How to Solve for \(x\) and \(y\) in a System of Linear ODEs?

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Discussion Overview

The discussion revolves around solving a system of linear ordinary differential equations (ODEs) involving two unknown functions, \(x(t)\) and \(y(t)\). Participants explore strategies for finding solutions given a single equation that relates the derivatives of these functions.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant notes that with two unknown functions and only one equation, it is impossible to solve for both \(x(t)\) and \(y(t)\) simultaneously, suggesting that one can express one function in terms of the other.
  • Another participant describes a method where they first solve for \(U\) and then for \(V\), indicating a preference for substituting \(U\) into the equation for \(V\) to simplify the process.
  • A request for clarification on the definitions of \(U\) and \(V\) is made, along with a call for more context regarding the problem's origin and additional information that may aid in finding a solution.
  • Links to external resources on similar problems are provided, indicating that previous attempts to seek help have been largely unproductive.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of solving for both functions simultaneously, with some agreeing on the limitations imposed by having only one equation. The discussion remains unresolved regarding the best approach to take in solving the system.

Contextual Notes

The discussion highlights the challenge of solving a system of equations with insufficient equations relative to the number of unknowns. There are references to external resources that may contain relevant information, but the specific assumptions and definitions used in the problem are not fully clarified.

kalish1
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I want to solve for $x$ and $y$ from the equation $$\frac{dx}{dt} + \frac{dy}{dt}=a-(b+c+d)y-bx.$$

What is the best strategy?
 
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kalish said:
I want to solve for $x$ and $y$ from the equation $$\frac{dx}{dt} + \frac{dy}{dt}=a-(b+c+d)y-bx.$$

What is the best strategy?

You have two unknown function x(*) and y(*) and only one equation. That means that You can find x(*) as function of y(*) and y'(*) or vice versa, not both x(*) and y(*)...

Kind regards

$\chi$ $\sigma$
 
chisigma said:
You have two unknown function x(*) and y(*) and only one equation. That means that You can find x(*) as function of y(*) and y'(*) or vice versa, not both x(*) and y(*)...

Kind regards

$\chi$ $\sigma$

$1.$ I solved for $U$.

$2.$ Then I solved the equation for $V$.

$3.$ Now I am plugging in the expression for $U$ into the equation for $V$ (because it looks easier than plugging in the other way around).

Shouldn't this give me a solvable equation for $V$, thus giving $V$, and then let me plug back into the main equation to get $U$??
 
kalish said:
$1.$ I solved for $U$.

$2.$ Then I solved the equation for $V$.

$3.$ Now I am plugging in the expression for $U$ into the equation for $V$ (because it looks easier than plugging in the other way around).

Shouldn't this give me a solvable equation for $V$, thus giving $V$, and then let me plug back into the main equation to get $U$??

Could you please explain what $U$ and $V$ are? I think it'd also be terrific if you could please provide more context for this problem. How did this problem come to you? Is there any other information you have?
 
Ackbach said:
Could you please explain what $U$ and $V$ are? I think it'd also be terrific if you could please provide more context for this problem. How did this problem come to you? Is there any other information you have?

Sure. I have provided the full details in the following 3 links:
differential equations - Solution to system of nonlinear ODEs - Mathematics Stack Exchange
differential equations - System of $2$ nonlinear DEs - Mathematics Stack Exchange
differential equations - Solving a system of linear ODEs - Mathematics Stack Exchange

People have largely been unable to help. I hope you can help.
 

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