- #1
roam
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Homework Statement
http://img444.imageshack.us/img444/9288/51927159.jpg
The Attempt at a Solution
(a) I'm mostly stuck on this part because I keep getting a complex number:
3x+y = 0 so x=-y/3
Substituting this in the second equation:
[itex]\frac{y^2}{9} +1 = 0[/itex]
[itex]\therefore \ y = \sqrt{-9} \implies y = 3 i[/itex]
Now putting this back into x=-y/3 we will get x=-i. So the only equilibrium point is (-i, 3i)? (b) We can write the first equation yx-y as y(x-1) = 0. So the solutions here would be y=0 and x=1.
And if we set the second equation equal to 0: cos(πx)=0, the solution would be
[itex]x=\frac{k \pi}{2}[/itex]
Where k is any odd integer (since cosine is zero at points like -π/2, π/2, 3π/2, etc).
So the equilibrium points would be (kπ/2, 0) and (1, 0). Is this right? And could there be any more solutions?
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