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**Final Exam / Re: FE-3**

« **on:**April 17, 2013, 06:14:16 PM »

i did this

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It appears that I have left out a negative sign in my previous scan. My apologies. For completion, I have fixed the error in red in the re scan.

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Also, the line y = 0 is not included in the basin of attraction, as we look at the original system for dy/dt, there clearly is a y term there, which will just stay on the line y = 0. Hence, the basin of attraction for the two critical points can be represented by the 2 rectangles (one above y = 0, the other below y = 0).

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We got saddle points (unstable) for both (0,0) and (-2,0). Also (-1,-1) and (2,2) are both asymptotically stable spirals of clockwise orientation. Solutions can be seen in the visual.

Note as t approaches infinity, all points satisfying x > -2 ultimately approach the particular critical points.

Note as t approaches infinity, all points satisfying x > -2 ultimately approach the particular critical points.

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I basically wrote what Brian did for the bonus. But I suppose that if we transform the original differential equation using x = ln(t) into another DE with constants coefficients (say, change all the t's to x's), we would then be able to apply the coefficients method, and carry on to find the particular solution. Can someone confirm the validity of this?

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I'm a lil late , but here's a quick way to find Y(particular). A slower "coefficients" method can also verify that A = 1/13

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General solution is the summation of the homogeneous and particular solutions. See attachment.

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Scans attached.

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Solution is attached.

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