Solving non-homogeneous system of ODE using matrix exponential

  • Thread starter Thread starter member 731016
  • Start date Start date
  • Tags Tags
    Ode
Click For Summary
The discussion centers on the confusion regarding the inclusion of constants of integration in the solution of a non-homogeneous system of ordinary differential equations (ODEs) using the matrix exponential method. Participants clarify that while the formula involves an integral, it inherently includes an arbitrary constant of integration, which is necessary when dealing with indefinite integrals. When the integral is defined with specific limits, the constant can be represented as the initial condition vector, denoted as x(t0). The conversation emphasizes the distinction between indefinite and definite integrals in the context of solving ODEs. Understanding these nuances is crucial for correctly applying the matrix exponential approach to non-homogeneous systems.
member 731016
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!
 
Physics news on Phys.org
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).
 
  • Love
Likes member 731016
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.
 
Last edited by a moderator:
  • Love
Likes member 731016