MHB Integration of a First order for physical application

[crevoise]

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

 

[Ackbach]

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.

 

[chisigma]

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

$\chi$ $\sigma$

 

[crevoise]

Thanks a lot for your two answers, really helpful!
I should have thought about the binomial theorem...
Thanks again