Re: [Maxima-discuss] Second order differential equation with boundary conditions

[serge de m. <sde...@gm...>]

Hello,

Indeed, bc2 seems to have an issue with the fact that the second boundary
condition is expressed as a function of y, or at least I wasn't able to use
bc2 to solve the equation.

But you can use maxima to assist you in doing what you did by hand:
   subst([x=0,y=0],sol);
gives you an equation in %k2, which you can solve:
   k2:solve(%)[1];
Now, for the second boundary condition, we substitute the solution for the
differential equation and k2 into it:
  subst([sol,k2],-y='diff(y,x));
This equation still has an unevaluated diff, which we can evaluate like
this:
  ev(%,nouns);
This is only valid for x=1 (as per the second boundary condition), so:
  subst(x=1,%);
This gives an equation in %k1, which we can solve:
  k2:solve(%)[1];

Finally, we can compute the solution to the boundary condition problem by
putting k1 and k2 back into sol:
  subst([k1,k2],sol);

  serge


On Thu, Jul 28, 2016 at 10:32 PM, Pablo Fonovich <Pab...@ho...>
wrote:

> Hi:
>
> I'm a begginer with maxima, and i want to solve this second order
> differential equation with boundary conditions:
>
>
> 'diff(y,x,2)+y+1=0
>
> y=0 for x=0
>
> 'diff(y,x)=-y for x=1
>
>
> Can i do this with bc2?
>
> i tried the following:
>
>
> eq: 'diff(y,x,2)+y+1=0;
>
> sol: ode2(eq,y,x);
>
> bc2: bc2(sol,x=0,y=0,x=1,'diff(y,x)=-y);
>
>
> but i'm getting a confusing result.
>
>
> By using the output of sol: ode2(eq,y,x); and applying the conditions by
> hand i arrived to the following solution:
>
> y=-1.2676*sen(x)+cos(x)-1
>
>
> How could i achieve this with maxima?
>
>
> Thanks a lot
>
>
>
>
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