Solving non-homogeneous system of ODE using matrix exponential

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The discussion focuses on solving a non-homogeneous system of ordinary differential equations (ODEs) using the matrix exponential method. The key formula under consideration is ##\vec x = e^{At}\vec x(0) + e^{At}\int_{0}^{t} e^{-As} f(s)~ds##. Participants clarify the necessity of including constants of integration, specifically ##c_1## and ##c_2##, when using indefinite integrals in the context of this formula. The distinction between indefinite and definite integrals is emphasized, with the latter eliminating the need for arbitrary constants by incorporating initial conditions.

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Homework Statement
Please see below
Relevant Equations
Please see below
For this problem,
1715737007574.png

I don't understand why they include the constants of integration ##c_1 and c_2##, since the formula that we are meant to be using is ##\vec x = e^{At}c + e^{At}\int_{t_0}^{t} e^{-As} F(s)~ds## so we already have the integration variables. Does anybody please know why?

Thanks!
 
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ChiralSuperfields said:
Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 345266
I don't understand why they include the constants of integration ##c_1 and c_2##, since the formula that we are meant to be using is ##\vec x = e^{At}c + e^{At}\int_{t_0}^{t} e^{-As} F(s)~ds## so we already have the integration variables. Does anybody please know why?

Thanks!
Sorry I made a mistake, the equation I'm using is ##\vec x = e^{At}\vec x(0)+ e^{At}\int_{0}^{t} e^{-As} f(s)~ds##
 
They are evidently using e^{At}\int^t e^{-As}f(s)\,ds with an indefinite integral, so that an arbitrary constant of integration must be included. If instead the integral is made definite with a lower limit of t_0, then the arbitrary constant becomes \vec{x}(t_0).
 
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Thank you for your reply @pasmith! Sorry do you mean ##\vec x = e^{At}\vec x(0) + \int e^{At}f(t)~dt## as the indefinite integral?

Thanks!
 
No. The indefinite integral already includes an arbitrary constant; you don't need to add an e^{At}x(0) term in this case. That leaves e^{At} \int e^{-At}f(t)\,dt.
 
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