Ordinary Differential Equation System with Variable Coefficients

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SUMMARY

The discussion focuses on solving the ordinary differential equation system represented by the matrix A with variable coefficients. The matrix A is defined as A = [[-1, 0, 0], [0, 2, 2t], [0, 0, 2]]. The solution approach involves using the formula x(t) = x_0 exp(∫ f(ξ) dξ) after diagonalizing the matrix A. Participants express uncertainty about applying methods used for constant coefficients to this variable coefficient scenario.

PREREQUISITES
  • Understanding of ordinary differential equations (ODEs)
  • Familiarity with matrix diagonalization techniques
  • Knowledge of variable coefficient systems
  • Proficiency in integration techniques
NEXT STEPS
  • Study the method of diagonalization for matrices with variable coefficients
  • Learn about the Lyapunov stability theory in relation to ODEs
  • Explore the application of the variation of parameters method for non-constant coefficients
  • Investigate numerical methods for solving ODEs with variable coefficients
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Students and researchers in mathematics, particularly those focusing on differential equations, as well as educators seeking to enhance their understanding of variable coefficient systems.

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Homework Statement


For t\in\mathbb{R}, let:

A=\left[\begin{array}{ccc}<br /> -1 &amp; 0 &amp; 0\\<br /> 0 &amp; 2 &amp; 2t\\<br /> 0 &amp; 0 &amp; 2\end{array}\right]

Get the solution for the general equation: X&#039;=A(t)X

Homework Equations



The Attempt at a Solution


I done many of these problems, all with constant coefficients, but I don't know how to do in this case.
 
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Did you know that the solution to
x&#039;(t) = f(t) x(t)
is
x(t) = x_0 \exp\left( \int_{t_0}^t f(\xi) \, d\xi \right)
?
 
Yes, I know. That's the formula I use after getting the exponential matrix, by "diagonalizing" the matrix A. My problem is that I'm not sure if I can do it as I do with constant coefficients, because supposedly it should be different.
 

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