Indefinite integral of function of x with boundary conditions
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From: Ash on 3 Jun 2010 12:51 Hi All, I have this differential equation y'(x) = C/p(x) where C is an unknown that I have to solve for. My boundary conditions are y(0) = a, y(1) = b and p(x) > 0 for all x in [0,1] to make life easier. I am trying to find y(x) and a value for C, but have no idea how to deal with the boundary conditions and the indefinite integral y(x) = INT C/p(x) that arises from integrating y'(x). Any help would be appreciated, Thanks, Ash
From: [Mr.] Lynn Kurtz on 3 Jun 2010 13:38
On Thu, 3 Jun 2010 09:51:36 -0700 (PDT), Ash <asriwi(a)googlemail.com> wrote: >Hi All, > >I have this differential equation > >y'(x) = C/p(x) > >where C is an unknown that I have to solve for. > >My boundary conditions are y(0) = a, y(1) = b and p(x) > 0 for all x >in [0,1] to make life easier. > >I am trying to find y(x) and a value for C, but have no idea how to >deal with the boundary conditions and the indefinite integral > >y(x) = INT C/p(x) > >that arises from integrating y'(x). > >Any help would be appreciated, > >Thanks, Ash Integrating from 0 to 1: b - a = y(1) - y(0) = C int[0..1] 1/p(u) du C = (b-a) / int[0..1] 1/p(u) du --Lynn http://math.asu.edu/~kurtz |