Astrophysics
  • Astrophysics
@ganeshie8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Astrophysics
  • Astrophysics
|dw:1449718570770:dw|
Astrophysics
  • Astrophysics
I need to find the critical values or values of alpha where the qualitative nature of the phase portrait changes the origin
Astrophysics
  • Astrophysics
|dw:1449718686257:dw| these are the eigenvalues

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Astrophysics
  • Astrophysics
This is like the last thing we did lol so I have no idea what it really means, have to find critical values then talk about degenerate nodes and such
Astrophysics
  • Astrophysics
It's like where the eigen values are negative and positive right
ganeshie8
  • ganeshie8
Yeah, I think you need to check when the eigenvalues change from real to complex or the real part changes from negative to positive etc..
Astrophysics
  • Astrophysics
So the critical values are from solving what is under the square root then
ganeshie8
  • ganeshie8
|dw:1449718963888:dw|
Astrophysics
  • Astrophysics
\[25+8 \alpha = 0 \implies \alpha = \frac{ -25 }{ 8 }\]
Astrophysics
  • Astrophysics
Are there more?
Astrophysics
  • Astrophysics
Would I have to solve for alpha if I set both eigen values = 0
Astrophysics
  • Astrophysics
Such a weird question, like how would i know where the stable spiral and such are
ganeshie8
  • ganeshie8
My knowledge is a bit shaky on these... @SithsAndGiggles knows these very well
ganeshie8
  • ganeshie8
Are you also asked to sketch the phase potrait ?
Astrophysics
  • Astrophysics
Nope
Astrophysics
  • Astrophysics
|dw:1449719545007:dw|
anonymous
  • anonymous
omg 2 much math tot lost
anonymous
  • anonymous
can one of you guys help me with: Find all solutions in the interval [0, 2π). (sin x)(cos x) = 0 I have no idea where to start
Astrophysics
  • Astrophysics
Use your trig identities
Astrophysics
  • Astrophysics
I think there maybe more critical values
UsukiDoll
  • UsukiDoll
I had these last semester, but I forgot... It has something to do with the eigenvalues and eigenvectors and there were five cases like a node, spiral, stable or unstable. yeahhh forgot >_<
IrishBoy123
  • IrishBoy123
you will have a spiral for complex e values. as the real part is negative, ie it is \(\color{red}{-} 1 \pm \beta i \), it will be a stable spiral. basically, you have exponential decay about that point. for repeated e values, ie where the radical is zero, \(\lambda_{1,2} = -1\), the trace of the matrix is -2, so it will be a stable improper node. the negative e value connote shrinkage or convergence on the critical point itself. lastly, you can look at real evalues that are not equal you will have a saddle if they are of different sign. so one e value s expanding in the direction of its e vector, the other contracting in the direction of its e vector. so you get that characteristic saddle shape. and a stable node if both neg, ie \(\beta \lt 1 \) they can't both be positive

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