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sympydsolve

Custom arbitrary constant names for sympy dsolve?


Is there a way to provide get dsolve() to use arbitrary constants named by a different sequence than C1, C2, etc?

cse() allows a symbols parameter that accepts an infinite generator of names, but I don't see a similar parameter to dsolve().

I'm solving circuit equations full of symbolic capacitances, and the dsolve constants are getting confusing.

Failing that, is there a quick way to replace the arbitrary constants with others such as k_1, k_2, etc? It looks like dsolve() is using C1, C2, etc, for its constants while my code uses C_1, C_2, etc for capacitances. I could change my naming everywhere to use a non-standard capacitance symbol, but I'd prefer not to.


With thanks to @Marshmallow, I've started using this routine which wraps dsolve lets me change the symbols after the fact. There's still the risk of collision, but it's good enough for now:

def newdsolve(eq,*args,**kwds):
    ss=kwds.get('symbols')
    sln=dsolve(eq,*args,**kwds)
    psyms=((newdsolve.csre.match(n.name),n) for n in eqVc.free_symbols)
    if ss and isinstance(ss,str):
        subsdict=dict([(n[1],'k_'+n[0].group(1)) for n in psyms if n[0]])
    elif ss:
        subsdict=dict([(n[1],next(ss)) for n in psyms if n[0]])
    else: subsdict={}
    return sln.subs(subsdict)
newdsolve.csre=re.compile(r'C(\d+)')

Solution

  • I don't see such an option in dsolve either, but you can use subs to replace the symbols C1,C2 with your own. The function sub_const replaces them with, e.g., k_1, k_2, and so on, until the expression stops changing. Another character can be used in place of k: it's a parameter.

    from sympy import *
    
    def sub_const(expr, char):
        i = 1
        while True:
            old = symbols('C{}'.format(i))
            new = symbols(char + '_{}'.format(i))
            new_expr = expr.subs(old, new)
            if new_expr == expr:
                return expr
            expr = new_expr
            i += 1
    
    x, f = symbols('x, f')
    print(sub_const(dsolve(Derivative(f(x), x, x) + 9*f(x), f(x)), 'k'))
    

    Output:

    Eq(f(x), k_1*sin(3*x) + k_2*cos(3*x))