Recent content by donifan

Just realized there is a typo (there is a d before y' that is no suppose to be there). So the whole problem is y'=\exp[-f(t)y] with y(0)=y_0 and f(t) = (a^{2} +b t)^{3/2} - a^3 Thanks.

The form of f(t) is f(t) = (a^{2} +b t)^{3/2} - a^3 a and b are constantly changing. The simplest case is a=0. In principle a look up table would work, but I was wondering if a more elegant solution can be reached. Thanks.

The answer is simple: time. An analytical approximation is faster. The solution is suppose to be part of a humongous code that cannot spare a microsecond. P.S. When I said error, I meant from the approximated analytical solution to the actual solution.

Analytical approximation would be something like a solution of a different but related problem, so the order of the error can be estimated. I know. The analytical solution by the Picard method is pretty much impossible. Maybe there is no solution however the equation seems so simple a...

Sorry about the confusion. Yes, I am looking for an analytical solution/approximation. I agree the numerical solution is actually simple. Thanks!

Thanks for the suggestion. However even for the simple case of f=t^3/2, y_2 is already a nasty function. I'll keep trying.

A "simple" nonlinear ode Hi Does anyone see a way to solve/approximate this ODE? dy'=exp(-f(t)y) with y(0)=yo f(t) can be as simple as c*t^3/2 but it may be more complex. This came out as the solution of a very complex problem. This is the final threshold. Thanks, Donifan

Sorry about the typo, that's what I meant. The solution for f is actually for the right equation (the one with the minus in the RHS).

Thanks guys! First of all, I had a mistake in the first post (nothing about the series). The equivalent problem must be: \frac{d^{n}f(0)}{dx^{n}} = e^{cn^{2}} So far I was able to check that f^{'}(x) = e^cf(e^{2c}x) (Very nice work Wizlem). I even tried making the substitution...

No biggie, x<0. To avoid confusion, let's write it better like \sum{\frac{(-1)^nt^n}{n!}e^{cn^2}} I also know from the physical process that: \displaystyle\lim_{t\to\infty} f(t)=0 Thanks!

No need to evaluate it. The integral is the first moment (mean) of an exponential distribution on x, so it is equal to \lambda^{-1}

Hello, I am trying to evaluate the series \sum{\frac{x^n}{n!}e^{cn^2}} where c is a constant. I think this problem is equivalent to find f(x) such that \frac{d^{n}f(0)}{dx^{n}} = \frac{e^{cn^{2}}}{n!} I believe this must be a modified exponential since for c=0, it reduces to...