Matrix Exponential: Solve Homework Equation w/ Initial Condition

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    Exponential Matrix
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Homework Help Overview

The discussion revolves around solving a differential equation involving a matrix exponential, specifically x' = Ax, where A is a 2x2 matrix. The original poster seeks to compute the matrix exponential and apply an initial condition to find the solution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to compute the matrix exponential and questions how to apply the initial condition. They consider whether to multiply the original matrix A by the initial condition vector or to use the computed matrix exponential directly.

Discussion Status

Some participants provide clarification on the relationship between the matrix exponential and the solution to the differential equation, noting that the solution can be expressed as x(t) = e^(At)x(0). There is a discussion about the necessity of differentiation, with some participants suggesting that it is not required to find the solution.

Contextual Notes

The original poster expresses uncertainty regarding their notes and the process of applying the initial condition, indicating a potential gap in understanding the relationship between the matrix exponential and the initial condition.

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Homework Statement



Given x' = Ax where A =

( 0 1 )
( -1 0 )

Compute the matrix exponential and then find the solution such that x(0) =

( 1 )
( 2 )

Homework Equations



The Attempt at a Solution



I computed the matrix exponential and obtained the matrix,

e^(A) =

( cos(t) sin(t) )
( -sin(t) cos(t) )

But I don't understand how to compute the initial condition. Am I supposed to compute the initial by multiplying the original A by x(0) and then compute the matrix exponential for the new A? Or multiple e^(A) by x(0)? My notes aren't very clear. But those are my only guesses..

Thanks for any help.
 
Last edited:
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What you computed was e^(At). And sure, your solution is then x(t)=e^(At)x(0). If you take d/dt of that then you get x'(t)=Ax(t), right?
 
Alright, I think I've got it. Compute x(t) then differentiate it?
 
You don't have to differentiate it, I was just pointing out why x(t)=e^(At)x(0) works as a solution. x'(t)=Ax(t) and x(0)=e^(0)x(0). It satisfies the ode and has the right initial condition.
 
Alright, thank you.
 

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