2019-02-19 03:00:06 +0100 received badge ● Notable Question (source) 2016-07-29 09:00:09 +0100 received badge ● Popular Question (source) 2015-12-17 19:19:49 +0100 received badge ● Supporter (source) 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... 2015-12-16 20:48:46 +0100 received badge ● Editor (source) 2015-12-15 19:10:35 +0100 received badge ● Scholar (source) 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. 2015-12-15 08:37:00 +0100 received badge ● Nice Question (source) 2015-12-14 15:41:29 +0100 asked a question Logarithms and desolve More questions about desolve results: Using  y = function('y', x) desolve(diff(y,x) == y*(100-y), y)  results in the solution: -1/100*log(y(x) - 100) + 1/100*log(y(x)) == _C + x  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: y = function('y', x) desolve(diff(y,x) == y*(100-y), y)  results in -1/100*log(y(x) - 100) + 1/100*log(y(x)) == _C + x  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 == 100e^(100C + 100x)/(e^(100C + 100*x) - 1) But I cannot get it to work in one calculation like this: y = function('y', x) de = diff(y,x) == y*(100-y) sol=desolve(de,y) print("Not so nice:") sol print("Better:") solve(sol, y, to_poly_solve=True)  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?