Differential Linear Operator Problem not making sense

[wadawalnut]

Homework Statement

I think there may be something wrong with a problem I'm doing for homework. The problem is:

Solve the IVP with the differential operator method:
[D^2 + 5D + 6D], 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 and y_m as \beta \rightarrow \infty.

Homework Equations

The Attempt at a Solution

So I solved the differential equation by simplifying the operator to [D^2 + 11D] (I thought it was weird that I had to do such a simple simplification). Then I wrote (D + 11)(D)y = 0. Next I said let z = [D]y so the problem simplifies to [D + 11]z = 0. From this I got that z = c_1e^{-11t} from separation of variables. Then, since z is just y', I integrated z to find y, which gave me y = \dfrac{-c_1}{11}e^{-11t} + c_2. According to Wolfram Alpha, so far, so good... (so what?).

Next, I solved the initial value problem.
y'(0) = \beta so \beta = c_1e^0 and c_1 = \beta.
y(0) = 2 so 2 = \dfrac{-\beta}{11} \beta e^0 + c_2 and c_2 = \dfrac{\beta}{11} + 2.

Now I have:
y = \dfrac{-\beta}{11} e^{-11t} + \dfrac{\beta}{11} + 2
y' = \beta e^{-11t}

Now for part a).
To find the maximum, I set y' to 0 and solved for t. I got:
0 = \beta e^{-11t}. This is not solveable. I'm assuming t \geq 0 because conventionally t is time and must be greater than 0. Since beta is positive, the maximum of y occurs when t = \infty, so (t_m,y_m) = (\infty, \dfrac{\beta}{11} + 2). This seems sketchy but is the most reasonable answer I can come up with.

And now the very strange part, part b).
How is it possible for me to find a value for beta such that y \geq 4 if THE PROBLEM STATES THAT y(0) = 2?
Isn't that a contradiction? I started this problem assuming that condition, and now my function of y prohibits the constraint that y \geq 4. Am I missing something?

I haven't really tried part c) because I'm assuming I already messed everything up, I just can't figure out where.
Any help is appreciated.

 

[Mark44]

Homework Statement

I think there may be something wrong with a problem I'm doing for homework. The problem is:

Solve the IVP with the differential operator method:
[D^2 + 5D + 6D], y(0) = 2, y'(0) = \beta > 0

Typos above? I think it should be $[D^2 + 5D + 6]y = 0$, and then the initial conditions.

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 and y_m as \beta \rightarrow \infty.

Homework Equations

The Attempt at a Solution

So I solved the differential equation by simplifying the operator to [D^2 + 11D] (I thought it was weird that I had to do such a simple simplification).

See my comment above. I'm almost certain that the DE is y'' + 5y' + 6y = 0. If so, it can be solved fairly easily.

Then I wrote (D + 11)(D)y = 0. Next I said let z = [D]y so the problem simplifies to [D + 11]z = 0. From this I got that z = c_1e^{-11t} from separation of variables. Then, since z is just y', I integrated z to find y, which gave me y = \dfrac{-c_1}{11}e^{-11t} + c_2. According to Wolfram Alpha, so far, so good... (so what?).

Next, I solved the initial value problem.
y'(0) = \beta so \beta = c_1e^0 and c_1 = \beta.
y(0) = 2 so 2 = \dfrac{-\beta}{11} \beta e^0 + c_2 and c_2 = \dfrac{\beta}{11} + 2.

Now I have:
y = \dfrac{-\beta}{11} e^{-11t} + \dfrac{\beta}{11} + 2
y' = \beta e^{-11t}

Now for part a).
To find the maximum, I set y' to 0 and solved for t. I got:
0 = \beta e^{-11t}. This is not solveable. I'm assuming t \geq 0 because conventionally t is time and must be greater than 0. Since beta is positive, the maximum of y occurs when t = \infty, so (t_m,y_m) = (\infty, \dfrac{\beta}{11} + 2). This seems sketchy but is the most reasonable answer I can come up with.

And now the very strange part, part b).
How is it possible for me to find a value for beta such that y \geq 4 if THE PROBLEM STATES THAT y(0) = 2?
Isn't that a contradiction? I started this problem assuming that condition, and now my function of y prohibits the constraint that y \geq 4. Am I missing something?

I haven't really tried part c) because I'm assuming I already messed everything up, I just can't figure out where.
Any help is appreciated.

 

[wadawalnut]

I did not make a typo, that was the actual question.

 

[Mark44]

I did not make a typo, that was the actual question.

I didn't say or even imply that you made the typo, but I am almost certain (99.44%) it is a typo.