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Astrophysics

  • one year ago

@empty and @ganeshie8

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  1. Astrophysics
    • one year ago
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    I have this question |dw:1443248793563:dw| the part in red I don't really understand, I solved the O.D.E and have \[y(t) = -2t-4/3-4e^t+(y_0+16/3)e^{3t/2},~~~C = y_0+16/3\]

  2. Astrophysics
    • one year ago
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    I mean it's pretty difficult to tell without a picture

  3. Empty
    • one year ago
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    Looks like you should plug in \(t=\infty\) and \(y=0\) and solve for \(y_0\)

  4. Astrophysics
    • one year ago
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    Ooh y = 0?

  5. Astrophysics
    • one year ago
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    Why exactly is y = 0?

  6. Astrophysics
    • one year ago
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    I don't think that's quite right

  7. Empty
    • one year ago
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    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) .

  8. Empty
    • one year ago
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    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.

  9. Empty
    • one year ago
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    Or hell, maybe they just want you to differentiate our solution and do what I described the step before.

  10. Astrophysics
    • one year ago
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    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

  11. ganeshie8
    • one year ago
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    Just to get some physical perspective, I think we may treat the differential equation as modeling the "money in the bank"

  12. ganeshie8
    • one year ago
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    \[\dfrac{dy}{dt} = 3/2y + 3t+2e^t\] 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 \(t\).

  13. ganeshie8
    • one year ago
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    Once we get a feel of DE, it becomes easy to see how the initial value decides the fate of the solution..

  14. ganeshie8
    • one year ago
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    Clearly, what ever the solution is, it is an increasing function (why ?)

  15. ganeshie8
    • one year ago
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    You can tell that w/o solviing, just by looking at the DE..

  16. Astrophysics
    • one year ago
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    I see that it relies on e^t

  17. ganeshie8
    • one year ago
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    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 \(t\to\infty\)

  18. Astrophysics
    • one year ago
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    I see, but how do we get the value of y_0?

  19. ganeshie8
    • one year ago
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    \[y(t) = -2t-4/3-4e^t+(y_0+16/3)e^{3t/2}\] 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

  20. Astrophysics
    • one year ago
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    Oh dang, so I sort of was on the right track haha

  21. Astrophysics
    • one year ago
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    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

  22. ganeshie8
    • one year ago
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    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...

  23. Astrophysics
    • one year ago
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    Ok I think I'm getting it, thanks

  24. Astrophysics
    • one year ago
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    Thanks for the analogies hehe

  25. ganeshie8
    • one year ago
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    \[f(x) = -1-x-2x^2 + cx^3\] It is easy to see that the asymptotic behavior of \(f(x)\) is decided "only" by the sign of the leading coefficient, \(c\)

  26. ganeshie8
    • one year ago
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    all other terms don't matter as \(x\to \pm \infty\)

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