Raul, where you have _1 * y you need _2 * y --Kip
On Sunday, March 15, 2015, Raul Miller <[email protected]> wrote: > On Sat, Mar 14, 2015 at 7:11 AM, Kip Murray <[email protected] > <javascript:;>> wrote: > > I recommend a matrix approach involving eigenvalues and eigenvectors, > > and assume you have a way to find eigenvalues and eigenvectors for an n > by > > n matrix which has n linearly independent eigenvectors. I confine my > > attention to linear differential equations with constant coefficients and > > right hand side 0, for example > > > > u''' - u'' - 4 u' + 4 u = 0 with initial conditions u(0) = 2 , u'(0) = > _1 > > , and u''(0) = 5 > > > > This has the solution u(t) = ( ^ _2 * t ) + ( ^ t ) where I use a mix > of > > conventional and J notation. > > I decided to try converting your notes here to J, and I ran into a snag. > > First, I tried expressing your initial constraints in J: > > constraint=: 1 :0 > if. 0 = y do. > assert. 2 = u y > assert. _1 = u D. 1 y > assert. 5 = u D. 2 y > end. > assert. 0= (u D.3 + (_1*u D.2) + (_4 * u D. 1) + 4 * u) y > ) > > And then I expressed your solution in J: > > solution=:3 :0 > (^ _1 * y) + ^ y > ) > > Or, alternatively: > > tacitsolution=: +&^ - > > And then I tested my code to see if I had gotten it right: > > solution constraint 0 > |assertion failure > | _1=u D.1 y > > Looking at this: > solution D.1 (0) > 9.76996e_8 > tacitsolution D.1 (0) > 9.76996e_8 > > working this through manually, > > dsolution=: 3 :0 > (_1 * ^ _1*y) + ^ y > ) > > dsolution 0 > 0 > > So J's implementation of D. isn't perfect, but I'm wondering if you > might not also have a typo somewhere in your presentation here? > > Thanks, > > -- > Raul > ---------------------------------------------------------------------- > For information about J forums see http://www.jsoftware.com/forums.htm > -- Sent from Gmail Mobile ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
