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Topic: Integration by substitution.
Replies: 5   Last Post: Dec 17, 2012 2:53 PM

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Torsten Hennig

Posts: 2,419
Registered: 12/6/04
Re: Integration by substitution.
Posted: Dec 17, 2012 7:07 AM
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> I'm sorry, I try to better explain my point.
> Let's suppose this is a ODEs I want to solve.
> png
> y' are derivatives w.r.t the time.
> I could use a Runge-Kutta method to solve them from
> t0 a the final time (please, see
> . Well done.
> Now let's suppose I would integrate my problem in
> auxiliary variable, let' say E. It is only
> theoretically.
> (e.g. E = t^2 or E = sin(t),...). I am able to expres
> t0 e in tf in terms of E.
> How can I solve my ODE now? It is sufficient to
> multiply y' with dt/dE and use my Runge Kutta?
> Thanks for your support

Let's say t~ = g(t).
Then dt~/dt = d/dt (g(t)) and thus - if y~(t~) = y(t) - :
dy/dt = dy~/dt~ * dt~/dt = dy~/dt~ * d/dt(g(t)) = dy~/dt~ * (d/dt(g(t))_evaluated at t=g^(-1)(t~)).
Thus you will have to solve the following ODE in t~:
dy~/dt~ = f(y~(t~))/((d/dt(g(t))_evaluated at g^(-1)(t~)).

E.g. if t~ = t^2,
dy/dt = f(y) transfroms to
dy~/dt~ = f(y~)/(2*sqrt(t~))

Best wishes

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