Comment #9 on issue 3148 by [email protected]: Too many constants from
dsolve()
http://code.google.com/p/sympy/issues/detail?id=3148
It just occurred to me that if you have too many constants, you can solve
for the excess ones in terms and substitute them out. e.g. in PR 1964 this
test is failing because it gives too many constants:
sol = Eq(f(x), C1/(x**2 + 6*x + 9) - S(3)/2)
eq = df + (3 + 2*f(x))/(x + 3)
dsolve(eq, hint='linear_coefficients')
f(x) == (C1 + C2*x - 3*x**3/2 - 27*x**2/2)/(x**3 + 9*x**2 + 27*x + 27)
eq.subs(f(x),_.rhs).doit()
(C2 - 9*x**2/2 - 27*x)/(x**3 + 9*x**2 + 27*x + 27) + (-3*x**2 - 18*x -
27)*(C1 +
C2*x - 3*x**3/2 - 27*x**2/2)/(x**3 + 9*x**2 + 27*x + 27)**2 + (2*(C1 +
C2*x - 3
*x**3/2 - 27*x**2/2)/(x**3 + 9*x**2 + 27*x + 27) + 3)/(x + 3)
We can get rid of C2 since we know that the above should be zero:
solve(_,C2)
[C1/3 - 27]
eq.subs(f(x),sol.rhs).doit().subs(C2,_[0])
C1*(-2*x - 6)/(x**2 + 6*x + 9)**2 + 2*C1/((x + 3)*(x**2 + 6*x + 9))
factor(_)
0
Is that something that constantsimp should be doing? Since C2 is
multiplying x it's not a simple issue that constants didn't combine that
should have (e.g. C1 + C2/2 should combine to C1 but in this case there is
an x on C2).
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