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

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To solve for \(x\) and \(y\) in the equation \(\frac{dx}{dt} + \frac{dy}{dt}=a-(b+c+d)y-bx\), it is noted that there are two unknown functions but only one equation, limiting the ability to solve for both simultaneously. The discussion suggests finding one variable as a function of the other, potentially using substitution to simplify the process. The user has attempted to define variables \(U\) and \(V\) and is working through a method of plugging one into the other to derive a solvable equation. Additional context and resources have been shared to clarify the problem, indicating a need for further assistance. The conversation highlights the challenges of solving systems of linear ordinary differential equations with limited equations.
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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