Question

@empty and @ganeshie8

Answer 1

I have this question the part in red I don't really understand, I solved the O.D.E and have \[y(t) = -2t-4/3-4e^t+(y_0+16/3)e^{3t/2},~~~C = y_0+16/3\]

Answer 2

I mean it's pretty difficult to tell without a picture

Answer 3

Looks like you should plug in \(t=\infty\) and \(y=0\) and solve for \(y_0\)

Answer 4

Ooh y = 0?

Answer 5

Why exactly is y = 0?

Answer 6

I don't think that's quite right

Answer 7

As \(t \to \infty\) if you find the value of \(y_0\) that causes \(y(\infty)=0\) then that means every other value of \(y_0\) will give you either positive or negative values (at least in this case) .

Answer 8

Yeah maybe this isn't quite right beats me.

Looking at this again, they're saying solutions that grow positively not that are necessarily positive. So really it looks like we should say

\(y'=0\) at t infinite. So we can try playing around with the differential equation itself.

Answer 9

Or hell, maybe they just want you to differentiate our solution and do what I described the step before.

Answer 10

Yeah haha, I was getting y0 = infinity at y(infinity) it doesn't really make sense, I'm not exactly sure, I think it may have to do with the \[(y_0+16/3)e^{3/2t}\] term itself, but I don't really know what that means xD

Answer 11

Just to get some physical perspective,
I think we may treat the differential equation as modeling the "money in the bank"

Answer 12

\[\dfrac{dy}{dt} = 3/2y + 3t+2e^t\]

the function \(y\) grows with \(y\) and time, \(t\)

Maybe, a better situation would be treating it as population growth in a particular city; it depends on :
1) existing population, \(y\)
2) people that migrate to the city each year, depends on \(t\).

Answer 13

Once we get a feel of DE, it becomes easy to see how the initial value decides the fate of the solution..

Answer 14

Clearly, what ever the solution is, it is an increasing function (why ?)

Answer 15

You can tell that w/o solviing, just by looking at the DE..

Answer 16

I see that it relies on e^t

Answer 17

Notice that \(\mathcal{O}(e^{3t/2})\gt \mathcal{O}(e^t)\gt \mathcal{O}(\text{any polynomial})\).
so we need to worry only about the sign of the coefficcient of \(e^{3t/2}\) as \(t\to\infty\)

Answer 18

I see, but how do we get the value of y_0?

Answer 19

\[y(t) = -2t-4/3-4e^t+(y_0+16/3)e^{3t/2}\]

the dominating term as \(t\to\infty\) is \(e^{3t/2}\); the solution approaches \((y_0+16/3)e^{3t/2}\) as \(t\to\infty\); so the critical value of \(y_0\) is \(-16/3\).

see the last problem

http://www.math.ucsd.edu/~hus003/math20D/HW0.pdf

Answer 20

Oh dang, so I sort of was on the right track haha

Answer 21

So the biggest term gives us the value that splits it from positive and negative values (makes sense) and then you're looking for the critical value at that position in a sense...hmm I hope I understood that correctly

Answer 22

Yes, in our earlier bank analogy, the function grows as long as the initial value is positive.
If the outstanding balance is negative, the function can decrease too...

Answer 23

Ok I think I'm getting it, thanks

Answer 24

Thanks for the analogies hehe

Answer 25

\[f(x) = -1-x-2x^2 + cx^3\]

It is easy to see that the asymptotic behavior of \(f(x)\) is decided "only" by the sign of the leading coefficient, \(c\)

Answer 26

all other terms don't matter as \(x\to \pm \infty\)