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Solve differential eq of separable variables that gives implicit solution?

Anxon Pués, Retired
Posted 9 years ago

Consider the following code:

DSolveValue[{\[DifferentialD]r[\[Theta]]/\[DifferentialD]\[Theta] == (
   r[\[Theta]] \[Theta] + r[\[Theta]])/(
   r [\[Theta]] \[Theta] + \[Theta]), r[1] == E}, 
 r[\[Theta]], \[Theta]]
by hand I can arrive to 

r[th] + Log[r[th]] = th + Log[th] + E

But not using DSolver neither DSolveValue,

and finally I do not get the way to plot , neither r[th] neither the vectorial ploting of field

thanks for advice and sorry if made mistakes in use of this facility
POSTED BY: Anxon Pués
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Welcome to Wolfram Community! Please make sure you know the rules and how to format your code properly, which you can find here: https://wolfr.am/READ-1ST If you do not format code, it may become corrupted and useless to other members. Please EDIT your posts and make sure code blocks start on a new paragraph and look framed and colored like this.

int = Integrate[1/(x^3 - 1), x];
Map[Framed, int, Infinity]

enter image description here
POSTED BY: EDITORIAL BOARD
Anxon Pués
Anxon Pués, Retired
Posted 9 years ago

Yessss really it's now all clear, I must improove a lot to do work like you! Thanks and I will try to post questions in this format in further opportunities. I can not plot this but this happens everytime I arrive to implicit functions, I look for examples like ellipses or other implicits but sonething I made bad when change to other functions...
POSTED BY: Anxon Pués
Updating Name
Updating Name
Posted 9 years ago

msol = DSolve[{y'[x] == (x y[x] + y[x])/(x y[x] + x), y[1] == E}, y[x], x]

Maybe this is better? anyway it returns as wrong, and a solution that is different from the hand done solution....
POSTED BY: Updating Name
Anxon Pués
Anxon Pués, Retired
Posted 9 years ago

msol = DSolve[{y'[x] == (x y[x] + y[x])/(x y[x] + x), y[1] == E}, y[x], x]

Maybe this is better? anyway it returns as wrong, and a solution that is different from the hand done solution....
POSTED BY: Anxon Pués

Your manual solution, if carried further, yields

Solve[r[th] + Log[r[th]] == th + Log[th] + E, r[th]]
(*  {{r[th] -> ProductLog[E^(E + th) th]}}  *)

which is equivalent to the DSolve solution.

You can also verify the DSolve solution with

{y'[x] == (x y[x] + y[x])/(x y[x] + x), y[1] == E} /. msol // FullSimplify // N[#, 80] &

(For some reason Mathematica does not simplify ProductLog[E^(1 + E)], so I verified it numerically out to 80 digits -- an arbitrary choice, but hopefully convincing.)
POSTED BY: Michael Rogers
POSTED BY: Michael Rogers
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