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Math Help - Diff Eq Translated Matrix?

  1. #1
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    Diff Eq Translated Matrix?

    This is probably a trivial problem, considering I have done most of the work already:

    -2x-y+y^2=x'
    -4x+y=y'

    This system is almost linear with Crit Points at 0,0 and 3/8,3/2

    My problem is that I cannot figure out for the life of me what the "translated matrix" of the point is 3/8,3/2 is in order to classify it. Does anyone know how to do this?

    Thank you in advance.
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  2. #2
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    Re: Diff Eq Translated Matrix?

    First do the translation: you want to "move" x and y so that x= 3/8, y= 3/2 is moved to 0, 0. Do that by taking u= x- 3/8 and v= y- 3/2. Then x= u+ 3/8, y= v+ 3/2, x'= u, y'= v', and y^2= (v+ 3/2)^2= v^2+ 3v+ 9/4 so the equations become x'= u'= -2(u+ 3/8)- (v+ 3/2)+ v^2+ 3v+ 9/2= -2u+ 2v+ v^2 and [tex]y'= v'= -4(u+ 3/8)+ (v+ 3/2)= -4u+ v.

    Now there are two, equivalent and very similar, ways to continue. The first is to recognize that the first equation is non-linear because of the " v^2". The only way to "linearize" that is to drop the y^2: u'= -2u+ 2v and v'= -4u+ v. The "translated matrix" is \begin{bmatrix}-2 & 2 \\ -4 & 1\end{bmatrix}.

    The other way is to form the "Jacobian". If x'= f(x,y) and y'= g(x,y) the Jacobian is \begin{bmatrix}\frac{\partial f}{\partial x} & \frac{\partial g}{\partial x} \\ \frac{\partial f}{\partial y} & \frac{\partial g}{\partial y}\end{bmatrix}

    In this case that would be [tex]\begin{bmatrix} \begin{bmatrix}-2 & 2+ 2v \\ -4 & 1\end{bmatrix}. Now let u and v be 0 which gives the same matrix as before. This second method is more "formal" and helps determine how to "linearize" a formula.
    Last edited by HallsofIvy; March 6th 2014 at 07:29 AM.
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