Dif.equation system

  • Thread starter prehisto
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  • #1
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Homework Statement


System1:
dx/dt=x+y
dy/dt=8x-y


Homework Equations





The Attempt at a Solution


detreminant=(1-λ)(-1-λ)=(λ-3)(λ+3);λ[itex]_{1}[/itex]=-3 and λ[itex]_{2}[/itex]=3

So system 2:
(1-λ)[itex]\alpha[/itex]+[itex]\beta[/itex]=0
8[itex]\alpha[/itex]+(-1-λ)[itex]\beta[/itex]=0

When i put λ[itex]_{1}[/itex]=-3 in system 2 -> [itex]\alpha[/itex] and [itex]\beta[/itex]=0.
the same goes for λ[itex]_{2}[/itex]

That menas that solution in form of y=C_1*[itex]\beta[/itex]_1*exp(λ[itex]_{1}[/itex]*t)+C_2*[itex]\beta[/itex]_2*exp(λ[itex]_{2}[/itex]*t) is equal to 0. Thats wrong.

Where is my mistake?
 
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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi prehisto! :smile:

so far so good! …
(1-λ)[itex]\alpha[/itex]+[itex]\beta[/itex]=0
8[itex]\alpha[/itex]+(-1-λ)[itex]\beta[/itex]=0

now solve either line to get β = 2α, so your eigenvector is any multiple of x + 2y :wink:
 
  • #3
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ok,that means that i can chose α1=1 β1=2 and
α2=1 β2=-4

y=C11*exp(λ1*t)+C22*exp(λ2*t)
x=C11*exp(λ1*t)+C22*exp(λ2*t)
Is this form of solution correct or I have to use something else?
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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i think it would be better if you checked by starting with the eigenvector equations

x + 2y = Ae3t
x - 4y = Ae-3t
 

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