## Astrophysics one year ago @empty and @ganeshie8

1. 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 $y(t) = -2t-4/3-4e^t+(y_0+16/3)e^{3t/2},~~~C = y_0+16/3$

2. Astrophysics

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

3. Empty

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

4. Astrophysics

Ooh y = 0?

5. Astrophysics

Why exactly is y = 0?

6. Astrophysics

I don't think that's quite right

7. 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) .

8. 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.

9. Empty

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

10. 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

11. ganeshie8

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

12. 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 $$t$$.

13. ganeshie8

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

14. ganeshie8

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

15. ganeshie8

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

16. Astrophysics

I see that it relies on e^t

17. 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 $$t\to\infty$$

18. Astrophysics

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

19. 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

20. Astrophysics

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

21. 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

22. 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...

23. Astrophysics

Ok I think I'm getting it, thanks

24. Astrophysics

Thanks for the analogies hehe

25. 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, $$c$$

26. ganeshie8

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