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.