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November 21st, 2012, 09:10 AM   #1
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stationary points

Find the stationary points of the dierential equation
d3x/dt3 +dx/dt+d2x/dt2 = x^3 + 0x^2 -39x -70
where x = x(t). Determine whether the stationary points are stable or unstable.

how would i approach this? thanks
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November 21st, 2012, 09:42 AM   #2
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Re: stationary points

I suggest you start by looking up the definition of "stationary point"! What is the definition?
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November 21st, 2012, 09:48 AM   #3
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Re: stationary points

a point where f '(x)=0
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November 21st, 2012, 10:22 AM   #4
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Re: stationary points

Which makes absolutely no sense here because there is no "f" in what you gave! But I suspect you are recalling that definition from Calculus. For differential equations, the definition is a little different. The basic definition really applies to systems of first order equations but the result is essentially that x and all of its derivatives are 0. And that is equivalent to setting the right side equal to 0. Then solve for x.

Now, what distinguishes "stable" and "unstable" equilibrium points?
(Once you have solved the equation, looking at the factors will answer your question.)
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November 21st, 2012, 10:27 AM   #5
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Re: stationary points

Quote:
Originally Posted by HallsofIvy
Which makes absolutely no sense here because there is no "f" in what you gave! But I suspect you are recalling that definition from Calculus. For differential equations, the definition is a little different. The basic definition really applies to systems of first order equations but the result is essentially that x and all of its derivatives are 0. And that is equivalent to setting the right side equal to 0. Then solve for x.

Now, what distinguishes "stable" and "unstable" equilibrium points?
(Once you have solved the equation, looking at the factors will answer your question.)
yes i was recalling the basic definition for it from last year calculus.

so make the rhs equal to 0 and solve from there ok thanks

and the difference is that a stable point is an attractor per say as the system we are working with alwys comes back to that point, with unstable its like repeller moves them away.
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November 21st, 2012, 12:51 PM   #6
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Re: stationary points

Which means you need to look at the sign on either side of the equilibrium points.
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