(************** Content-type: application/mathematica **************
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Notebook[{

Cell[CellGroupData[{
Cell[TextData[{
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " Tutorial"
}], "Title"],

Cell[CellGroupData[{

Cell["Algebraic Manipulations", "Section"],

Cell[BoxData[
    \(\(Expression1 = 
        EnzS[t]\ \((1 + \(k1b + k2f\)\/\(\(k1f\)\(\ \)\))\) \[Equal] 
          S[t];\)\)], "Input"],

Cell[TextData[{
  "In ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ", every object (from equations, to numerical solutions and graphics) is \
represented as an expression. All expressions have a 'Head', that tells you \
what type of expression the object is. In ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ", one can find out information of commands by typing ",
  StyleBox["?Command",
    FontWeight->"Bold"],
  ". So, to find out what head does, type and evaluate:"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(?Head\)\)], "Input"],

Cell[BoxData[
    RowBox[{"\<\"Head[expr] gives the head of expr.\"\>", " ", 
      ButtonBox[
        StyleBox["More\[Ellipsis]",
          "SR"],
        ButtonData:>"Head",
        Active->True,
        ButtonStyle->"RefGuideLink"]}]], "Print",
  CellTags->"Info3384087771-5332558"]
}, Open  ]],

Cell[TextData[{
  "Now find out what head ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  " has."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Head[Expression1]\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    \(Equal\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "In ",
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  ", functions can be written in the ",
  StyleBox["Prefix",
    FontWeight->"Bold"],
  " form as well as the ",
  StyleBox["Postfix",
    FontWeight->"Bold"],
  " form. As an example for showing the two (equivalent) ways to input \
expressions, evaluate and compare the following:"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(\(Head[Expression1]\)\(\[IndentingNewLine]\)
    \)\), "\[IndentingNewLine]", 
    \(Expression1 // Head\)}], "Input"],

Cell[BoxData[
    \(Equal\)], "Output"],

Cell[BoxData[
    \(Equal\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "What does it mean that objects are expressions? To look at how ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  " is represented, use the command ",
  StyleBox["FullForm",
    FontWeight->"Bold"],
  ":"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Expression1 // FullForm\)], "Input"],

Cell[BoxData[
    TagBox[
      StyleBox[\(Equal[
          Times[Plus[1, Times[Power[k1f, \(-1\)], Plus[k1b, k2f]]], EnzS[t]], 
          S[t]]\),
        ShowSpecialCharacters->False,
        ShowStringCharacters->True,
        NumberMarks->True],
      FullForm]], "Output"]
}, Open  ]],

Cell[TextData[{
  "As another (visual) representation of the same expression, use ",
  StyleBox["TreeForm",
    FontWeight->"Bold"]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Expression1 // TreeForm\)], "Input"],

Cell[BoxData[
    InterpretationBox[
      RowBox[{"\<\"Equal\"\>", "[", 
        RowBox[{
          InterpretationBox[GridBox[{
                {"\<\"|\"\>"},
                {
                  RowBox[{"\<\"Times\"\>", "[", 
                    RowBox[{GridBox[{
                          {"\<\"|\"\>"},
                          {
                            RowBox[{"\<\"Plus\"\>", "[", 
                              RowBox[{"1", ",", GridBox[{
                                    {"\<\"|\"\>"},
                                    {
                                      RowBox[{"\<\"Times\"\>", "[", 
                                        RowBox[{GridBox[{
                                        {"\<\"|\"\>"},
                                        {\("Power"[k1f, \(-1\)]\)}
                                        },
                                        GridBaseline->{Baseline, {1, 1}},
                                        ColumnAlignments->{Left}], ",", 
                                        GridBox[{
                                        {"\<\"|\"\>"},
                                        {\("Plus"[k1b, k2f]\)}
                                        },
                                        GridBaseline->{Baseline, {1, 1}},
                                        ColumnAlignments->{Left}]}], "]"}]}
                                    },
                                  GridBaseline->{Baseline, {1, 1}},
                                  ColumnAlignments->{Left}]}], "]"}]}
                          },
                        GridBaseline->{Baseline, {1, 1}},
                        ColumnAlignments->{Left}], ",", GridBox[{
                          {"\<\"|\"\>"},
                          {\("EnzS"[t]\)}
                          },
                        GridBaseline->{Baseline, {1, 1}},
                        ColumnAlignments->{Left}]}], "]"}]}
                },
              GridBaseline->{Baseline, {1, 1}},
              ColumnAlignments->{Left}],
            ColumnForm[ {"|", 
              "Times"[ 
                ColumnForm[ {"|", 
                  "Plus"[ 1, 
                    ColumnForm[ {"|", 
                      "Times"[ 
                        ColumnForm[ {"|", 
                          "Power"[ k1f, -1]}], 
                        ColumnForm[ {"|", 
                          "Plus"[ k1b, k2f]}]]}]]}], 
                ColumnForm[ {"|", 
                  "EnzS"[ t]}]]}],
            Editable->False], ",", 
          InterpretationBox[GridBox[{
                {"\<\"|\"\>"},
                {\("S"[t]\)}
                },
              GridBaseline->{Baseline, {1, 1}},
              ColumnAlignments->{Left}],
            ColumnForm[ {"|", 
              "S"[ t]}],
            Editable->False]}], "]"}],
      TreeForm[ 
        Equal[ 
          Times[ 
            Plus[ 1, 
              Times[ 
                Power[ k1f, -1], 
                Plus[ k1b, k2f]]], 
            EnzS[ t]], 
          S[ t]]],
      Editable->False]], "Output"]
}, Open  ]],

Cell[TextData[{
  "How do we extract elements of expressions? Use commands ",
  StyleBox["Last",
    FontWeight->"Bold"],
  " and ",
  StyleBox["First",
    FontWeight->"Bold"],
  ". Find out what they do below, and find out what are the ",
  StyleBox["Last",
    FontWeight->"Bold"],
  " and ",
  StyleBox["Head",
    FontWeight->"Bold"],
  " of ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  "."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[{
    \(Expression1 // First\[IndentingNewLine]\), "\[IndentingNewLine]", 
    \(Expression1 // Last\)}], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    \(\((1 + \(k1b + k2f\)\/k1f)\)\ EnzS[t]\)], "Output"],

Cell[BoxData[
    \(S[t]\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "The command ",
  StyleBox["Together",
    FontWeight->"Bold"],
  " is useful in cancelling factors in a common denominator. Find out what it \
does, and factor the left-hand-side of ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  ", using combination of ",
  StyleBox["First",
    FontWeight->"Bold"],
  " and ",
  StyleBox["Together",
    FontWeight->"Bold"],
  ". "
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(?Together\)\[IndentingNewLine]\), "\[IndentingNewLine]", 
    \(Together[Expression1 // First]\)}], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    RowBox[{"\<\"Together[expr] puts terms in a sum over a common \
denominator, and cancels factors in the result.\"\>", " ", 
      ButtonBox[
        StyleBox["More\[Ellipsis]",
          "SR"],
        ButtonData:>"Together",
        Active->True,
        ButtonStyle->"RefGuideLink"]}]], "Print",
  CellTags->"Info3384087850-7840154"],

Cell[BoxData[
    \(\(\((k1b + k1f + k2f)\)\ EnzS[t]\)\/k1f\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "For comparison, we see that applying ",
  StyleBox["Together",
    FontWeight->"Bold"],
  " on ",
  StyleBox["Expression 1",
    FontWeight->"Bold"],
  " fails to get the desired result."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Expression1 // Together\)], "Input"],

Cell[BoxData[
    \(\((1 + \(k1b + k2f\)\/k1f)\)\ EnzS[t] \[Equal] S[t]\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "The problem arises from the fact that the term ",
  Cell[BoxData[
      \(\((1 + \(k1b + k2f\)\/k1f)\)\)]],
  "is a subexpression, i.e., not at the first level of ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  ". To obtain the desired result, we should map ",
  StyleBox["Together",
    FontWeight->"Bold"],
  " to all subexpressions. Find out what ",
  StyleBox["MapAll",
    FontWeight->"Bold"],
  " does, and apply it to ",
  StyleBox["Expression1",
    FontWeight->"Bold"],
  ". Complete the following:"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(?MapAll\)\)], "Input"],

Cell[BoxData[
    RowBox[{"\<\"MapAll[f, expr] or f //@ expr applies f to every \
subexpression in expr.\"\>", " ", 
      ButtonBox[
        StyleBox["More\[Ellipsis]",
          "SR"],
        ButtonData:>"MapAll",
        Active->True,
        ButtonStyle->"RefGuideLink"]}]], "Print",
  CellTags->"Info3384087862-1699713"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(\(Together\)\(//@\)\(\ \ \)\(Expression1\)\(\ \ \ \)\)\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    \(\(\((k1b + k1f + k2f)\)\ EnzS[t]\)\/k1f \[Equal] S[t]\)], "Output"]
}, Open  ]],

Cell["\<\
The command Thread is useful in writing down a system of equations, \
equating a list of left-hand-side to a list of right-hand-side. Simply \
equating the lists do not work:\
\>", "Text"],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(LHS = {a, b, c};\)\), "\[IndentingNewLine]", 
    \(\(\(RHS = {1, 2, 3};\)\(\[IndentingNewLine]\)
    \)\), "\[IndentingNewLine]", 
    \(LHS \[Equal] RHS\)}], "Input"],

Cell[BoxData[
    \({a, b, c} \[Equal] {1, 2, 3}\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "To equate the individual terms, one needs to apply ",
  StyleBox["Thread",
    FontWeight->"Bold"],
  "; see what happens:"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Thread[{a, b, c} == {1, 2, 3}]\)], "Input"],

Cell[BoxData[
    \({a \[Equal] 1, b \[Equal] 2, c \[Equal] 3}\)], "Output"]
}, Open  ]],

Cell[CellGroupData[{

Cell["Perturbation Analysis", "Subsection"],

Cell["\<\
Suppose one has algebraic equation obtained from perturbative \
analysis, e.g.,\
\>", "Text"],

Cell[BoxData[
    \(\(PertEqn = \((La\  + \ Lb\ \[Epsilon])\)*\((Lc + Ld\ \[Epsilon])\) + 
            Lf\ \[Epsilon]\^2 \[Equal] \ 
          Ra\  + \ \((Rb + Ra)\)*\[Epsilon]\  + \ 
            Rc\ \[Epsilon]\^2;\)\)], "Input"],

Cell["\<\
One would like to equate equal powers of left and right-hand sides.\
\
\>", "Text"],

Cell["Firstly, obtain the LHS of PertEqn below:", "Text"],

Cell[CellGroupData[{

Cell[BoxData[{
    \(PertEqnLHS\  = \ \ \ \ PertEqn // 
        First\[IndentingNewLine]\), "\[IndentingNewLine]", 
    \(PertEqnRHS\  = \ \ \ \ PertEqn // Last\)}], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    \(Lf\ \[Epsilon]\^2 + \((La + Lb\ \[Epsilon])\)\ \((Lc + 
            Ld\ \[Epsilon])\)\)], "Output"],

Cell[BoxData[
    \(Ra + \((Ra + Rb)\)\ \[Epsilon] + Rc\ \[Epsilon]\^2\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "The terms of each order in ",
  "\[Epsilon] ",
  "can be obtained using the command ",
  StyleBox["CoefficientList",
    FontWeight->"Bold"]
}], "Text"],

Cell[BoxData[
    \(\(?CoefficientList\)\)], "Input"],

Cell[TextData[{
  "Using ",
  StyleBox["CoefficientList",
    FontWeight->"Bold"],
  ", obtain a list of equations matching LHS and RHS, for each order in ",
  StyleBox["\[Epsilon]",
    FontWeight->"Bold"],
  ". Hint: use ",
  StyleBox["Thread",
    FontWeight->"Bold"],
  " to equate each term in the LHS and RHS lists. Complete the following:"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[{
    \(LHSCoefficientList = 
      CoefficientList[\ PertEqnLHS, \ \[Epsilon]]\), "\[IndentingNewLine]", 
    \(RHSCoefficientList = 
      CoefficientList[\ 
        PertEqnRHS, \ \[Epsilon]]\ \[IndentingNewLine]\), \
"\[IndentingNewLine]", 
    \(OrderEqn\  = \ \ Thread[\ \ LHSCoefficientList\  \[Equal] \ \ \
RHSCoefficientList\ ]\)}], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    \({La\ Lc, Lb\ Lc + La\ Ld, Lb\ Ld + Lf}\)], "Output"],

Cell[BoxData[
    \({Ra, Ra + Rb, Rc}\)], "Output"],

Cell[BoxData[
    \({La\ Lc \[Equal] Ra, Lb\ Lc + La\ Ld \[Equal] Ra + Rb, 
      Lb\ Ld + Lf \[Equal] Rc}\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "\nNow, let's extract the RHS unknowns, based on the following hint using \
the command ",
  StyleBox["Cases",
    FontWeight->"Bold"]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(?Cases\)\)], "Input"],

Cell[BoxData[
    RowBox[{"\<\"Cases[{e1, e2, ... }, pattern] gives a list of the ei that \
match the pattern. Cases[{e1, ... }, pattern -> rhs] gives a list of the \
values of rhs corresponding to the ei that match the pattern. Cases[expr, \
pattern, levspec] gives a list of all parts of expr on levels specified by \
levspec which match the pattern. Cases[expr, pattern -> rhs, levspec] gives \
the values of rhs which match the pattern. Cases[expr, pattern, levspec, n] \
gives the first n parts in expr which match the pattern.\"\>", " ", 
      ButtonBox[
        StyleBox["More\[Ellipsis]",
          "SR"],
        ButtonData:>"Cases",
        Active->True,
        ButtonStyle->"RefGuideLink"]}]], "Print",
  CellTags->"Info3384088001-7593916"]
}, Open  ]],

Cell[TextData[{
  "As an example on using ",
  StyleBox["Cases",
    FontWeight->"Bold"],
  " to extract all symbols from a expression, evaluate below:"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(Cases[{a + b, \ c*d/f + f, \ 2.0*5 + g}, \ _Symbol, \ 
      Infinity]\)], "Input"],

Cell[BoxData[
    \({a, b, c, d, f, f, g}\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "Try a call to Cases, apply it to ",
  StyleBox["RHSCoefficientList",
    FontWeight->"Bold"]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(\(RHS\[UnderBracket]Unknowns\)\(=\)\(\ \ \ \)\(Cases[
        RHSCoefficientList\ , \ _Symbol, \ Infinity]\)\(\ \)\)\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    \({Ra, Ra, Rb, Rc}\)], "Output"]
}, Open  ]],

Cell["Remove redundant elements, by using the command Union:", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(?Union\)\)], "Input"],

Cell[BoxData[
    RowBox[{"\<\"Union[list1, list2, ... ] gives a sorted list of all the \
distinct elements that appear in any of the listi. Union[list] gives a sorted \
version of a list, in which all duplicated elements have been dropped.\"\>", 
      " ", 
      ButtonBox[
        StyleBox["More\[Ellipsis]",
          "SR"],
        ButtonData:>"Union",
        Active->True,
        ButtonStyle->"RefGuideLink"]}]], "Print",
  CellTags->"Info3384088126-3864133"]
}, Open  ]],

Cell[TextData[{
  "Apply ",
  StyleBox["Union ",
    FontWeight->"Bold"],
  "to ",
  StyleBox["RHS\[UnderBracket]Unknowns",
    FontWeight->"Bold"],
  StyleBox[", and over-write the result:",
    FontVariations->{"CompatibilityType"->0}]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(RHS\[UnderBracket]Unknowns = \ \ \(\(RHS\[UnderBracket]Unknowns\)\(//\)\
\(Union\)\(\ \ \ \)\)\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    \({Ra, Rb, Rc}\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "Let's now solve ",
  StyleBox["OrderEqn",
    FontWeight->"Bold"],
  ", with respect to ",
  StyleBox["RHS\[UnderBracket]Unknowns",
    FontWeight->"Bold"],
  ", using the command ",
  StyleBox["Solve",
    FontWeight->"Bold"],
  ":"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(?Solve\)\)], "Input"],

Cell[BoxData[
    RowBox[{"\<\"Solve[eqns, vars] attempts to solve an equation or set of \
equations for the variables vars. Solve[eqns, vars, elims] attempts to solve \
the equations for vars, eliminating the variables elims.\"\>", " ", 
      ButtonBox[
        StyleBox["More\[Ellipsis]",
          "SR"],
        ButtonData:>"Solve",
        Active->True,
        ButtonStyle->"RefGuideLink"]}]], "Print",
  CellTags->"Info3384088145-4635598"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[{
    \(\(RHS\[UnderBracket]Soln = \ 
        Solve[\ OrderEqn, \ \ RHS\[UnderBracket]Unknowns\ ];\)\), "\
\[IndentingNewLine]", 
    \(\(\(%\)\(//\)\(ColumnForm\)\(\ \ \ \ \ \ \ \)\)\)}], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    InterpretationBox[GridBox[{
          {\({Ra \[Rule] La\ Lc, Rb \[Rule] \(-La\)\ Lc + Lb\ Lc + La\ Ld, 
              Rc \[Rule] Lb\ Ld + Lf}\)}
          },
        GridBaseline->{Baseline, {1, 1}},
        ColumnAlignments->{Left}],
      ColumnForm[ {{Ra -> Times[ La, Lc], Rb -> Plus[ 
          Times[ -1, La, Lc], 
          Times[ Lb, Lc], 
          Times[ La, Ld]], Rc -> Plus[ 
          Times[ Lb, Ld], Lf]}}],
      Editable->False]], "Output"]
}, Open  ]]
}, Open  ]]
}, Open  ]],

Cell[CellGroupData[{

Cell["Numerical Integration", "Section"],

Cell[TextData[{
  "To numerically integrate ordinary differential equations, the function ",
  StyleBox["NDSolve",
    FontWeight->"Bold"],
  " is used."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(?NDSolve\)\)], "Input"],

Cell[BoxData[
    RowBox[{"\<\"NDSolve[eqns, y, {x, xmin, xmax}] finds a numerical solution \
to the ordinary differential equations eqns for the function y with the \
independent variable x in the range xmin to xmax. NDSolve[eqns, y, {x, xmin, \
xmax}, {t, tmin, tmax}] finds a numerical solution to the partial \
differential equations eqns. NDSolve[eqns, {y1, y2, ... }, {x, xmin, xmax}] \
finds numerical solutions for the functions yi.\"\>", " ", 
      ButtonBox[
        StyleBox["More\[Ellipsis]",
          "SR"],
        ButtonData:>"NDSolve",
        Active->True,
        ButtonStyle->"RefGuideLink"]}]], "Print",
  CellTags->"Info3384088179-1846413"]
}, Open  ]],

Cell[TextData[{
  "Lets now solve the ODE system\n\n",
  Cell[BoxData[
      FormBox[
        StyleBox[\(dy\/dt = \(-y\) + 1\),
          FontColor->RGBColor[0, 0, 1]], TraditionalForm]]],
  "\n\nsubject to the initial condition ",
  StyleBox["y[0]=2",
    FontColor->RGBColor[0, 0, 1]],
  ". \n\nComplete the following"
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(?D\)\)], "Input"],

Cell[BoxData[
    RowBox[{"\<\"D[f, x] gives the partial derivative of f with respect to x. \
D[f, {x, n}] gives the nth partial derivative of f with respect to x. D[f, \
x1, x2, ... ] gives a mixed derivative.\"\>", " ", 
      ButtonBox[
        StyleBox["More\[Ellipsis]",
          "SR"],
        ButtonData:>"D",
        Active->True,
        ButtonStyle->"RefGuideLink"]}]], "Print",
  CellTags->"Info3384088181-7830048"]
}, Open  ]],

Cell[CellGroupData[{

Cell[BoxData[
    \(ODE = 
      D[\ \ y[t], \ 
          t]\  \[Equal] \ \(\(-\ 
            y[t]\)\(\ \)\(+\)\(\ \)\(1\)\(\ \ \ \)\)\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    RowBox[{
      RowBox[{
        SuperscriptBox["y", "\[Prime]",
          MultilineFunction->None], "[", "t", "]"}], 
      "\[Equal]", \(1 - y[t]\)}]], "Output"]
}, Open  ]],

Cell["\<\

Also, we need the initial condition. Complete below:\
\>", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(IC\  = \ \ \(\(y[0]\)\(\[Equal]\)\(2\)\(\ \)\)\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    \(y[0] \[Equal] 2\)], "Output"]
}, Open  ]],

Cell[TextData[{
  "\nNow, combine ",
  StyleBox["ODE",
    FontWeight->"Bold"],
  " and ",
  StyleBox["IC",
    FontWeight->"Bold"],
  " into a single List, and call it ",
  StyleBox["ODE\[UnderBracket]IC",
    FontWeight->"Bold"]
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(ODE\[UnderBracket]IC = \ \ {ODE, \ IC}\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    RowBox[{"{", 
      RowBox[{
        RowBox[{
          RowBox[{
            SuperscriptBox["y", "\[Prime]",
              MultilineFunction->None], "[", "t", "]"}], 
          "\[Equal]", \(1 - y[t]\)}], ",", \(y[0] \[Equal] 2\)}], 
      "}"}]], "Output"]
}, Open  ]],

Cell[TextData[{
  "Now, using ",
  StyleBox["NDSolve",
    FontWeight->"Bold"],
  ", compute the solution from t=0 to t=5."
}], "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(NumericSoln = \ \ NDSolve[ODE\[UnderBracket]IC, \ 
        y, \ {t, 0, 5}\ ]\)], "Input",
  Background->RGBColor[0, 1, 1]],

Cell[BoxData[
    RowBox[{"{", 
      RowBox[{"{", 
        RowBox[{"y", "\[Rule]", 
          TagBox[\(InterpolatingFunction[{{0.`, 5.`}}, "<>"]\),
            False,
            Editable->False]}], "}"}], "}"}]], "Output"]
}, Open  ]],

Cell["Lets now plot the solution.", "Text"],

Cell[CellGroupData[{

Cell[BoxData[
    \(\(?Plot\)\)], "Input"],

Cell[BoxData[
    RowBox[{"\<\"Plot[f, {x, xmin, xmax}] generates a plot of f as a function \
of x from xmin to xmax. Plot[{f1, f2, ... }, {x, xmin, xmax}] plots several \
functions fi.\"\>", " ", 
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