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:
[itex] [D^2 + 5D + 6D], y(0) = 2, y'(0) = \beta > 0[/itex]
a) Determine the coordinates [itex](t_m,y_m)[/itex] of the maximum point of the solution as a function of [itex]\beta[/itex].
b) Determine the smallest value of [itex]\beta[/itex] for which [itex]y \geq 4 [/itex].
c) Determine the behavior of [itex] t_m [/itex] and [itex]y_m[/itex] as [itex]\beta \rightarrow \infty [/itex].

Homework Equations

The Attempt at a Solution

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

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

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

Now for part a).
To find the maximum, I set y' to 0 and solved for t. I got:
[itex] 0 = \beta e^{-11t} [/itex]. This is not solveable. I'm assuming [itex] t \geq 0 [/itex] because conventionally t is time and must be greater than 0. Since beta is positive, the maximum of y occurs when [itex] t = \infty [/itex], so [itex] (t_m,y_m) = (\infty, \dfrac{\beta}{11} + 2)[/itex]. 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 [itex] y \geq 4[/itex] if THE PROBLEM STATES THAT [itex]y(0) = 2[/itex]?
Isn't that a contradiction? I started this problem assuming that condition, and now my function of y prohibits the constraint that [itex]y \geq 4 [/itex]. 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:
[itex] [D^2 + 5D + 6D], y(0) = 2, y'(0) = \beta > 0[/itex]

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

a) Determine the coordinates [itex](t_m,y_m)[/itex] of the maximum point of the solution as a function of [itex]\beta[/itex].
b) Determine the smallest value of [itex]\beta[/itex] for which [itex]y \geq 4 [/itex].
c) Determine the behavior of [itex] t_m [/itex] and [itex]y_m[/itex] as [itex]\beta \rightarrow \infty [/itex].

Homework Equations

The Attempt at a Solution

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

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

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

Now for part a).
To find the maximum, I set y' to 0 and solved for t. I got:
[itex] 0 = \beta e^{-11t} [/itex]. This is not solveable. I'm assuming [itex] t \geq 0 [/itex] because conventionally t is time and must be greater than 0. Since beta is positive, the maximum of y occurs when [itex] t = \infty [/itex], so [itex] (t_m,y_m) = (\infty, \dfrac{\beta}{11} + 2)[/itex]. 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 [itex] y \geq 4[/itex] if THE PROBLEM STATES THAT [itex]y(0) = 2[/itex]?
Isn't that a contradiction? I started this problem assuming that condition, and now my function of y prohibits the constraint that [itex]y \geq 4 [/itex]. 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.