Re: ODE in mathematica. what is wrong?

• To: mathgroup at smc.vnet.net
• Subject: [mg91329] Re: ODE in mathematica. what is wrong?
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Fri, 15 Aug 2008 06:56:22 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <g813bb\$4r\$1@smc.vnet.net>

```hoi-su jung wrote:

> I am a freshman of Mathematica, I need help here...
>
> I met such problem as:
>
> NDSolve[{ y'[x] == -\[Rho]y[x], y[0] == Subscript[y, 0],
>   I\[HBar] y'[x] == \[Epsilon]y[x], y[0] == Subscript[y, 0],
>   y''[x] + 2 \[Rho]y'[x] + k^2 == 0, y'[0] == Subscript[g, 0]}, y, x]
> NDSolve::dvnoarg: The function y appears with no arguments.
>
> How to Handle???

Few remarks:

. Do not use subscripted functions/variables or use the Symbolize
command from the Notation package

. Spaces between variables are important since they are interpreted as
implicit multiplication (a b == a*b != ab) otherwise you just have a
name two ore more symbols

. Capital I is a reserved/built in symbol that denote the imaginary unit
Sqrt[-1], here I believe you want i

. Your system is overdetermined, which usually mean that you have
conflicting equations

. NDSolve requires to have numeric values for every parameter, otherwise
you may want to use DSolve, which is a symbolic solver

In[1]:= NDSolve[{(y^\[Prime])[x]==-\[Rho] y[x],y[0]==y0,i \[HBar]
(y^\[Prime])[x]==\[Epsilon] y[x],y[0]==y0,(y^\[Prime]\[Prime])[x]+2
\[Rho] (y^\[Prime])[x]+k^2==0,(y^\[Prime])[0]==g0},y,x]

During evaluation of In[1]:= NDSolve::overdet: There are fewer
dependent variables, {y[x]}, than equations, so the system is
overdetermined.

Out[1]= NDSolve[{(y^\[Prime])[x]==-\[Rho] y[x],y[0]==y0,i \[HBar]
(y^\[Prime])[x]==\[Epsilon] y[x],y[0]==y0,k^2+2 \[Rho]
(y^\[Prime])[x]+(y^\[Prime]\[Prime])[x]==0,(y^\[Prime])[0]==g0},y,x]

In[2]:= DSolve[{(y^\[Prime])[x]==-\[Rho] y[x],y[0]==y0},y,x]

Out[2]= {{y->Function[{x},E^(-x \[Rho]) y0]}}

Regards,
-- Jean-Marc

```

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