(************** Content-type: application/mathematica ************** CreatedBy='Mathematica 5.2' Mathematica-Compatible Notebook This notebook can be used with any Mathematica-compatible application, such as Mathematica, MathReader or Publicon. The data for the notebook starts with the line containing stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). 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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 15619, 582]*) (*NotebookOutlinePosition[ 16686, 617]*) (* CellTagsIndexPosition[ 16642, 613]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Ordinary Differential Equations", "Title", Evaluatable->False, TextAlignment->Center, AspectRatioFixed->True], Cell["Simplified Version", "Subsubtitle", TextAlignment->Center], 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"], "\n", "This work is distributed under the GNU FDL. See ", ButtonBox["license.nb ", ButtonData:>{"license.nb", None}, ButtonStyle->"Hyperlink"], "for details." }], "Text", Evaluatable->False, AspectRatioFixed->True], Cell[CellGroupData[{ Cell["First Order Differential Equations", "Section"], Cell[TextData[{ "Use the ", StyleBox["DSolve[]", "Input"], " function to solve differential equations." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?DSolve\)\)], "Input"], Cell[TextData[{ "The first argument is a list of differential equations. The second \ argument is a list of dependent variables, (the functions to solve for); the \ final argument is the independent variable. Recall that in ", StyleBox["Mathematica", FontSlant->"Italic"], " equations have a ", Cell[BoxData[ \(TraditionalForm\` == \)]], " instead of the single ", Cell[BoxData[ \(TraditionalForm\` = \)]], " used for assignment. Below we obtain the general solution for a first \ order differential equation." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Remove[y]; \nDSolve[\(y'\)[x] + \(1\/x\) y[x] == x\^2, y[x], x]\)], "Input"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " uses ", Cell[BoxData[ \(TraditionalForm\`C[1], \ C[2], \ \[Ellipsis]\)]], " for the constants of integration. To impose an initial condition, add \ the constraint to the list of equations." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(DSolve[{\(y'\)[x] + \(1\/x\) y[x] == x\^2, y[1] == 1}, y[x], x]\)], "Input"], Cell["\<\ We can set a function equal to the solution to work with the \ result.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(f[x_] = y[x] /. First[%]\)], "Input"], Cell[BoxData[ \(Plot[f[x], {x, 0, 5}]; \nClear[f]; \)], "Input"], Cell[CellGroupData[{ Cell["Exercise 1", "Subsubsection"], Cell["Solve the differential equation,", "Text"], Cell[BoxData[ \(TraditionalForm\`y' + y\/\(sin\ x\) = 0. \)], "DisplayFormula", TextAlignment->Center] }, Open ]], Cell[CellGroupData[{ Cell["Exercise 2", "Subsubsection"], Cell["Solve the differential equation,", "Text"], Cell[BoxData[ \(TraditionalForm\`y' - \(1\/x\) y = \(\(x\^\[Alpha]\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Higher Order Differential Equations", "Section"], Cell["Suppose we wish to solve,", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ FormBox[ RowBox[{\(y'' - 2 y' + 2 y == 0\), ",", " ", FormBox[\(y(0) = 0\), "TraditionalForm"], ",", " ", FormBox[\(y' \((0)\) = \(\(1\)\(.\)\)\), "TraditionalForm"]}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell["\<\ We could first obtain the general solution of the differential \ equation\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Remove[y]; \)\)], "Input"], Cell[BoxData[ \(DSolve[\(\(y'\)'\)[x] - 2 \( y'\)[x] + 2 y[x] == 0, y[x], x]\)], "Input"], Cell[BoxData[ \(f[x_] = y[x] /. First[%]\)], "Input"], Cell[TextData[{ "and then solve for ", Cell[BoxData[ \(TraditionalForm\`C[1]\)]], " and ", Cell[BoxData[ \(TraditionalForm\`C[2]\)]], " by applying the initial conditions." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(Solve[{f[0] == 0, \(f'\)[0] == 1}]\)], "Input"], Cell[BoxData[ \(f[x_] = f[x] /. First[%]\)], "Input"], Cell[TextData[{ "Alternatively, we can include the initial conditions in the list of \ equations we give to ", Cell[BoxData[ FormBox[ StyleBox[\(DSolve[]\), "Input"], TraditionalForm]]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(DSolve[{\(\(y'\)'\)[x] - 2 \( y'\)[x] + 2 y[x] == 0, y[0] == 0, \(y'\)[0] == 1}, y[x], x]\)], "Input"], Cell[BoxData[ \(f[x_] = y[x] /. First[%]\)], "Input"], Cell[BoxData[ \(Plot[f[x], {x, 0, 5}, PlotRange -> All]; \nClear[f]\)], "Input"], Cell[CellGroupData[{ Cell["Exercise 3", "Subsubsection"], Cell["Find the general solution of:", "Text"], Cell[BoxData[ \(TraditionalForm\`\(\(y'\)'\)' - \(y'\)' + y' - y = 0. \)], "DisplayFormula", TextAlignment->Center] }, Open ]], Cell[CellGroupData[{ Cell["Exercise 4", "Subsubsection"], Cell["Find the fundamental set of solutions for:", "Text"], Cell[BoxData[ \(TraditionalForm\`\(y'\)' + 2 a\ y' + \(a\^2\) y = 0\)], "DisplayFormula", TextAlignment->Center], Cell[TextData[{ "at the point ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], "." }], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["Vector Field of First Order Equations", "Section"], Cell["\<\ You can visualize the solutions of a first order differential \ equation by plotting its vector field. First load the PlotField \ package.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(<< Graphics`PlotField`\)], "Input"], Cell[TextData[{ "We will use the ", StyleBox["PlotVectorField[]", FontWeight->"Bold"], " function." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?PlotVectorField\)\)], "Input"], Cell["\<\ The optional arguments (xu) and (yu) specify the spacing between \ vectors. Consider the problem\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ FormBox[ RowBox[{\(y' = 2 y - y\^2\), ",", " ", FormBox[\(y(0) = \(\(y\_0\)\(.\)\)\), "TraditionalForm"]}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "The vector ", Cell[BoxData[ \(TraditionalForm\`\((x, 2 y - y\^2)\)\)]], " is tangent to a solution passing through the point ", Cell[BoxData[ \(TraditionalForm\`\((x, y)\)\)]], ". Below we plot this vector in the domain ", Cell[BoxData[ \(TraditionalForm\`\([0, 8]\)\[Times]\([\(-4\), 4]\)\)]], ". By default, ", StyleBox["PlotVectorField", FontWeight->"Bold"], " does not draw an axes or a frame around the graph. If you want a frame, \ you need to specify that in the options. You can make the graph easier to \ read by clicking on the graph and dragging a corner to enlarge it." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(PlotVectorField[{x, 2 y - y\^2}, {x, 0, 8}, {y, \(-4\), 4}, Frame -> True];\)\)], "Input"], Cell[TextData[{ "Notice that the vectors are of different lengths. The lengths of the \ vectors is proportional to the length of ", Cell[BoxData[ \(TraditionalForm\`\((x, 2 y - y\^2)\)\)]], ". When drawing the vectors, ", StyleBox["Mathematica", FontSlant->"Italic"], " first uses a scale function to determine the length, by default this is \ the Cartesian length of the vector. Then ", StyleBox["Mathematica", FontSlant->"Italic"], " multiplies this length by a scale factor so that all the vectors fit in \ the graph.\nThe graph above is a little hard to read. For our purposes it \ would be more useful if the vectors all had the same length. We can \ accomplish this by changing the default scale function and scale factor. \ Below we specify that all vectors should be of length ", Cell[BoxData[ \(TraditionalForm\`1\/3\)]], " and there should be no rescaling of the length by a scale factor. In \ addition, we draw a vector at every half integer." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(vecField = PlotVectorField[{x, 2 y - y\^2}, {x, 0, 8, 1/2}, {y, \(-4\), 4, 1/2}, Frame -> True, ScaleFunction -> \((1\/3\ &)\), ScaleFactor -> None];\)\)], "Input"], Cell[TextData[{ "From the above vector field it appears that if the initial value, ", Cell[BoxData[ \(TraditionalForm\`y\_0\)]], " is in the ranges ", Cell[BoxData[ \(TraditionalForm\`\((0, 2)\)\)]], " or ", Cell[BoxData[ \(TraditionalForm\`\((2, 4)\)\)]], " then the solution will quickly approach ", Cell[BoxData[ \(TraditionalForm\`y = 2\)]], ". If the initial value is negative, then the solution stays negative and \ grows rapidly." }], "Text", TextAlignment->Left, TextJustification->1], Cell[TextData[{ "We can use ", StyleBox["DSolve[]", FontWeight->"Bold"], " to find the general solution of the differential equation." }], "Text"], Cell[BoxData[ \(DSolve[\(y'\)[x] == 2 y[x] - y[x]\^2, y[x], x]\)], "Input"], Cell[TextData[{ "Multiplying the numerator and denominator by ", Cell[BoxData[ \(TraditionalForm\`\[ExponentialE]\^\(\(-2\) x\)\)]], " and renaming the constant of integration, we see that the solution can be \ written" }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm \`y = \(2\/\(1 - c\ \[ExponentialE]\^\(\(-2\) x\)\) . \)\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell["The solutions", "Text"], Cell[BoxData[ FormBox[ RowBox[{ \(y = 2\/\(1 - \(1\/3\) \[ExponentialE]\^\(\(-2\) x\)\)\), ",", " ", FormBox[\(y = 2\/\(1 + \[ExponentialE]\^\(\(-2\) x\)\)\), "TraditionalForm"], ",", " ", FormBox[\(y = 2\/\(1 - 3 \[ExponentialE]\^\(\(-2\) x\)\)\), "TraditionalForm"], ","}], TraditionalForm]], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "satisfy the initial conditions ", Cell[BoxData[ \(TraditionalForm\`y(0) = 3, 1, \(-1\)\)]], ", respectively. These solutions are plotted below. Note that the third \ solution is singular at ", Cell[BoxData[ \(TraditionalForm\`x = \(log\ 3\)\/2\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(sol = Plot[{2\/\(1 - \(1\/3\) \[ExponentialE]\^\(\(-2\) x\)\), 2\/\(1 + \[ExponentialE]\^\(\(-2\) x\)\), 2\/\(1 - 3 \[ExponentialE]\^\(\(-2\) x\)\)}, {x, 0, 8}, PlotRange -> {\(-4\), 4}];\)\)], "Input"], Cell["\<\ By superimposing the vector field and the solutions we see how the \ solutions \"follow\" the vector field.\ \>", "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(Show[vecField, sol, PlotRange -> {\(-4.5\), 4.5}];\)\)], "Input"], Cell[BoxData[ \(Clear[vecField, sol]\)], "Input"] }, Closed]], Cell[CellGroupData[{ Cell["Solutions", "Section"], Cell[CellGroupData[{ Cell["Solution 1", "Subsubsection"], Cell[BoxData[ \(TraditionalForm\`y' + y\/\(sin\ x\) = 0. \)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(DSolve[\(y'\)[x] + y[x]\/Sin[x] == 0, y[x], x]\)], "Input"], Cell[BoxData[ \(% // Simplify\)], "Input"], Cell["The solution is", "Text"], Cell[BoxData[ \(TraditionalForm\`y = c\ \(cot(x\/2) . \)\)], "DisplayFormula", TextAlignment->Center] }, Closed]], Cell[CellGroupData[{ Cell["Solution 2", "Subsubsection"], Cell[BoxData[ \(TraditionalForm\`y' - \(1\/x\) y = \(\(x\^\[Alpha]\)\(.\)\)\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(DSolve[\(y'\)[x] - \(1\/x\) y[x] == x\^\[Alpha], y[x], x]\)], "Input"], Cell[TextData[{ "For ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] \[NotEqual] 0\)]], " the solution is" }], "Text"], Cell[BoxData[ \(TraditionalForm\`y = x\^\(\[Alpha] + 1\)\/\[Alpha] + c\ \(x . \)\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(DSolve[\(y'\)[x] - \(1\/x\) y[x] == 1, y[x], x]\)], "Input"], Cell[TextData[{ "For ", Cell[BoxData[ \(TraditionalForm\`\[Alpha] = 0\)]], " the solution is" }], "Text"], Cell[BoxData[ \(TraditionalForm\`y = x\ log\ x + c\ \(x . \)\)], "DisplayFormula", TextAlignment->Center] }, Closed]], Cell[CellGroupData[{ Cell["Solution 3", "Subsubsection"], Cell[BoxData[ \(TraditionalForm\`y''' - y'' + y' - y = 0. \)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(DSolve[\(\(\(y'\)'\)'\)[x] - \(\(y'\)'\)[x] + \(y'\)[x] - y[x] == 0, y[x], x]\)], "Input"], Cell["The general solution is:", "Text"], Cell[BoxData[ \(TraditionalForm \`y(x) = \(c\_1\) sin\ x + \(c\_2\) cos\ x + \(c\_3\) \(\[ExponentialE]\^x . \)\)], "DisplayFormula", TextAlignment->Center] }, Closed]], Cell[CellGroupData[{ Cell["Solution 4", "Subsubsection"], Cell[BoxData[ \(TraditionalForm\`y'' + 2 a\ y' + \(a\^2\) y = 0\)], "DisplayFormula", TextAlignment->Center], Cell[BoxData[ \(DSolve[{\(\(y'\)'\)[x] + 2 a\ \(y'\)[x] + a\^2\ y[x] == 0, y[0] == 1, \(y'\)[0] == 0}, y[x], x]\)], "Input"], Cell[BoxData[ \(DSolve[{\(\(y'\)'\)[x] + 2 a\ \(y'\)[x] + a\^2\ y[x] == 0, y[0] == 0, \(y'\)[0] == 1}, y[x], x]\)], "Input"], Cell[TextData[{ "The fundamental set of solutions at the point ", Cell[BoxData[ \(TraditionalForm\`x = 0\)]], " is:" }], "Text"], Cell[BoxData[ \(TraditionalForm \`\({\((1 + a\ x)\) \[ExponentialE]\^\(\(-a\)\ x\), x\ \[ExponentialE]\^\(\(-a\)\ x\)} . \)\)], "DisplayFormula", TextAlignment->Center] }, Closed]] }, Closed]] }, Open ]] }, FrontEndVersion->"5.2 for Microsoft Windows", ScreenRectangle->{{0, 1680}, {0, 963}}, WindowToolbars->"EditBar", CellGrouping->Automatic, WindowSize->{773, 769}, WindowMargins->{{2, Automatic}, {Automatic, 2}}, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PaperSize"->{612, 792}, "PaperOrientation"->"Portrait", "Magnification"->1}, PrivateNotebookOptions->{"ColorPalette"->{RGBColor, 128}}, ShowCellLabel->True, ShowCellTags->False, RenderingOptions->{"ObjectDithering"->True, "RasterDithering"->False}, CharacterEncoding->"XAutomaticEncoding", Magnification->1.5 ] (******************************************************************* Cached data follows. 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