I have the following problem in an assignment that seems to have some issues.

The problem is:

Solve the IVP with the differential operator method:

$ [D^2 + 5D + 6], y(0) = 2, y'(0) = \beta > 0 $

a) Determine the coordinates $ (t_m,y_m) $ of the maximum point of the solution as a function of $\beta $

b) Determine the smallest value of $ \beta $ for which $ y \geq 4 $

c) Determine the behavior of $t_m, y_m $ as $ \beta $ approaches infinity.

So I got the general solution of the ODE by saying $ [(D + 2)(D + 3)]y = 0 $, my result is $ y = A * e^{-2t} + B * e^{-3t} $. Next I had to solve the IVP, so I first found y':

$ y' = -2tAe^{-2t} -3tBe^{-3t} $.

Now for y(0) = 2: $2 = A * 1 + B * 1$ or $ A + B = 2 $.

And for y'(0) = $ \beta $: $ \beta = -2A -3B $.

I solved for A and B by system of equations and I got: $A = \beta + 6, B = -(\beta + 4) $

So $y = (\beta + 6)e^{-2t} - (\beta + 4)e^{-3t} $

And $y' = -2(\beta + 6)e^{-2t} + 3(\beta - 4)e^{-3t}$.

I solved part a) by setting y' = 0 and solving for t, then plugging t into y. Both were fairly complicated functions of beta but that's not a huge problem.

EDIT: I hadn't simplified my result for $ y_m $. Now that I did, I find that $ y_m $ simplifies to 0. Since this is the only critical point, does this mean that I must test the boundary points of t to find $t_m, y_m$? If so, assuming $ t \geq 0$, $(t_m,y_m) = (\infty,2)$! Any way I look at this, I can't see how question b) is possible.

Part b) is where it gets weird. I don't understand how any possible beta can assure that y is greater than 4 for all values of x. As t approaches infinity, $ e^{-2t}, e^{-3t} = 0$, therefore y tends to 0 as t approaches infinity, regardless of the value of beta. Is that not true?

I haven't tried part c) yet, I'm mostly worried about part b). Have I done something wrong?