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Friday, 24 February 2017

Existence theorem for planar curves (II)

We follow with the above entry taking an 'easy' function k: find a planar curve as a   y=f(x) where k(x)=2x.

By the given method in the last entry, we integrate k between a and x. Take a=0, obtaining  g(x)=x2 (here m=0). Then the solution is
y(x)=x0g(t)1g(t)2dt.

In this case,
x0t21t4dt.

This integral is not possible to compute!

Other example. Take k(x)=ex. Now
x0etdt=ex1.

If we take m=1, then g(x)=ex and finally f(x)=:0xet1e2tdt=arcsin(ex).
In order to make sense in the integrand, we are assuming that  x<0. The solution is y(x)=arcsin(ex) whose graphic is

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