t^2}=Le^{t^2}$ for some constant $L$ that could be negative. Solving for $x$ we get $x=\frac{1}{2}\left(Le^{t^2}-1\right)$.
A: $$\frac{dx}{2x+1}=tdt$$
$$0.5\log|2x+1|=\frac{t^2}{2}+C$$
usr $x(0)=1$
$$0.5\log|2*1+1|=\frac{0^2}{2}+C$$
$$C=0.5\log |3|=0.5\log 3$$
$$0.5\log|2x+1|=\frac{t^2}{2}+0.5\log 3$$
$$0.5\log|2x+1|-0.5\log 3=\frac{t^2}{2}$$
$$\log|2x+1|-\log 3=t^2$$
$$\frac{2x+1}{3}=e^{t^2}$$
