PG,
For 0 ≤ x ≤ 3, y'+2y=1, so the integrating factor is e^∫2dx = e^2x. That gives y'e^2x+2e^2x = e^2x, or (ye^2x)' = e^2x. Integrating both sides gives ye^2x = e^2x/2 + c or y = ½ + ce^-2x. Substituting the initial conditions gives 0 = ½+c ==> c = -½. So we have y = ½ -½e^-2x for 0 ≤ x ≤ 3.
If x is outside the above range, y'+2y=0, the variables are separable, and the solution is easy.
Question. Did you use Snagit to write your request. If not, what processor did you use.
Ratch