First Order Linear Nonhomogenous Differential Equation

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Homework Help Overview

The discussion revolves around solving a nonhomogenous first order linear differential equation, specifically examining the methods applicable to such equations and the conditions under which they can be applied.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore whether the usual nonhomogenous methods can be applied without using the integrating factor. There is a mention of variation of parameters and the method of superposition, with some questioning the necessity of linearity for these methods. One participant provides a specific example of a differential equation and discusses the solution process, including the characteristic equation and the application of variation of parameters.

Discussion Status

The discussion is active, with participants sharing different approaches and questioning the assumptions related to the linearity of the equation. There is no explicit consensus on the applicability of certain methods, but various lines of reasoning are being explored.

Contextual Notes

There is a noted ambiguity regarding the linearity of the differential equation in question, which affects the applicability of certain solution methods. The original poster's inquiry suggests a need for clarification on the methods suitable for nonhomogenous equations.

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A question, if you have a nonhomogenous first order differential equation, can it be solved by using the usual nonhomogenous way, with the arbitrary constants, instead of using the integrating factor?
 
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You can use variation of parameters or method of superposition
 
The question did NOT state that the equation was linear and both "variation of parameters" and the "method of superposition" require that the equation be linear.
 
So, let's take the following:
[tex]Ay'+By=C[/tex]

I know its solution is: [tex]\frac{C}{B}(1-e^{-t\frac{B}{A}})[/tex]

For the homogenous part:
[tex]Ay'+By=0[/tex]
Characteristic equation is:
[tex]Ak+B=0[/tex]
[tex]k=-\frac{B}{A}[/tex]

So the solution for the homogenous part is:
[tex]y=e^{-t\frac{B}{A}}[/tex]

Now, from what I have read about variation of parameters, there is a function v such that:
[tex]Y=vy[/tex]
[tex]Y'=v'y+vy'[/tex]

So, would that be:
[tex]vBe^{-t\frac{B}{A}}=Y[/tex]
[tex]v'Be^{-t\frac{B}{A}}-v\frac{B^2}{A}e^{-t\frac{B}{A}}=Y'[/tex]

Err... Correct?
 
Last edited:

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