Date: Jul 20, 2013 5:54 AM
Author: Bob Hanlon
Subject: Re: f'[0]=0.5 is True?


Try starting with a fresh kernel.


Clear[f, x, sol];


eqn = f''[x] + 2 f'[x] + 30 f[x] == 0;
eqn1 = f[0] == 1;
eqn2 = f'[0] == 0.5;


sol[x_] = f[x] /. DSolve[{eqn, eqn1, eqn2}, f[x], x][[1]]


(1.*(1.*Cos[Sqrt[29]*x] + 0.2785430072655778*
Sin[Sqrt[29]*x]))/E^x


sol[0]


1.


sol'[0]


0.5


Using exact numbers


eqn2 = f'[0] == 1/2;


sol[x_] = f[x] /. DSolve[{eqn, eqn1, eqn2}, f[x], x][[1]]


((1/58)*(58*Cos[Sqrt[29]*x] + 3*Sqrt[29]*
Sin[Sqrt[29]*x]))/E^x


sol[0]


1


sol'[0]


1/2



Bob Hanlon




On Thu, Jul 18, 2013 at 3:00 AM, mariusz sapinski <
mariusz.sapinski@gmail.com> wrote:

> Dear All,
>
> I'm trying a simple exercise:
>
> eqn = f''[x] + 2 f'[x] + 30 f[x] == 0;
> Clear[f];
> Clear[x];
> eqn1 = f[0] == 1;
> eqn2 = f'[0] == 0.5;
> DSolve[{eqn, eqn1, eqn2}, f[x], x]
>
>
> and I get:
> DSolve::deqn: Equation or list of equations expected instead of True in
> the first argument {30 f[x]+2
> (f^\[Prime])[x]+(f^\[Prime]\[Prime])[x]==0,f[0]==1,True}. >>
>
> so f'[0]=0.5 is True for Mathematica?
>
> How can it be?
>
> If I remove eqn2 from DSolve then I get a solution with a parameter of
> course.
>
> Cheers,
>
> Mariusz
>
>