3
$\begingroup$

I give Mathematica 10 the following command:

DSolve[I D[P[x], {x, 2}] f[x] - (P'[x])^2 f[x] + 2 I P'[x] f'[x] == 0, P, x]

and I have the following output:

{{P -> Function[{x}, C[2] - I Log[C[1] - Integrate[-(I/f[K[1]]^2), {K[1], 1, x},
 Assumptions -> True]]]}}

What does Assumptions -> True mean? Which assumptions? I didn't make any assumption.

$\endgroup$
1
  • $\begingroup$ It's probably artifact due to the lack of information about the integrability of 1/f^2 $\endgroup$ Commented Oct 16, 2015 at 15:45

1 Answer 1

3
$\begingroup$

Assumptions -> True is the default for no assumptions. Setting this may seem unnecessary, but it overrides $Assumptions. One can see this in the following:

Integrate[Sin[k x], {x, 0, 2 Pi}]

Block[{$Assumptions = k ∈ Integers},
 Integrate[Sin[k x], {x, 0, 2 Pi}]
 ]

Block[{$Assumptions = k ∈ Integers},
 Integrate[Sin[k x], {x, 0, 2 Pi}, Assumptions -> True]
 ]
(*
  (2 Sin[k π]^2)/k
  0
  (2 Sin[k π]^2)/k
*)

So DSolve is protecting the solution it returned by using the option.

Ref: Usage of Assuming for Integration

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.