The Kitchin Research Group

from sympy import solve, symbols, pprint

a,b,c,x = symbols('a,b,c,x')

f = a*x**2 + b*x + c

solution = solve(f, x)
print solution
pprint(solution)

>>> >>> >>> >>> >>> >>> [(-b + (-4*a*c + b**2)**(1/2))/(2*a), -(b + (-4*a*c + b**2)**(1/2))/(2*a)]
_____________   /       _____________\ 
        /           2    |      /           2 | 
 -b + \/  -4*a*c + b    -\b + \/  -4*a*c + b  / 
[---------------------, -----------------------]
          2*a                     2*a

The solution you should recognize in the form of \(\frac{b \pm \sqrt{b^2 - 4 a c}}{2 a}\) although python does not print it this nicely!

you might find this helpful!

from sympy import diff

print diff(f, x)
print diff(f, x, 2)

print diff(f, a)

>>> 2*a*x + b
2*a
>>> x**2

from sympy import integrate

print integrate(f, x)          # indefinite integral
print integrate(f, (x, 0, 1))  # definite integral from x=0..1

>>> a*x**3/3 + b*x**2/2 + c*x
a/3 + b/2 + c

from sympy import Function, Symbol, dsolve
f = Function('f')
x = Symbol('x')
fprime = f(x).diff(x) - f(x) # f' = f(x)

y = dsolve(fprime, f(x))

print y
print y.subs(x,4)
print [y.subs(x, X) for X in [0, 0.5, 1]] # multiple values

>>> >>> >>> >>> >>> >>> f(x) == exp(C1 + x)
f(4) == exp(C1 + 4)
[f(0) == exp(C1), f(0.5) == exp(C1 + 0.5), f(1) == exp(C1 + 1)]

It is not clear you can solve the initial value problem to get C1.

The symbolic math in sympy is pretty good. It is not up to the capability of Maple or Mathematica, (but neither is Matlab) but it continues to be developed, and could be helpful in some situations.