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MathGroup Archive 2001

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Problem with InverseFunction

  • To: mathgroup at smc.vnet.net
  • Subject: [mg28199] Problem with InverseFunction
  • From: Low Choon Song <eng81288 at nus.edu.sg>
  • Date: Wed, 4 Apr 2001 04:13:22 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

Hello,
I have this problem that I have been trying to solve for weeks now. But I
still can't succeed. 
I use DSolve in Mathematica to solve a Differential Eqn and in the output by
Mathematica, it always come out with the form of InverseFunction.
How can I express the results( output ) in a more readable form?
How can I get rid of the InverseFunction?
How can I get rid of the "#1" that appear in the solution?

The problem I am solving is below and I have attached it too, if you are
willing to take a quick look.
Thanks in advance.
Low


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Cell[CellGroupData[{
Cell[BoxData[{
    \(\(Clear[Derivative, \ Rg, \ T, \ P, \ no, \ \ k, \ d, \ R, \ Ro, 
\ n, \
        C\[Infinity], \ t];\)\), "\[IndentingNewLine]",
    \(DSolve[{{\ \(R'\)[
              t]\  \[Equal] \ \(Rg\ T\)\/P\ \((\@\(2\ \((\(Rg\ T\ \
C\[Infinity]\)\/P)\)\ \ d\ k\ 1\/R[t]\  + \ k\^2\/4\)\ \  - \
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\[Infinity]\)\/\(\(\ \)\(d\)\))\)\ \((R[
                              t])\)\  + \ \(\(\(1\)\(\ 
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\(\(1\
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                        t]\ )\))\), \ R[0] \[Equal] \ 0}, \ R[t], \ t}, 
\
      InverseFunctions\  \[Rule] \ False]\)}], "Input"],

Cell[BoxData[
    \(DSolve::"argct" \(\(:\)\(\ \)\)
      "\!\(DSolve\) called with \!\(2\) arguments."\)], "Message"],

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        RowBox[{"{",
          RowBox[{
            RowBox[{"{",
              RowBox[{
                RowBox[{
                  RowBox[{
                    SuperscriptBox["R", "\[Prime]",
                      MultilineFunction->None], "[", "t", "]"}],
                  "==", \(\(Rg\ T\ \((\(-2\)\ k + \@\(k\^2\/4 + 
\(2\ C\
\[Infinity]\ d\ k\ Rg\ T\)\/\(P\ R[t]\)\))\)\ \((\(-\(\(k\ P\ R[
                                      t]\)\/\(4\ d\ Rg\ T\)\)\) + 
\@\(\(C\
\[Infinity]\ k\ P\ R[t]\)\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ 
R[t]\^2\)\/\(16\ d\
\^2\ Rg\^2\ T\^2\)\))\)\)\/P\)}], ",", \(R[0] == 0\)}], "}"}], ",", 
\(R[t]\),
            ",", "t"}], "}"}], ",", \(InverseFunctions \[Rule] 
False\)}],
      "]"}]], "Output"]
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              "==", \(\(Rg\ T\ \((\(-2\)\ k + \@\(k\^2\/4 + \(2\ 
C\[Infinity]\
\ d\ k\ Rg\ T\)\/\(P\ R[t]\)\))\)\ \((\(-\(\(k\ P\ R[
                                  t]\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\
\[Infinity]\ k\ P\ R[t]\)\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ R[t]\^2\)\/\(
16\ d\
\^2\ Rg\^2\ T\^2\)\))\)\)\/P\)}], ",", \(R[0] == 0\)}], "}"}], ",", 
\(R[t]\),
        ",", "t"}], "]"}]], "Input"],

Cell[BoxData[
    \({R[t] \[Rule] \(InverseFunction[
            C[1] - \((32\ d\ Log[
                        8\ C\[Infinity]\ d\ Rg\ T -
                          15\ k\ P\ #1]\ \@\(\(8\ C\[Infinity]\ d\ k\ 
Rg\ T + \
k\^2\ P\ #1\)\/\(P\ #1\)\)\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ #1 + 
k\^2\ \
P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\)\ \((\(-k\)\ P\ #1 +
                          d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ 
T\ #1 \
+ k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\))\)/\((225\ k\^3\ 
\((8\ C\
\[Infinity]\ d\ Rg\ T +
                          k\ P\ #1)\)\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ 
Rg\ \
T\)\)\) + \@\(\(C\[Infinity]\ k\ P\ #1\)\/\(2\ d\ Rg\ T\) + \(k\^2\ 
P\^2\ \
#1\^2\)\/\(16\ d\^2\ Rg\^2\ T\^2\)\))\))\) + \(1\/\(Rg\ T\)\) \((P\ \
\((\(-\(\(8\ Log[\(-8\)\ C\[Infinity]\ d\ Rg\ T +
                                      15\ k\ P\ #1]\ \((\(-k\)\ P\ #1 + 

                                      d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ 
d\ k\ \
P\ Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ 
T\^2\)\))\)\)\/\(225\ k\^2\
\ P\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\ 
P\ #1\
\)\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ 
T\^2\)\))\)\)\
\)\) - \(17\ Log[2\ \((4\ C\[Infinity]\ d\ Rg\ T + k\ P\ #1)\) + 2\ P\ 
#1\ \@\
\(\(8\ C\[Infinity]\ d\ k\ Rg\ T + k\^2\ P\ #1\)\/\(P\ #1\)\)]\ 
\((\(-k\)\ P\ \
#1 + d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ #1 + k\^2\ P\^2\ 
\
#1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\)\)\/\(225\ k\^2\ P\ \((\(-\(\(k\ P\ 
#1\)\/\
\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\ P\ #1\)\/\(2\ d\ Rg\ T\) + 
\
\(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ T\^2\)\))\)\) + \((8\ \
Log[\(-\(\(3375\ k\^2\ P\^2\ #1\ \@\(\(8\ C\[Infinity]\ d\ k\ Rg\ T + 
k\^2\ P\
\ #1\)\/\(P\ #1\)\)\)\/\(128\ C\[Infinity]\ d\ Rg\ T\ \((\(-8\)\ 
C\[Infinity]\
\ d\ Rg\ T +
                                        15\ k\ P\ #1)\)\)\)\) + \(3375\ 
\((8\ \
C\[Infinity]\ d\ k\^2\ P\ Rg\ T + 17\ k\^3\ P\^2\ #1)\)\)\/\(1024\ C\
\[Infinity]\ d\ Rg\ T\ \((8\ C\[Infinity]\ d\ Rg\ T - 15\ k\ P\ 
#1)\)\)]\ \((\
\(-k\)\ P\ #1 +
                                    d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ 
k\ P\ \
Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\))\)/\((225\ 
k\^2\ \
P\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\ 
P\ #1\)\
\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ 
T\^2\)\))\))\) \
- \((68\ Log[2\ \((4\ C\[Infinity]\ d\ Rg\ T + k\ P\ #1)\) +
                                    2\ d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ 
d\ k\ \
P\ Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\)]\ 
\((\(-k\)\ P\ \
#1 + d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ #1 + k\^2\ P\^2\ 
\
#1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\))\)/\((225\ k\^2\ P\ \((\(-\(\(k\ P\ 
\
#1\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\ P\ #1\)\/\(2\ d\ Rg\ 
T\) + \
\(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ T\^2\)\))\))\) + \((32\ 
Log[\(3375\ \
\((8\ C\[Infinity]\ d\ k\^2\ P\ Rg\ T + 17\ k\^3\ P\^2\ 
#1)\)\)\/\(4096\ C\
\[Infinity]\ d\ Rg\ T\ \((8\ C\[Infinity]\ d\ Rg\ T - 15\ k\ P\ #1)\)\) 
- \
\(3375\ k\^2\ P\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ #1 + k\^2\ 
P\^2\ \
#1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\)\)\/\(512\ C\[Infinity]\ \((\(-8\)\ C\
\[Infinity]\ d\ Rg\ T + 15\ k\ P\ #1)\)\)]\ \((\(-k\)\ P\ #1 +
                                    d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ 
k\ P\ \
Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\))\)/\((225\ 
k\^2\ \
P\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\ 
P\ #1\)\
\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ 
T\^2\)\))\))\) \
- \(#1\ \((\(-k\)\ P\ #1 + d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ 
Rg\ T\ \
#1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\)\)\/\(15\ 
C\[Infinity]\ d\
\ k\ Rg\ T\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ Rg\ T\)\)\) + 
\@\(\(C\[Infinity]\ k\
\ P\ #1\)\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ 
T\^2\)\
\))\)\) - \(#1\ \@\(\(8\ C\[Infinity]\ d\ k\ Rg\ T + k\^2\ P\ 
#1\)\/\(P\ #1\)\
\)\ \((\(-k\)\ P\ #1 + d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ 
#1 + \
k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\))\)\)\/\(60\ C\[Infinity]\ 
d\ \
k\^2\ Rg\ T\ \((\(-\(\(k\ P\ #1\)\/\(4\ d\ Rg\ T\)\)\) + 
\@\(\(C\[Infinity]\ \
k\ P\ #1\)\/\(2\ d\ Rg\ T\) + \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ 
\
T\^2\)\))\)\) - \(\@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ T\ #1 + k\^2\ 
P\^2\ \
#1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\)\ \((\(-k\)\ P\ #1 + d\ Rg\ T\ \@\(\(8\ 
C\
\[Infinity]\ d\ k\ P\ Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ \
T\^2\)\))\)\)\/\(15\ C\[Infinity]\ k\^2\ P\ \((\(-\(\(k\ P\ #1\)\/\(4\ 
d\ Rg\ \
T\)\)\) + \@\(\(C\[Infinity]\ k\ P\ #1\)\/\(2\ d\ Rg\ T\) + \(k\^2\ 
P\^2\ \
#1\^2\)\/\(16\ d\^2\ Rg\^2\ T\^2\)\))\)\) + \((\@\(\(8\ C\[Infinity]\ 
d\ k\ \
Rg\ T + k\^2\ P\ #1\)\/\(P\ #1\)\)\ \@\(\(8\ C\[Infinity]\ d\ k\ P\ Rg\ 
T\ #1 \
+ k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ T\^2\)\)\ \((\(-\(1\/\(60\ 
C\[Infinity]\
\ k\^3\ P\)\)\) + \(2\ d\ Rg\ T\)\/\(15\ k\^3\ P\ \((8\ C\[Infinity]\ 
d\ Rg\ \
T + k\ P\ #1)\)\))\)\ \((\(-k\)\ P\ #1 +
                                    d\ Rg\ T\ \@\(\(8\ C\[Infinity]\ d\ 
k\ P\ \
Rg\ T\ #1 + k\^2\ P\^2\ #1\^2\)\/\(d\^2\ Rg\^2\ 
T\^2\)\))\))\)/\((\(-\(\(k\ P\
\ #1\)\/\(4\ d\ Rg\ T\)\)\) + \@\(\(C\[Infinity]\ k\ P\ #1\)\/\(2\ d\ 
Rg\ T\) \
+ \(k\^2\ P\^2\ #1\^2\)\/\(16\ d\^2\ Rg\^2\ T\^2\)\))\))\))\) &]\)[
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