Comment #4 on issue 2802 by [email protected]: solve error with expression with derivative
http://code.google.com/p/sympy/issues/detail?id=2802

How does this look?

sympy\solvers\solvers.py:541: UserWarning:
Note: there is an ode module that offers several functions for
handling this sort of equation, including a dsolve function that can
be used rather than solve.
  '''))
[C1]

f(x) == C1

[(2*C2 - 3*exp(2*t))*exp(-2*t)/2]

The solution could not be validated: when <[C1*exp(-2*t)]> was
substituted into the original equation, <g(t)> was obtained. If you
want the solution anyway use the flag 'check=False'.
[]

The solution could not be validated: when <[C1*exp(-2*t)]> was
substituted into the original equation, <g(t)> was obtained. If you
want the solution anyway use the flag 'check=False'.
[]

[C1*exp(-2*t)]

f(t) == C1*exp(-2*t)

[-sqrt((x + 2)**2*exp(-2*C1) + 1), sqrt((x + 2)**2*exp(-2*C1) + 1)]

-log(f(x)**2 - 1)/2 == C1 - log(x + 2)


btw, for some reason sometimes I get a different solution for the
equation you gave, e.g.:

[exp(2*C1 - 2*t)/2 - 3/2]

The solution could not be validated: when <f(t) == (C1 -
g(t)*exp(2*t)/2)*exp(-2*t)> was substituted into the original
equation, <-Derivative(g(t), t)> was obtained. If you want the
solution anyway use the flag 'check=False'.

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