Show that the differential equation $\displaystyle x' = \frac{x \sin(e^x+t)}{1+(e^t \cos x+x)^2}$ has a nontrivial solution $\displaystyle \phi(t)$ defined on

[0,2] such that $\displaystyle 0 < \phi(t) < \frac{\pi}{2}$ for all t in [0,2]

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- Nov 9th 2012, 04:53 PMalphabeta89existence of non-trivial solution!!!
Show that the differential equation $\displaystyle x' = \frac{x \sin(e^x+t)}{1+(e^t \cos x+x)^2}$ has a nontrivial solution $\displaystyle \phi(t)$ defined on

[0,2] such that $\displaystyle 0 < \phi(t) < \frac{\pi}{2}$ for all t in [0,2] - Nov 10th 2012, 12:52 AMchiroRe: existence of non-trivial solution!!!
Hey alphabeta89.

Try looking at the sign of the derivative in that region and the values of the function at the initial value boundary. - Nov 10th 2012, 03:24 AMalphabeta89Re: existence of non-trivial solution!!!
Hi, I know that the denominator is always positive. But how do we deal with the numerator?

- Nov 10th 2012, 04:00 PMchiroRe: existence of non-trivial solution!!!
The way I was thinking was to consider that if you have a bound on the derivative in magnitude and you can show where it increases and decreases, then if you have a minimum value at your initial point and your derivative bound is |x'| then you know that f(x) <= I + |x'|*(b-a) where (a,b) is the interval you are considering.

Now you have x' so you can calculate the magnitude of |x'|, and if you find the turning points (when x' = 0) as well as upper bounds for when these happen (and show they are not points of inflexion), then you also get the minimums and maximums for the function.

If those points are always bounded in the region, then that is also a proof that the solution is in that interval.

These are just some suggestions.