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2015-12-17 19:19:28 +0100 | commented answer | Logarithms and desolve Thanks! It even worked in Sage Cell Server. |

2015-12-17 09:36:44 +0100 | commented answer | Logarithms and desolve I found that it is possible to set a option variable in Maxima to get the log(abs(x)) answer when doing integrals http://maxima.sourceforge.net/docs/ma... . Can you set that variable to True in Sage. |

2015-12-17 09:26:43 +0100 | commented answer | logistic differential equation I hope it will improve! Something like substitute(old variable or function, new variable or function) would be much easier to remember and makes a lot more sense than knowing that you have to rewrite y(x) as yx and use python dictionary notation... |

2015-12-16 20:51:09 +0100 | commented question | Logarithms and desolve Testing: assume(x<0) integral(1/x,x) And the result is log(x). Hm... |

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2015-12-15 19:10:30 +0100 | commented answer | logistic differential equation Yes, Thanks again. I will use your answer. I think there should be a straight-forward way to go from a solution of a differential equation to using that solution in further calculations. But there seems not to be that way (yet?) |

2015-12-15 13:54:18 +0100 | commented answer | logistic differential equation Thanks! I have now been searching for how to rewrite without getting the "DeprecationWarning: Substitution using function-call syntax and unnamed arguments ..." warning from sol.substitute({y(x):z}) But without finding the answer. |

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2015-12-14 15:41:29 +0100 | asked a question | Logarithms and desolve More questions about desolve results: Using results in the solution: But that solution does not work for y<100. But it is correct if you assume that it should be log(abs(y(x)-100)) in the first term. Is the abs() implicit? (The log terms actually comes from the integral of 1/y and 1/(100-y) so it should be absolute values.) |

2015-12-07 22:30:45 +0100 | received badge | ● Student (source) |

2015-12-07 19:10:44 +0100 | asked a question | logistic differential equation Using: results in This is correct but not very helpful. I found out that "manually" feeding the answer above to solve() changing _C to a defined variable and the function y(x) for a variable and using the argument to_poly_solve=True gives the expected result: y == 100 But I cannot get it to work in one calculation like this: Would like to know: a) Is there a way to get desolve or similar solver to return the y= answer directly? b) If not a) how do I make the desolve -> solve calculation without manually rewriting the solution? |

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