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Notebook[{

Cell[CellGroupData[{
Cell["ODE/First Order", "Title",
  Evaluatable->False,
  TextAlignment->Center,
  AspectRatioFixed->True],

Cell["Exact Equations", "Subtitle"],

Cell[TextData[{
  "Sean Mauch\nsean@caltech.edu\n",
  ButtonBox["http://www.its.caltech.edu/~sean",
    ButtonData:>{
      URL[ "http://www.its.caltech.edu/~sean"], None},
    ButtonStyle->"Hyperlink"],
  "\n01/09/2001\nAnti-Copyright @ 2000 by Mauch Publishing Company, \
un-Incorporated"
}], "Text",
  Evaluatable->False,
  AspectRatioFixed->True],

Cell[CellGroupData[{

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`\[DifferentialD]\ y\/\[DifferentialD]\ x = \(x\^2 + x\ \
y + y\^2\)\/x\^2\)]]], "Section"],

Cell["Find the general solution of the equation", "Text"],

Cell[BoxData[
    \(TraditionalForm\`\[DifferentialD]y\/\[DifferentialD]x = \(\(\(x\^2 + 
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Cell[CellGroupData[{

Cell[BoxData[
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Cell[BoxData[
    \({{y[x] \[Rule] x\ Tan[C[1] + Log[x]]}}\)], "Output"]
}, Open  ]],

Cell["The solution of the differential equation is", "Text"],

Cell[BoxData[
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}, Closed]],

Cell[CellGroupData[{

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`\((4  y - 3  x)\) \[DifferentialD]x + \((y - 
              2  x)\) \[DifferentialD]y = 0\)]]], "Section"],

Cell["Find the general solution of the equation", "Text"],

Cell[BoxData[
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Cell[BoxData[
    \(DSolve[\((4  y[x] - 3  x)\) + \((y[x] - 2  x)\) \(y'\)[x] \[Equal] 0, 
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Cell[BoxData[
    \(x + 243\ x\^5\ C[1] - #1 + 405\ x\^4\ C[1]\ #1 + 
        270\ x\^3\ C[1]\ #1\^2 + 90\ x\^2\ C[1]\ #1\^3 + 15\ x\ C[1]\ #1\^4 + 
        C[1]\ #1\^5 /. {#1 \[Rule] y, C[1] \[Rule] c}\)], "Input"],

Cell["The solution of the differential equation is given by", "Text"],

Cell[BoxData[
    \(TraditionalForm\`x + 243\ c\ x\^5 - y + 405\ c\ x\^4\ y + 
        270\ c\ x\^3\ y\^2 + 90\ c\ x\^2\ y\^3 + 15\ c\ x\ y\^4 + c\ y\^5 = 
      0. \)], "DisplayFormula"],

Cell["We simplify this equation.", "Text"],

Cell[BoxData[
    \(TraditionalForm\`243\ x\^5 + 405\ x\^4\ y + 270\ x\^3\ y\^2 + 
        90\ x\^2\ y\^3 + 15\ x\ y\^4 + y\^5 = 
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Cell[BoxData[
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        15\ x\ y\^4 + y\^5]\)], "Input"],

Cell["Now we have a nice implicit form for the solution.", "Text"],

Cell[BoxData[
    \(TraditionalForm\`\((y + 3  x)\)\^5\/\(y - x\) = c\)], "DisplayFormula"]
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Cell[CellGroupData[{

Cell[TextData[Cell[BoxData[
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              2)\) \[DifferentialD]x + \((6  y\^2 - x\^2 + 
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Cell["Find the general solution of the equation", "Text"],

Cell[BoxData[
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              2)\) \[DifferentialD]x + \((6  y\^2 - x\^2 + 
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Cell["First we try using DSolve[].", "Text"],

Cell[BoxData[
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  "The answer is a little messy because ",
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  " gives an explicit equation for ",
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  ".  Since this is an exact equation, we try integrating the equation \
instead."
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Cell[BoxData[
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Cell[TextData[Cell[BoxData[
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Cell[BoxData[
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Cell["First we try using DSolve[].", "Text"],

Cell[BoxData[
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Cell[TextData[{
  "We see that ",
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  " satisfies a quadratic equation.  We write the equation in exact form and \
try integrating the equation."
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Find the solution of the following differential equation which \
satisfies the given initial condition.  Where is the solution defined?\
\>", \
"Text"],

Cell[BoxData[
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Cell["The solution of the differential equation is", "Text"],

Cell[BoxData[
    \(TraditionalForm\`y = \(\(1\/\(\((x + 2)\) \((x - 
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Cell[TextData[{
  "The solution is defined on the interval ",
  Cell[BoxData[
      \(TraditionalForm\`\((\(\(-2\)\  ... \)\ 3)\)\)]],
  ".  Below is a plot of the solution on this interval."
}], "Text"],

Cell[BoxData[
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Find the solution of the following differential equation which \
satisfies the given initial condition.  Where is the solution defined?\
\>", \
"Text"],

Cell[BoxData[
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Cell["The solution of the differential equation is", "Text"],

Cell[BoxData[
    \(TraditionalForm\`y = \(\(\@\(2 \((1 - x)\) \[ExponentialE]\^x - 
              1\)\)\(.\)\)\)], "DisplayFormula"],

Cell["\<\
We try to determine where the argument of the square root is \
non-negative.\
\>", "Text"],

Cell[BoxData[
    \(Solve[2 \((1 - x)\) Exp[x] - 1 \[Equal] 0, x]\)], "Input"],

Cell[BoxData[
    \(% // N\)], "Input"],

Cell[TextData[{
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  " gives us a solution in terms of the roots of an algebraic/transcendental \
equation."
}], "Text"],

Cell[BoxData[
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Cell[TextData[{
  "The argument of the square root in the solution is non-negative on the \
interval ",
  Cell[BoxData[
      \(TraditionalForm\`\((\(\(-1.67835\)\  ... \)\ 0.768039)\)\)]],
  "."
}], "Text"],

Cell[BoxData[
    \(\(Plot[2 \((1 - x)\) Exp[x] - 1, {x, \(-5\), 1.5}];\)\)], "Input"],

Cell["Below is a plot of the solution.", "Text"],

Cell[BoxData[
    \(\(Plot[\@\(2 \((1 - x)\) Exp[x] - 1\), {x, \(-1.678346990016661`\), 
          0.7680390470134655`}];\)\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`\[DifferentialD]\ y\/\[DifferentialD]\ t = 
      f(y\/t)\)]]], "Section"],

Cell[CellGroupData[{

Cell["Part a", "Subsection"],

Cell[BoxData[
    \(y[t_] := t\ v[t]\)], "Input"],

Cell[BoxData[
    \(\(y'\)[t] \[Equal] f[y[t]/t]\)], "Input"]
}, Open  ]],

Cell[CellGroupData[{

Cell["Part b", "Subsection"],

Cell[BoxData[
    \(\(y'\)[t] \[Equal] 2\ y[t]\/t + \((y[t]\/t)\)\^2\)], "Input"],

Cell[BoxData[
    \(DSolve[%, v[t], t]\)], "Input"],

Cell[BoxData[
    \(Clear[y]\)], "Input"],

Cell[BoxData[
    \(DSolve[\(y'\)[t] \[Equal] 2\ y[t]\/t + \((y[t]\/t)\)\^2, y[t], 
      t]\)], "Input"]
}, Open  ]]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`y\^2\ sin\ t + 
        y\ \(f(t)\)\ \[DifferentialD]y\/\[DifferentialD]t = 0\)]]], "Section"],

Cell[BoxData[
    \(D[\(y\^2\) Sin[t], y] \[Equal] D[y\ f[t], t]\)], "Input"],

Cell[BoxData[
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Cell[BoxData[
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Cell[BoxData[
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Cell[BoxData[
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Cell[BoxData[
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Cell[CellGroupData[{

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`\[DifferentialD]y\/\[DifferentialD]x + x\ y = 
        x\^\(2  n + 1\)\)]],
  ",    ",
  Cell[BoxData[
      \(TraditionalForm\`y(1) = 1\)]],
  ",    ",
  Cell[BoxData[
      \(TraditionalForm\`n \[Element] Z\)]]
}], "Section"],

Cell[BoxData[
    \(DSolve[{\(y'\)[x] + x\ y[x] \[Equal] x\^\(2  n + 1\), y[1] \[Equal] 1}, 
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Cell[BoxData[
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Cell[TextData[{
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " gives the answer in terms of the Exponential Integral function."
}], "Text"],

Cell[BoxData[
    \(\(?ExpIntegralE\)\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[{
  Cell[BoxData[
      \(TraditionalForm\`\[DifferentialD]y\/\[DifferentialD]x - 2  x\ y = 
        1\)]],
  ",    ",
  Cell[BoxData[
      \(TraditionalForm\`y(0) = 1\)]]
}], "Section"],

Cell[BoxData[
    \(DSolve[{\(y'\)[x] - 2  x\ y[x] \[Equal] 1, y[0] \[Equal] 1}, y[x], 
      x]\)], "Input"],

Cell[TextData[{
  StyleBox["Mathematica",
    FontSlant->"Italic"],
  " gives the answer in terms of the error function."
}], "Text"],

Cell[BoxData[
    \(\(?Erf\)\)], "Input"]
}, Closed]],

Cell[CellGroupData[{

Cell[TextData[Cell[BoxData[
    \(TraditionalForm\`\[DifferentialD]y\/\[DifferentialD]x + \[Alpha]\ y = \
\[Beta]\ \[ExponentialE]\^\(\(-\[Lambda]\)\ x\)\)]]], "Section"],

Cell[TextData[{
  "Show that if ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] > 0\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`\[Lambda] > 0\)]],
  ", then for any real ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta]\)]],
  ", every solution of"
}], "Text"],

Cell[BoxData[
    \(TraditionalForm\`\[DifferentialD]y\/\[DifferentialD]x + \[Alpha]\ y = \
\[Beta]\ \[ExponentialE]\^\(\(-\[Lambda]\)\ x\)\)], "DisplayFormula"],

Cell[TextData[{
  "satisfies ",
  Cell[BoxData[
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  ".  (The case ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] = \[Lambda]\)]],
  " requires special treatment.)  Find the solution for ",
  Cell[BoxData[
      \(TraditionalForm\`\[Beta] = \(\[Lambda] = 1\)\)]],
  " which satisfies ",
  Cell[BoxData[
      \(TraditionalForm\`y(0) = 1\)]],
  ".  Plot this solution for ",
  Cell[BoxData[
      \(TraditionalForm\`0 \[LessEqual] x < \[Infinity]\)]],
  " for several values of ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha]\)]],
  ".  In particular, show what happens when ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] \[Rule] 0\)]],
  " and ",
  Cell[BoxData[
      \(TraditionalForm\`\[Alpha] \[Rule] \[Infinity]\)]],
  "."
}], "Text"],

Cell[BoxData[
    \(DSolve[\(y'\)[
            x] + \[Alpha]\ y[x] \[Equal] \[Beta]\ Exp[\(-\[Lambda]\)\ x], 
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Cell[BoxData[
    \(% // Simplify\)], "Input"],

Cell["The general solution is", "Text"],

Cell[BoxData[
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        c\ \(\(\[ExponentialE]\^\(\(-\[Alpha]\)\ x\)\)\(.\)\)\)], \
"DisplayFormula"],

Cell[TextData[{
  "Clearly this solution vanishes as ",
  Cell[BoxData[
      \(TraditionalForm\`x \[Rule] \[Infinity]\)]],
  "."
}], "Text"],

Cell[TextData[{
  "The solution is not valid for ",
  Cell[BoxData[
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  ".  We consider this case separately."
}], "Text"],

Cell[BoxData[
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Cell[TextData[{
  "The general solution for ",
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Cell[BoxData[
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Cell[TextData[{
  "One could use L'Hospital's rule to show that this solution vanishes as ",
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  "."
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Cell[TextData[{
  "Now we solve the problem for ",
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  " subject to the initial condition ",
  Cell[BoxData[
      \(TraditionalForm\`y(0) = 1\)]],
  "."
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Cell[BoxData[
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Cell[TextData[{
  "We plot the solution for ",
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      \(TraditionalForm\`\[Alpha] = 1\/16, 1\/8,  ... , 16\)]],
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Cell[BoxData[
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Cell[TextData[{
  "and as ",
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  "."
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Cell[BoxData[
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*)




(***********************************************************************
End of Mathematica Notebook file.
***********************************************************************)