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.

$$y'(t)=1+\frac{y(t)}{2}$$
$$\frac{dy}{dt}=\frac{2+y(t)}{2}$$
$$dy=\frac{(2+y(t))dt}{2}$$
$$\frac{dy}{y(t)+2}=\frac{dt}{2}$$
$$\int{\frac{dy}{y(t)+2}}=\int{\frac{dt}{2}$$
$$\ln{(y(t)+2)}=\frac{t}{2}+C$$
$$y(t)+2=Ce^{t/2}$$
$$y(t)=Ce^{t/2}-2$$
$$y(0)=.5=Ce^{0/2}-2$$
$$C=2.5$$
$$y(t)=2.5e^{t/2}-2$$