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Astrophysics
 one year ago
@empty and @ganeshie8
Astrophysics
 one year ago
@empty and @ganeshie8

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Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I have this question dw:1443248793563:dw the part in red I don't really understand, I solved the O.D.E and have \[y(t) = 2t4/34e^t+(y_0+16/3)e^{3t/2},~~~C = y_0+16/3\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I mean it's pretty difficult to tell without a picture

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Looks like you should plug in \(t=\infty\) and \(y=0\) and solve for \(y_0\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Why exactly is y = 0?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I don't think that's quite right

Empty
 one year ago
Best ResponseYou've already chosen the best response.1As \(t \to \infty\) 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) .

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Or hell, maybe they just want you to differentiate our solution and do what I described the step before.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Yeah 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Just to get some physical perspective, I think we may treat the differential equation as modeling the "money in the bank"

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\dfrac{dy}{dt} = 3/2y + 3t+2e^t\] 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 \(t\).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Once we get a feel of DE, it becomes easy to see how the initial value decides the fate of the solution..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Clearly, what ever the solution is, it is an increasing function (why ?)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2You can tell that w/o solviing, just by looking at the DE..

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I see that it relies on e^t

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Notice 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 \(t\to\infty\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1I see, but how do we get the value of y_0?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[y(t) = 2t4/34e^t+(y_0+16/3)e^{3t/2}\] 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

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh dang, so I sort of was on the right track haha

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1So 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes, 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...

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Ok I think I'm getting it, thanks

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Thanks for the analogies hehe

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[f(x) = 1x2x^2 + cx^3\] It is easy to see that the asymptotic behavior of \(f(x)\) is decided "only" by the sign of the leading coefficient, \(c\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2all other terms don't matter as \(x\to \pm \infty\)
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