## Separation of Variables

1. The problem statement, all variables and given/known data
Using separation of variables determine if the solution escapes to infinity in finite time or infinite time?

$$y'(t)=1+\frac{y(t)}{2}$$
$$y(0)=.5$$

2. Relevant equations
Knowing how to do separation of variables.

3. The attempt at a solution
Here is my attempt, but I get stuck...
$$y'(t)=1+\frac{y(t)}{2}$$
$$y'(t)-\frac{y(t)}{2}=1$$
$$\int_0^t{y'(x)-\frac{y(x)}{2}dx}=\int_0^t{1dx}$$
The next step I'm not sure of...
$$(y(t)-y(0))-(\frac{y(t)^2}{4}-\frac{y(0)^2}{4})=t$$
$$y(t)-\frac{y(t)^2}{4}=t+y(0)-\frac{y(0)^2}{4}$$
Now solving for $$y(t)$$ becomes a problem if the above step is correct... I'm sure I'm doing something wrong.
 Sorry about that... I'm learning DiffEq through Mathematica and needless to say, it's poop. Anyway, I figured out how to do it. $$y'(t)=1+\frac{y(t)}{2}$$ $$\frac{dy}{dt}=\frac{2+y(t)}{2}$$ $$dy=\frac{(2+y(t))dt}{2}$$ $$\frac{dy}{y(t)+2}=\frac{dt}{2}$$ $$\int{\frac{dy}{y(t)+2}}=\int{\frac{dt}{2}$$ $$\ln{(y(t)+2)}=\frac{t}{2}+C$$ $$y(t)+2=Ce^{t/2}$$ $$y(t)=Ce^{t/2}-2$$ $$y(0)=.5=Ce^{0/2}-2$$ $$C=2.5$$ $$y(t)=2.5e^{t/2}-2$$