Escaping to Infinity: Solving a Separable Differential Equation

[lylos]

Homework Statement

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

Homework Equations

Knowing how to do separation of variables.

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.

 

[lylos]

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