# Thread: Differential Operators Problem doesn't seem to make sense

1. ## Differential Operators Problem doesn't seem to make sense

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?

2. ## Re: Differential Operators Problem doesn't seem to make sense

a) you have $y^\prime$ as a function of $t$ and $\beta$. Hold $\beta$ fixed and differentiate w/respect to t. Set this equal to zero and solve for t.

That's the $t_m$, they want. Evaluate $y_m = y(t_m)$ and you are done.

b) pretty sure they want $y_m > 4$

c) this is pretty straightforward once you get (b) squared away.

3. ## Re: Differential Operators Problem doesn't seem to make sense

a) Ok thanks, I actually found a mistake in my work and then did the exact procedure you mentioned. However, the only result that produced was the local minimum, not maximum. I got $\displaystyle (t_m, y_m) = (\infty, \frac{\beta}{11} + 2)$. Is it possible for tm to be infinity?
b) At least I'm not crazy. I sent an email to the TA to hopefully get that cleared up.

4. ## Re: Differential Operators Problem doesn't seem to make sense

a) Ok thanks, I actually found a mistake in my work and then did the exact procedure you mentioned. However, the only result that produced was the local minimum, not maximum. I got $\displaystyle (t_m, y_m) = (\infty, \frac{\beta}{11} + 2)$. Is it possible for tm to be infinity?
b) At least I'm not crazy. I sent an email to the TA to hopefully get that cleared up.

$\displaystyle{\lim_{\beta->\infty}}t_m = \ln\left(\dfrac 3 2 \right)$

$\displaystyle{\lim_{\beta->\infty}}y_m =\infty$

5. ## Re: Differential Operators Problem doesn't seem to make sense

Thanks, but that's the answer to c). I was talking about a).

6. ## Re: Differential Operators Problem doesn't seem to make sense

$t_m(b) = \log \left(\dfrac{3 (b+4)}{2 (b+6)}\right)$
$y_m(b) = \dfrac{4 (b+6)^3}{27 (b+4)^2}$