# Diff Eq Translated Matrix?

• Mar 6th 2014, 06:36 AM
stealth4933
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?

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.