?How to know if equilibrium points are stable or not. Is my solution correct

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  • #26
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ok, i am very grateful for your help. i have my exam in 3 days.

can you please kindly tell me how do i answer part e. wether there is a unique solution through the point (1,2).

i know that to prove uniqueness, we must have y'=f(t,y), and differentiate w.r.t to y. i.e find (f(t,y))'

and then show that (f(t,y))' is continuous.

to show whether or not it has a unique solution at (1,2), should i sub t=1 and y=2 into (f(t,y))', and if i get a unique solution. does that mean there is a unique solution through the point (1,2)??
Aren't you skipping a step? You have y' = f(t, y) = (t2 - y - 2)(1 - y2).

For there to be a unique solution at (1, 2), both f and fy have to be defined and continous on some rectangle around (1, 2). The part that you seem to have skipped is finding
[tex]f_y = \frac{\partial }{\partial y}f(t, y)[/tex]
 
  • #27
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Aren't you skipping a step? You have y' = f(t, y) = (t2 - y - 2)(1 - y2).

For there to be a unique solution at (1, 2), both f and fy have to be defined and continous on some rectangle around (1, 2). The part that you seem to have skipped is finding
[tex]f_y = \frac{\partial }{\partial y}f(t, y)[/tex]
First y’’=-2(t^2-2)y+3y^2-1,

http://www.wolframalpha.com/input/?i=-%28t^2-+y-+2%29%28y-+1%29%28y%2B+1%29

at t=1 and y=2, we get that the only solution is 15. So the point (1,2) gives a unique solution, as required

http://www.wolframalpha.com/input/?i=-2+%28t^2-2%29+y%2B3+y^2-1%2C+t%3D1%2Cy%3D2

is my solution correct?
 
  • #28
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First y’’=-2(t^2-2)y+3y^2-1,
No, this isn't even very close.
y' = f(t, y) = (t2 - y - 2)(1 - y2). Here, y' means dy/dt.
To calculate fy, which in different notation is
[tex]\frac{\partial f}{\partial y}[/tex]
you need to use the product rule. I think you might have done this, but what you got is incorrect.

http://www.wolframalpha.com/input/?i=-%28t^2-+y-+2%29%28y-+1%29%28y%2B+1%29

at t=1 and y=2, we get that the only solution is 15. So the point (1,2) gives a unique solution, as required
This isn't about solutions or unique solutions. All you need to do is confirm that f(t, y) and fy(t, y) are continuous in some rectangle around the point (1, 2).

http://www.wolframalpha.com/input/?i=-2+%28t^2-2%29+y%2B3+y^2-1%2C+t%3D1%2Cy%3D2

is my solution correct?
No. See above.
 
  • #29
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No, this isn't even very close.
y' = f(t, y) = (t2 - y - 2)(1 - y2). Here, y' means dy/dt.
To calculate fy, which in different notation is
[tex]\frac{\partial f}{\partial y}[/tex]
you need to use the product rule. I think you might have done this, but what you got is incorrect.

This isn't about solutions or unique solutions. All you need to do is confirm that f(t, y) and fy(t, y) are continuous in some rectangle around the point (1, 2).


No. See above.
I did differentaite w.r.t y, see wolfram

http://www.wolframalpha.com/input/?i=-%28t^2-+y-+2%29%28y-+1%29%28y%2B+1%29

then i sub t=1 y=2 and got 15
 
  • #30
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Apologies, your derivative was correct. Your notation was not so good, though.
If y' means dy/dt, then y'' means d2y/dt2, and that's not what is called for here. Instead of y'', you should use fy or other notation to indicate that you're taking the partial with respect to y.

The object isn't merely to evaluate f(1, 2) and fy(1, 2). You need to show that f(t, y) and fy(t, y) are continuous in some rectangle around (1, 2). Can you do that?
 
Last edited:
  • #31
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Apologies, your derivative was correct. Your notation was not so good, though.
If y' means dy/dt, then y'' means d2y/dt2, and that's not what is called for here. Instead of y'', you should use fy or other notation to indicate that you're taking the partial with respect to y.

The object isn't merely to evaluate f(1, 2) and fy(1, 2). You need to show that f(t, y) and fy(t, y) are continuous in some rectangle around (1, 2). Can you do that?
Fy=-2(t^2-2)y+3y^2-1

i have no idea how to show this is continuous?

is there a method. if so i dont know what it is, cany you give me a link to a website which possibly teaches you how to know if somethig is continous.

i know how to prove a function is 1-1 though
 
  • #32
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Both functions are defined at all point (t, y) in the plane. Since the only operations involved in their formulas are multiplication and addition, both functions are also continuous at every point (t, y) in the plane.

The basis ideas involved here are these: if g(x) and h(x) are continous functions, the the sum (g + h)(x) and product gh(x) are continous. both f(t, y) and fy(t, y) can be split up into the sum or product of simple linear or quadratic functions, which is enough to show that f(t, y) and fy(t, y) are continuous. This is enough to show that the differential equation y' = f(t, y) has a unique solution that contains the point (1, 2). That's what you wanted to show in part e, so you're done.
 

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