[sage-support] Re: Solving a 2nd order linear ODE with initial conditions

2009-02-06 Thread Robert Bradshaw
On Feb 1, 2009, at 3:42 PM, David Joyner wrote: > On Sun, Feb 1, 2009 at 5:34 PM, Christophe Deroulers > wrote: >> > >> >> When one looks at what Sage sends to Maxima when "desolve(diff(y,x, >> 2)+y(x)==0,y,[0,3,2])" is called, it turns out that Maxima receives >> something like >> >> my_ode: d

[sage-support] Re: Solving a 2nd order linear ODE with initial conditions

2009-02-01 Thread David Joyner
On Sun, Feb 1, 2009 at 5:34 PM, Christophe Deroulers wrote: > > > When one looks at what Sage sends to Maxima when "desolve(diff(y,x, > 2)+y(x)==0,y,[0,3,2])" is called, it turns out that Maxima receives > something like > > my_ode: diff('y(x),x,2) + 'y(x) = 0; > my_sol: ode2(my_ode, 'y(x), x);

[sage-support] Re: Solving a 2nd order linear ODE with initial conditions

2009-02-01 Thread Christophe Deroulers
On Jan 21, 3:05 am, David Joyner wrote: > (However, for other problems, this trick will not work so well.) > I know about the problem with desolve and don't know how > to "fix" it (though, it is actually documented that way, as you will see if > you type desolve?). Sorry. I guess you know all t

[sage-support] Re: Solving a 2nd order linear ODE with initial conditions

2009-01-21 Thread Jason Grout
calcp...@aol.com wrote: > In a message dated 1/20/2009 9:37:38 P.M. Eastern Standard Time, > wdjoy...@gmail.com writes: > > (The notation diff(y,t,t) comes from the notation for partial > derivatives, > I think.) diff(y,t,t) is just a shortcut for diff(diff(y,t),t). If you do diff?

[sage-support] Re: Solving a 2nd order linear ODE with initial conditions

2009-01-21 Thread CalcPage
In a message dated 1/20/2009 9:37:38 P.M. Eastern Standard Time, wdjoy...@gmail.com writes: (The notation diff(y,t,t) comes from the notation for partial derivatives, I think.) That's odd since the given DiffEqu to solve was and ODE, not a PDE, right? TIA, A. Jorge Garcia calcp...@a

[sage-support] Re: Solving a 2nd order linear ODE with initial conditions

2009-01-20 Thread David Joyner
On Tue, Jan 20, 2009 at 9:30 PM, wrote: > In a message dated 1/20/2009 9:05:58 P.M. Eastern Standard Time, > wdjoy...@gmail.com writes: > > sage: t = var('t') > sage: x = function('x',t) > sage: DE = lambda y: diff(y,t,t) + y > sage: f = eval(desolve_laplace(DE(x(t)), ["t","x"], [0,1,0])) > sage

[sage-support] Re: Solving a 2nd order linear ODE with initial conditions

2009-01-20 Thread CalcPage
In a message dated 1/20/2009 9:05:58 P.M. Eastern Standard Time, wdjoy...@gmail.com writes: sage: t = var('t') sage: x = function('x',t) sage: DE = lambda y: diff(y,t,t) + y sage: f = eval(desolve_laplace(DE(x(t)), ["t","x"], [0,1,0])) sage: f(t) cos(t) OK, I saw this in the DiffEqu te

[sage-support] Re: Solving a 2nd order linear ODE with initial conditions

2009-01-20 Thread David Joyner
In this case, the following works: sage: t = var('t') sage: x = function('x',t) sage: DE = lambda y: diff(y,t,t) + y sage: f = eval(desolve_laplace(DE(x(t)), ["t","x"], [0,1,0])) sage: f(t) cos(t) (However, for other problems, this trick will not work so well.) I know about the problem with deso