Finding the fundamental matrix where psi(0) = the identity matrix

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SUMMARY

The discussion focuses on finding the fundamental matrix ψ(t) for a system of first-order linear equations, specifically where ψ(0) equals the identity matrix. The proposed solution for ψ(t) is ψ(t) = <, <-e^{-3t}, e^{-t}>>. However, this does not yield the identity matrix at t=0. The conversation suggests using linear combinations of the basis vectors to construct the required fundamental matrix that satisfies the initial condition.

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Homework Statement



If I have a solution to a system of first order linear equations: [itex]<x,y> = c_1 e^{-3t} <1,-1> + c_2 e^{-t} <1,1>[/itex] , how do I find the fundamental matrix psi(t) so that psi(0) = I ?

Homework Equations





The Attempt at a Solution



[itex]psi(t) = <<e^{3t}, e^{-t}>, <-e^{-3t}, e^{-t}>>[/itex]
[itex]psi(0) = <<1, 1>, <-1, 1>>[/itex]

This is clearly not the identity matrix.
Now what?
 
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just an idea but you have two linearly independent basis vectors call them u1, u2

could you take a linear comibination of your vectors such that
v1=au1+bu2 gives v1(0) = <1,0>
and
v2=cu1+du2 gives v2(0) = <0,1>?

then could you use those basis vectors to write the reuired fundamental matrix?
 

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