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Question Consider $x^{'}$ = ax and $y^{'}$ = $-y$

Where a is a real number strictly less then -1

Show that all Trajectories become parallel to the y-direction as t goes to Infinity and parallel to the x direction as t goes to negative Infinity.

So im a bit Confused here, i think that if we take $y^{'}/x^{'}$ = $-y/ax$ where a is always negative. so i want to re write it as b=$-a$

Yielding $y^{'}/x^{'}$ = $y/bx$ what im confused about is there no time in here? doesn't $dy/dt$ * $dt/dx$ = $dy/dx$ ? if it does there's no time parameter.

Clearly i have no idea what im doing, and this is nothing like the other examples we have done in class or in our homework. can someone please point me in the right direction?

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1 Answer 1

up vote 2 down vote accepted

If you write this as a 2x2 matrix and find the characteristic polynomial, you get:

$$\lambda^2 + (1-a)\lambda -a = 0$$

This gives us two real and distinct negative eigenvalues (since your conditions do not allow $a = -1$).

So, for example, solve the CP for $a = \{-2, -3, -4\}$. The eigenvalues are negative, real and distinct.

How does that make the phase portrait behave? Plot the solutions and see the behavior for all three.

Update

If you write,

$x' = ax +0y$

$y' = 0x -1y$

This can be written as a matrix $A$:

$$A = \pmatrix{a & 0 \\ 0 & -1} $$

Then, solve $|A -\lambda I|= 0$.

Got it?

Regards

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Ah, a stable node with all ivp flowing to 0,0 along parabolas with the only fp at 0,0? ah nvm i see it now ty im so stupid haha. Math anything notation always confuses me. –  Faust7 Feb 28 '13 at 6:34
    
See update for how to get the CP. Make sense? –  Amzoti Feb 28 '13 at 6:43
    
Haha yeah i figured it out when i wrote out the matrix n solved with lambda in it. tyvm man! –  Faust7 Feb 28 '13 at 6:48
    
You are very welcome. It is good to think about what happens when eigenvalues are positive, negative, complex with all variants and know the behaviors. This goes a long way! Regards –  Amzoti Feb 28 '13 at 6:51
    
Great job, again, and nice to have such enthusiastic feedback, too! That's reward in and of itself! –  amWhy Apr 28 '13 at 0:25

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