Thread: Application to Differential Equations

A particle moving in a planar force field has a position vector x that satisfies x' = Ax. The 2x2 matrix A has eigenvalues 4 and 2, with corresponding eigenvectors
v1 = [-3
1] and
v2 = [-1
1] Find the position of the particle at time t, assuming that
x(0) = [-6
1].

I hate to ask like this..but I really have no clue on where to start this. I think I am trying to find x(t), but I don't know how I would go about doing that. Can anyone explain this to me or give me some hints/starting points explained? Thank you for any help.

First, since web browsers don't respect tabs, lets clean that up using LaTex:
A has eigenvalues 4 with corresponding eigenvector and 2 with corresponding eigenvector . That means that, if we used those two eigenvectors as basis vectors, we can write the equation as which is equivalent to the two equations x'= 4x and y'= 2y which have solutions and .

Using the eigenvectors as basis had the nice property of separating the equation into two independent equations but now we should also put the intial value in terms of those: which gives the two equations -3a+ b= -6 and a+ b= 1. Subtracting the second equation from the first, -4a= -7 so a= 7/4. Then 7/4+ b= 1 so b= 1- 7/4= -3/4.

That is, must satisfy so C= 7/4, D= -3/4, and .

But, once again, that is in terms of the eigenvectors as basis- those numbers are the coefficients of the eigenvectors, not the "standard basis". To put them in terms of the standard basis,
.