It would help if you'd post the code you are having problems with. I don't experience the problem you seem to run into:

```
sage: var('a,b,c')
(a, b, c)
sage: function('f')
f
sage: function('g')
g
sage: E=a*f(x)+b*diff(f(x),x)+c*diff(f(x),x,x)
sage: E
a*f(x) + b*D[0](f)(x) + c*D[0, 0](f)(x)
sage: E.substitute_function(f,g)
a*g(x) + b*D[0](g)(x) + c*D[0, 0](g)(x)
```

f gets beautifully replaced by g. It seems from your question that you might be trying

```
sage: E.substitute_function(f(x),g(x))
a*f(x) + b*D[0](f)(x) + c*D[0, 0](f)(x)
```

which is not according to the documentation of substitute_function and indeed doesn't work: the arguments should be functions, not expressions.

**edit:** Thank you for clarifying (editing your question would have been better than posting the additional information as an answer)

You're a victim of erroneous documentation. The command `f=function('f',x)`

doesn't give you a function, it gives you an expression "function f evaluated at x". This is misused in a lot of places, particularly the DE code. It means that the argument that you give to `substitute_function`

isn't a function. Try and print it:

```
sage: f=function('f',x) # This doesn't do what you might think it does. Don't use it.
sage: f
f(x)
sage: f(x)
DeprecationWarning: Substitution using function-call syntax and unnamed arguments is deprecated and will be removed from a future release of Sage; you can use named arguments instead, like EXPR(x=..., y=...)
See http://trac.sagemath.org/5930 for details.
f(x)
```

The DeprecationWarning is already an indication that typing in `f(x)`

isn't what you'd expect it to be.

If instead you do the following, things works as you'd expect.

```
sage: var('x,a,b')
(x, a, b)
sage: function('f')(x)
f(x)
sage: assume(a>0); assume(b<0)
sage: eqn = a*diff(f(x),x,2) == b*f(x)
sage: g(x) = desolve_laplace(eqn, f(x), ivar=x)
sage: eqn.substitute_function(f,g)
b*cos(sqrt(-a*b)*x/a)*f(0) - sqrt(-a*b)*sin(sqrt(-a*b)*x/a)*D[0](f)(0) == (a^2*sin(sqrt(-a*b)*x/a)*D[0](f)(0)/sqrt(-a*b) + a*cos(sqrt(-a*b)*x/a)*f(0))*b/a
```

You just did what is in the documentation of `desolve_laplace`

, but as you found out, the code there doesn't quite do what it claims, so that `substitute_function`

doesn't work as expected. You're not the first one to fall in this trap. There are some tickets about it, see http://trac.sagemath.org/ticket/17447 for instance.