# Differential Equations not solvable

• A
In the first image it shows the ##\alpha^2-w_0^2<0## situation whereas in the second image the situation is when ##\alpha^2-w_0^2=0##.The problem is the book says to use ##T_0=2\pi/w_0## to determine diabetes but you can't do that when ##\alpha^2-w_0^2=0## because it can't be put into a cosine function. What do I do in this situation?

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BvU
Homework Helper
I don't see the second image. Fig 2. has ##\alpha^2-\omega_0^2>0##, not = 0. But the picture doesn't look credible at all. It looks more like underdamped to me.

Your book makes a mess of things with "three types, depending on ##\alpha^2-\omega_0^2>0##, < 0 or zero. These three types correspond to overdamped, critically damped and underdamped cases".

Confusing, to put it mildly:
##\alpha^2-\omega_0^2=0 \qquad \Rightarrow ## critically damped

Check out a better text, e.g. here

And yes, in the case of critical damping there are no oscillations.

By the way, I don't hear the book saying to use ##T_0##. Did you quote correctly ?

I meant Exercise 2 not the Figure 2, sorry for not pointing it out. And ##T_0## is on a different page, uploaded below.

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BvU
Homework Helper
It looks as if they want you to make quite some assumptions for this exercise: reliable measurements to begin with. G0 = 70. And no additional glucose dosing in between, ...

If glucose concentration goes from 95 via 65 to 75 that means something significant for T0 . Agree ?

( Could such measurements be consistent with critical damping ? )

The measurements can be consistent if the function is of the form ##(a-bt)e^{ct}## where ##a##,##b## and ##c## are positive. I don't know the actual meaning of ##T_0## but from the hint I guess it has got something to do with the level going down to normal again.

BvU