Integration of a First order for physical application

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Discussion Overview

The discussion revolves around finding an analytic solution to the first-order differential equation given by x'(t) = [A*exp(B*t)-C]^(m). Participants explore methods for integration and the implications of the equation's structure.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant seeks hints for solving the differential equation analytically, having already obtained a numerical solution using Matlab.
  • Another participant notes that the right-hand side (RHS) of the equation is independent of x, suggesting that integration can be performed directly, although it may lead to a complex integral involving hypergeometric functions.
  • A further elaboration provides a specific expression derived from the RHS, indicating that it can be integrated term by term using the binomial theorem.
  • A participant expresses gratitude for the insights, acknowledging the utility of the binomial theorem in the context of the problem.

Areas of Agreement / Disagreement

Participants appear to agree on the approach to integration and the involvement of hypergeometric functions, but no consensus on the final analytic solution is reached.

Contextual Notes

The discussion does not clarify certain assumptions regarding the parameters A, B, C, and m, nor does it resolve the complexity of the integral or the conditions under which the proposed methods are valid.

crevoise
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Hello everyone.

I wish to get the solution to the following:

x'(t) = [A*exp(B*t)-C]^(m)

I can get the plotted solution by Matlab, but I wish to find the analytic solution by myself.
Does anyone has some hints to help me in this?

Thanks a lot for your help

/Crevoise
 
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The RHS is independent of $x$. You can integrate both sides directly, although it's not a pretty integral. You've got yourself a hypergeometric function in there.
 
Ackbach said:
The RHS is independent of $x$. You can integrate both sides directly, although it's not a pretty integral. You've got yourself a hypergeometric function in there.

With simple steps You obtain...

$\displaystyle f(t)= (A\ e^{B\ t}-C)^{m}= \{-C\ (1-\frac{A}{C}\ e^{B\ t})\}^{m}= (-1)^{m}\ C^{m}\ \sum_{n=0}^{m} (-1)^{n}\ \binom{m}{n}\ (\frac{A}{C})^{n}\ e^{n\ B\ t}$ (1)

... and (1) can be integrated 'term by term'...

Kind regards

$\chi$ $\sigma$
 
Last edited:
Thanks a lot for your two answers, really helpful!
I should have thought about the binomial theorem...
Thanks again
 

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