final notecard

# final notecard - if x(t)=0 no prey dy/dt=-by (exp model) if...

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Trajectories: parameterized solutions x(t) & y(t) of given Types of Critical Points: If direction is going in = stable If direction is going out = unstable Spiral: Proper Node: Improper Node: Center (always stable): Saddle(always unstable): λ 1 , λ 2 >0 Improper node unstabl e λ 1 , λ 2 <0 Improper node a. stable λ 1 <0< λ 2 Saddle unstabl e λ 1 = λ 2 >0 Node/saddl e unstabl e λ 1 = λ 2 <0 Node/saddl e a. stable λ 1 =a ± bi a >0 Spiral unstabl e a <0 Spiral a. stable a = 0 Center/spir al u-spiral Given x’=f(x,y) & y’=g(x,y); cp=(x 0 , y 0 )=(0,0) If not: u=x-x 0 0 system becomes… Linearization: given x’=f(x,y) & y’=g(x,y) u’=f x (x 0 ,y 0 )u + f y (x 0 ,y 0 )v v’=g x (x 0 ,y 0 )u + g y (x 0 ,y 0 )v if written in matrix form, coefficient matrix known as Jacobian . Eigenvalues tell type Predator/Prey Models: if y(t)=0 dx/dt=ax (exp model)
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Unformatted text preview: if x(t)=0 no prey dy/dt=-by (exp model) if x(t) & y(t) ≠ 0 -pxy=decline of prey qxy=growth of predator dx/dt = ax-pxy & dy/dt=-by+qxy Laplace Transform: £ (f(t))=F(s) = ∫ ∞ e-st f(t) dt Translation & Partial Fractions: £ P(s)/Q(s) where degree of P(s) > Q(s) if Q(s) =(s-a) n-- A/(s-a) + B/(s-a) 2 + … +C/(s-a) n if Q(s)=[(s-a) 2 + b 2 ] n-- (As+B)/[(s-a) 2 +b 2 ] + (Cs+D)/[(s-a) 2 + b 2 ] 2 + … + (Es+F)/[(s-a) 2 + b 2 ] n Transform of Product & Convolution: £ (f(t)g(t)) = £ (f * g) f * g = ∫ t f(z)g(t-z)dz £-1 (F(s)G(s)) = f(t) * g(t) Useful Trig identity Sin(A)Cos(B) = ½(sin(A+B)-sin(A-B)) Unit Step Function: £ (u a (t) f(t-a)) = e-as F(s) Delta Function: ∫ ∞ g(t) δ (t-a) dt = g(a)...
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## This note was uploaded on 10/06/2008 for the course MATH 216 taught by Professor Stenstones? during the Spring '07 term at University of Michigan.

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