Astrophysics
  • Astrophysics
@empty and @ganeshie8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Astrophysics
  • Astrophysics
I have this question |dw:1443248793563:dw| the part in red I don't really understand, I solved the O.D.E and have \[y(t) = -2t-4/3-4e^t+(y_0+16/3)e^{3t/2},~~~C = y_0+16/3\]
Astrophysics
  • Astrophysics
I mean it's pretty difficult to tell without a picture
Empty
  • Empty
Looks like you should plug in \(t=\infty\) and \(y=0\) and solve for \(y_0\)

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Astrophysics
  • Astrophysics
Ooh y = 0?
Astrophysics
  • Astrophysics
Why exactly is y = 0?
Astrophysics
  • Astrophysics
I don't think that's quite right
Empty
  • Empty
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) .
Empty
  • Empty
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.
Empty
  • Empty
Or hell, maybe they just want you to differentiate our solution and do what I described the step before.
Astrophysics
  • Astrophysics
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
ganeshie8
  • ganeshie8
Just to get some physical perspective, I think we may treat the differential equation as modeling the "money in the bank"
ganeshie8
  • ganeshie8
\[\dfrac{dy}{dt} = 3/2y + 3t+2e^t\] 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
  • ganeshie8
Once we get a feel of DE, it becomes easy to see how the initial value decides the fate of the solution..
ganeshie8
  • ganeshie8
Clearly, what ever the solution is, it is an increasing function (why ?)
ganeshie8
  • ganeshie8
You can tell that w/o solviing, just by looking at the DE..
Astrophysics
  • Astrophysics
I see that it relies on e^t
ganeshie8
  • ganeshie8
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 \(t\to\infty\)
Astrophysics
  • Astrophysics
I see, but how do we get the value of y_0?
ganeshie8
  • ganeshie8
\[y(t) = -2t-4/3-4e^t+(y_0+16/3)e^{3t/2}\] 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
Astrophysics
  • Astrophysics
Oh dang, so I sort of was on the right track haha
Astrophysics
  • Astrophysics
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
ganeshie8
  • ganeshie8
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...
Astrophysics
  • Astrophysics
Ok I think I'm getting it, thanks
Astrophysics
  • Astrophysics
Thanks for the analogies hehe
ganeshie8
  • ganeshie8
\[f(x) = -1-x-2x^2 + cx^3\] It is easy to see that the asymptotic behavior of \(f(x)\) is decided "only" by the sign of the leading coefficient, \(c\)
ganeshie8
  • ganeshie8
all other terms don't matter as \(x\to \pm \infty\)

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