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This is unnecessary in ", StyleBox["Mathematica", FontSlant->"Italic"], " 4.0." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(<< Calculus`LaplaceTransform`\)], "Input"], Cell[TextData[{ "The ", Cell[BoxData[ FormBox[ StyleBox[\(LaplaceTransform[]\), "Input"], TraditionalForm]]], " function gives the Laplace transform of a function." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm\`F( s) = \[Integral]\_0\%\[Infinity]\( \[ExponentialE]\^\(\(-s\)\ t\)\) \ \(f(t)\) \[DifferentialD]t\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[BoxData[ \(\(?LaplaceTransform\)\)], "Input"], Cell[TextData[{ "The Laplace transform of ", Cell[BoxData[ \(TraditionalForm\`t\ \[ExponentialE]\^t\)]], " is" }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(LaplaceTransform[t\ Exp[t], t, s]\)], "Input"], Cell[TextData[{ "The Laplace transform of ", Cell[BoxData[ \(TraditionalForm\`\(f'\)' \((t)\)\)]], " is" }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(LaplaceTransform[\(\(f'\)'\)[t], t, s]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["Inverse Laplace Transform", "Subsection"], Cell[TextData[{ "The ", Cell[BoxData[ FormBox[ StyleBox[\(InverseLaplaceTransform[]\), "Input"], TraditionalForm]]], " function gives the inverse Laplace transform." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(TraditionalForm\`f( t) = \(1\/\(2 \[Pi]\ \[ImaginaryI]\)\) \(\[Integral]\_\(c - \ \[ImaginaryI]\[Infinity]\)\%\(c + \[ImaginaryI]\[Infinity]\)\(\[ExponentialE]\ \^\(s\ t\)\) \(F(s)\) \(\(\[DifferentialD]s\)\(.\)\)\)\)], "DisplayFormula", TextAlignment->Center, TextJustification->0], Cell[TextData[{ "Here ", Cell[BoxData[ \(TraditionalForm\`c\)]], " is to the right of the singularities of ", Cell[BoxData[ \(TraditionalForm\`F(s)\)]], ". ", StyleBox["Mathematica", FontSlant->"Italic"], " has a function for doing inverse Laplace transforms." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(\(?InverseLaplaceTransform\)\)], "Input"], Cell[TextData[{ "Note that ", StyleBox["InverseLaplaceTransform[]", "Input"], " returns a result that is valid only for ", Cell[BoxData[ \(TraditionalForm\`t \[GreaterEqual] 0\)]], ". The inverse Laplace transform of ", Cell[BoxData[ \(TraditionalForm\`1\/s\^2\)]], " is ", Cell[BoxData[ \(TraditionalForm\`t\ \(H(t)\)\)]], ". ", StyleBox["Mathematica", FontSlant->"Italic"], " returns a result that is valid for positive ", Cell[BoxData[ \(TraditionalForm\`t\)]], "." }], "Text", TextAlignment->Left, TextJustification->1], Cell[BoxData[ \(InverseLaplaceTransform[1\/s\^2, s, t]\)], "Input"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\[ScriptCapitalL][Log\ t]\)]]], "Section"], Cell[BoxData[ \(LaplaceTransform[Log[t], t, s]\)], "Input"], Cell[BoxData[ \(\(?EulerGamma\)\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\(\[ScriptCapitalL]\^\(-1\)\)[\(\@\(\[Pi]\/s\)\) \ \[ExponentialE]\^\(\(-2\) \@\(a\ s\)\)]\)]]], "Section"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " is not able to do the inverse transform." }], "Text"], Cell[BoxData[ \(InverseLaplaceTransform[\(\@\(\[Pi]\/s\)\) Exp[\(-2\) \@\(a\ s\)], s, t]\)], "Input"], Cell["However, it can do the transform.", "Text"], Cell[BoxData[ \(LaplaceTransform[Exp[\(-a\)/t]\/\@t, t, s]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[Cell[BoxData[ \(TraditionalForm\`y\^\((4)\) - y = t\)]]], "Section"], Cell["We can solve the differential equation directly.", "Text"], Cell[BoxData[ \(DSolve[{\(y''''\)[t] - y[t] \[Equal] t, y[0] \[Equal] 0, \(y'\)[0] \[Equal] 0, \(y''\)[0] \[Equal] 0, \(y'''\)[0] \[Equal] 0}, y[t], t]\)], "Input"], Cell[BoxData[ \(% // FullSimplify\)], "Input"], Cell["We can also use the Laplace transform.", "Text"], Cell[BoxData[ \(LaplaceTransform[\(y''''\)[t] - y[t], t, s] \[Equal] LaplaceTransform[t, t, s]\)], "Input"], Cell[BoxData[ \(% /. {LaplaceTransform[y[t], t, s] \[Rule] Y[s], y[0] \[Rule] 0, \(y'\)[0] \[Rule] 0, \(y''\)[0] \[Rule] 0, \(y'''\)[0] \[Rule] 0}\)], "Input"], Cell[BoxData[ \(Solve[%, Y[s]]\)], "Input"], Cell[BoxData[ \(InverseLaplaceTransform[1\/\(s\^2\ \((\(-1\) + s\^4)\)\), s, t]\)], "Input"], Cell[BoxData[ \(% // FullSimplify\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\[DifferentialD]y\/\[DifferentialD]t = sin\ t + \[Integral]\_0\%t\( y(\[Tau])\) \(cos( t - \[Tau])\) \[DifferentialD]\[Tau]\)]]], "Section"], Cell[BoxData[ \(LaplaceTransform[\(y'\)[t], t, s] \[Equal] LaplaceTransform[ Sin[t] + \[Integral]\_0\%t y[\[Tau]] Cos[t - \[Tau]] \[DifferentialD]\[Tau], t, s]\)], "Input"], Cell[BoxData[ \(% /. {LaplaceTransform[y[t], t, s] \[Rule] Y[s], y[0] \[Rule] 0}\)], "Input"], Cell[BoxData[ \(Solve[%, Y[s]]\)], "Input"], Cell[BoxData[ \(InverseLaplaceTransform[1\/s\^3, s, t]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[Cell[BoxData[ \(TraditionalForm\`\[DifferentialD]u\/\[DifferentialD]t + u(t) - u(t - 1) = 0\)]]], "Section"], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " will not \"simplify\" the Laplace transform of ", Cell[BoxData[ \(TraditionalForm\`u(t - 1)\)]], "." }], "Text"], Cell[BoxData[ \(LaplaceTransform[u[t - 1], t, s]\)], "Input"], Cell["However, we can do another part of the problem.", "Text"], Cell[BoxData[ \(u0[0]\/\(1 + s - Exp[\(-s\)]\) + \(Exp[\(-s\)]\/\(1 + s - Exp[\(-s\)]\)\) \(\[Integral]\_\(-1\)\%0 Exp[\(-s\)\ t] u0[t] \[DifferentialD]t\) /. u0[t_] \[Rule] 1\)], "Input"], Cell[BoxData[ \(% // Simplify\)], "Input"] }, Open ]] }, Open ]] }, FrontEndVersion->"4.0 for X", ScreenRectangle->{{0, 1152}, {0, 864}}, WindowSize->{520, 600}, WindowMargins->{{0, Automatic}, {Automatic, 0}}, Magnification->1.25 ] (*********************************************************************** Cached data follows. 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