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2015-12-17 19:19:28 +0200 commented answer Logarithms and desolve

Thanks! It even worked in Sage Cell Server.

2015-12-17 09:36:44 +0200 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 +0200 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 +0200 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 +0200 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 +0200 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 +0200 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.)

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2015-12-07 19:10:44 +0200 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?