- #1
stunner5000pt
- 1,461
- 2
For this system of differential equations
[tex] \frac{dx}{dt} = x(0.8 -\frac{0.8x}{3129} - \frac{y}{x+1000}) [/tex]
[tex] \frac{dy}{dt} = y(-0.5 + \frac{x}{1000+x})[/tex]
for all [tex] x,y \geq 0 [/tex]
now the equilibrium points are
(x,y) = (0,0), (1000,1088), (3129,0)
i need to linearize this system such taht i can figure out the eigenvalues of this system and figure out whether this system is a spiral sink, source or whatever
if i use x = u -1000 and y = v-1088
i get answers like this
[tex] \frac{dx}{dt} = (1.054u +2.55*10^{-4} u^2 -v) \frac{u+1000}{u+2000} [/tex]
[tex] \frac{dy}{dt} = \frac{0.5uv - 544u}{2000 + u} [/tex]
but hte denominator throws things off, doesn't it??
but if i just accept it like it is and assumethat as u,v approach zero the non linear terms get insignificant (ya you not proper math language)
and the matrix becomes
[tex] \left(\begin{array}{cc}0.527&-0.5\\0.272&0\end{array}\right) [/tex]
is this right so far?? Any help would be appreciated, greatly!
[tex] \frac{dx}{dt} = x(0.8 -\frac{0.8x}{3129} - \frac{y}{x+1000}) [/tex]
[tex] \frac{dy}{dt} = y(-0.5 + \frac{x}{1000+x})[/tex]
for all [tex] x,y \geq 0 [/tex]
now the equilibrium points are
(x,y) = (0,0), (1000,1088), (3129,0)
i need to linearize this system such taht i can figure out the eigenvalues of this system and figure out whether this system is a spiral sink, source or whatever
if i use x = u -1000 and y = v-1088
i get answers like this
[tex] \frac{dx}{dt} = (1.054u +2.55*10^{-4} u^2 -v) \frac{u+1000}{u+2000} [/tex]
[tex] \frac{dy}{dt} = \frac{0.5uv - 544u}{2000 + u} [/tex]
but hte denominator throws things off, doesn't it??
but if i just accept it like it is and assumethat as u,v approach zero the non linear terms get insignificant (ya you not proper math language)
and the matrix becomes
[tex] \left(\begin{array}{cc}0.527&-0.5\\0.272&0\end{array}\right) [/tex]
is this right so far?? Any help would be appreciated, greatly!
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