(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of 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). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. 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[ 96180, 3712]*) (*NotebookOutlinePosition[ 96858, 3737]*) (* CellTagsIndexPosition[ 96814, 3733]*) (*WindowFrame->Normal*) Notebook[{ Cell["Math 325: Differential Equations\tAssignment \tFall 1998", "Subsection"], Cell["Name:", "Subsection"], Cell["Section: ", "Subsection"], Cell["Professor: ", "Subsection"], Cell["\<\ This assignment is due Monday, Nov 23, 1998. Print out your \ answer to the problems. This assignment will be counted as one homework.\ \>", "Subsection"], Cell[CellGroupData[{ Cell["Initialization", "Subsubsection"], Cell["<True] }, Open ]], Cell[CellGroupData[{ Cell["Problem 1", "Subsubsection"], Cell[TextData[{ "Find the indefinite integral of the following functions\n", StyleBox["\tE^t Cos[t]^3 Sin[t]^5; (t-1)^10 Log[t+1]", "Input"] }], "Text"], Cell["Solution", "Subsubsection"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{"Integrate", "[", StyleBox[\(E^t\ Cos[t]^3\ Sin[t]^5, t\), "Input"], StyleBox["]", "Input"]}], "\n"}]], "Input"], Cell[BoxData[ \(\(-\(1\/5233280\(( E\^t\ \(( 98124\ Cos[2\ t] - 19240\ Cos[4\ t] - 13260\ Cos[6\ t] + 5032\ Cos[8\ t] - 49062\ Sin[2\ t] + 4810\ Sin[4\ t] + 2210\ Sin[6\ t] - 629\ Sin[8\ t])\))\)\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Integrate[(", StyleBox["t-1)^10 Log[t+1]", "Input"], ",t]" }], "Input"], Cell[BoxData[ \(\(-\(\(2047\ t\)\/11\)\) + \(1018\ t\^2\)\/11 - \(1981\ t\^3\)\/33 + \(454\ t\^4\)\/11 - \(1486\ t\^5\)\/55 + \(512\ t\^6\)\/33 - \(562\ t\^7\)\/77 + \(29\ t\^8\)\/11 - \(67\ t\^9\)\/99 + \(6\ t\^10\)\/55 - t\^11\/121 + 2047\/11\ Log[1 + t] + 1\/11\ t\ \(( 11 - 55\ t + 165\ t\^2 - 330\ t\^3 + 462\ t\^4 - 462\ t\^5 + 330\ t\^6 - 165\ t\^7 + 55\ t\^8 - 11\ t\^9 + t\^10)\)\ Log[1 + t]\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 2", "Subsubsection"], Cell[TextData[{ "Verify that the functions\n", StyleBox[ "\ty1[t_] = 1; y2[t_] = t;\n\ty3[t_] = E^(-t); y4[t_] = t*E^(-t);\n", "Input"], "are solutions of the differential equation\n", StyleBox["\ty''''[t] + 2y'''[t] + y''[t] == 0\n", "Input"], "and calculate their Wronskian." }], "Text"], Cell["Solution", "Subsubsection"], Cell[CellGroupData[{ Cell[TextData[StyleBox[ "y1[t_] = 1; y2[t_] = t;\ny3[t_] = E^(-t); y4[t_] = t*E^(-t);", "Input"]], "Input"], Cell[CellGroupData[{ Cell[BoxData[ RowBox[{ RowBox[{\(L[y_]\), ":=", StyleBox[\(D[y, {t, 4}]\ + \ 2\ D[y, {t, 3}]\ + \ D[y, {t, 2}]\), "Input"]}], ";", "\n", \({L[y1[t]], L[y2[t]], L[y3[t]], L[y4[t]]} // Simplify\)}]], "Input"], Cell[BoxData[ \({0, 0, 0, 0}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(w[f1_, f2_, f3_, f4_]\ := \ Det[\n{{\ \ f1, \ \ \ \ \ \ \ \ \ f2, \ \ \ \ \ \ \ \ \ f3, \ \ \ \ \ \ \ \ \ f4\ \ \ }, \n \ {D[f1, t], \ \ \ \ D[f2, t], \ \ \ \ D[f3, t], \ \ \ \ D[f4, t]}, \n\ {D[f1, {t, 2}], D[f2, {t, 2}], D[f3, {t, 2}], D[f4, {t, 2}]}, \n \ {D[f1, {t, 3}], D[f2, {t, 3}], D[f3, {t, 3}], D[f4, {t, 3}]}}]; \n w[y1[t], y2[t], y3[t], y4[t]]\)], "Input"], Cell[BoxData[ \(E\^\(\(-2\)\ t\)\)], "Output"] }, Open ]] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 3", "Subsubsection"], Cell[TextData[{ "Show that the Wronskian of ", StyleBox["Cos[t]", "Input"], ", ", StyleBox["Cos[t]^3", "Input"], ", ", StyleBox["Cos[t]^5", "Input"], ", and ", StyleBox["Cos[5t]", "Input"], " is ", StyleBox["0", "Input"], ".\nEstablish this result without direct evaluation of the Wronskian." }], "Text"], Cell["Solution", "Subsubsection"], Cell[CellGroupData[{ Cell["TrigExpand[Cos[5t]]", "Input"], Cell[BoxData[ \(Cos[t]\^5 - 10\ Cos[t]\^3\ Sin[t]\^2 + 5\ Cos[t]\ Sin[t]\^4\)], "Output"], Cell[CellGroupData[{ Cell[BoxData[ \(\(% /. {Sin[t]^2 -> 1 - Cos[t]^2, \ Sin[t]^4 -> \((1 - Cos[t]^2)\)^2} // ExpandAll\n\)\)], "Input"], Cell[BoxData[ \(5\ Cos[t] - 20\ Cos[t]\^3 + 16\ Cos[t]\^5\)], "Output"] }, Open ]] }, Open ]] }, Open ]], Cell[BoxData[""], "Input"], Cell[CellGroupData[{ Cell["Problem 4", "Subsubsection"], Cell[TextData[{ "Find all the 10th roots of ", StyleBox["1024", "Input"], ", including complex numbers. 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+ 4y[t]==Sin[2t] + t*E^(-t)\n", "Input"], "with the initial conditions\n", StyleBox["\ty[0] == y0, y'[0] == 0, y''[0] == 0, y'''[0] == 0\n", "Input"], "Superimpose plots of the particular solutions", StyleBox[" ", "Input"], "for several values of ", StyleBox["y0", "Input"], ". Explain why all these solutions seem to pass through the ", StyleBox["t", "Input"], "-axis at the same points and at regular intervals for large values of ", StyleBox["t", "Input"], ". 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Verify that ", StyleBox["Phi'[t]==A.Phi[t]", "Input"], ". Then use ", StyleBox["Phi[t]", "Input"], " to find a solution satisfying the given initial condition.\n\n", StyleBox["a)", FontSlant->"Italic"], StyleBox[" A =", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", "Input", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[ "{{-7,-8, 2},\n { 9,11,-3},\n { 9, 8, 0}}", "Input"], StyleBox[";\n X[0]=={3,-1,1};", FontFamily->"Courier", FontWeight->"Bold"], "\n\n", StyleBox["b)", FontSlant->"Italic"], StyleBox[" A =", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", "Input", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[ "{{-4,-2,-1},\n { 2, 6,-3},\n { 2, 4,-1}}", "Input"], StyleBox[";\n X[0]=={1,1,-2};\n \n", FontFamily->"Courier", FontWeight->"Bold"], StyleBox["c)", FontSlant->"Italic"], StyleBox[" A =", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", "Input", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[ "{{ 5, 3,-1},\n {-16,-6,-1},\n { -4,-3, 2}}", "Input"], StyleBox[";\n X[0]=={1,1,1}; \n \n", FontFamily->"Courier", FontWeight->"Bold"] }], "Text"], Cell["Solution", "Subsubsection"], Cell[TextData[{ StyleBox["a", FontSlant->"Italic"], ")" }], "Text"], Cell[TextData[{ StyleBox["A =", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", "Inpute", FontFamily->"Courier", FontWeight->"Bold"], StyleBox["{{-7,-8, 2},\n { 9,11,-3},\n { 9, 8, 0}}", "Inpute"], StyleBox[";", FontFamily->"Courier", FontWeight->"Bold"] }], "Input"], Cell[CellGroupData[{ Cell["Eigensystem[A]", "Input", AspectRatioFixed->True], Cell[BoxData[ \({{\(-1\), 2, 3}, {{\(-1\), 1, 1}, {\(-2\), 3, 3}, {\(-2\), 3, 2}}}\)], "Output"] }, Open ]], Cell["Fundamental solutions:", "Text"], Cell["\<\ X1[t_] = {-1,1,1}E^(-t); X2[t_] = {-2,3,3}E^(2t); X3[t_] = {-2,3,2}E^(3t);\ \>", "Input"], Cell[TextData[{ "A fundamental matrix has columns given by ", StyleBox["X1[t]", "Input"], ", ", StyleBox["X2[t]", "Input"], ", and ", StyleBox["X2[t]", "Input"], ". We use ", StyleBox["Transpose[]", "Input"], " to covert these vectors from rows into columns." }], "Text"], Cell[CellGroupData[{ Cell["\<\ Psi[t_] = Transpose[{X1[t], X2[t], X3[t]}]; MatrixForm[Psi[t]]\ \>", "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {\(-E\^\(-t\)\), \(\(-2\)\ E\^\(2\ t\)\), \(\(-2\)\ E\^\(3\ t\)\)}, {\(E\^\(-t\)\), \(3\ E\^\(2\ t\)\), \(3\ E\^\(3\ t\)\)}, {\(E\^\(-t\)\), \(3\ E\^\(2\ t\)\), \(2\ E\^\(3\ t\)\)} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell[TextData[{ "To get a fundamental matrix ", StyleBox["Phi[t]", "Input"], " such that ", StyleBox["Phi[0]", "Input"], " is the identity matrix, we multiply ", StyleBox["Psi[t]", "Input"], " on the right by the inverse of ", StyleBox["Psi[0]", "Input"], "." }], "Text"], Cell[CellGroupData[{ Cell["\<\ Phi[t_] = Psi[t].Inverse[Psi[0]]; Phi[t]//MatrixForm\ \>", "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {\(3\ E\^\(-t\) - 2\ E\^\(2\ t\)\), \(2\ E\^\(-t\) - 2\ E\^\(3\ t\)\), \(\(-2\)\ E\^\(2\ t\) + 2\ E\^\(3\ t\)\)}, {\(\(-3\)\ E\^\(-t\) + 3\ E\^\(2\ t\)\), \(\(-2\)\ E\^\(-t\) + 3\ E\^\(3\ t\)\), \(3\ E\^\(2\ t\) - 3\ E\^\(3\ t\)\)}, {\(\(-3\)\ E\^\(-t\) + 3\ E\^\(2\ t\)\), \(\(-2\)\ E\^\(-t\) + 2\ E\^\(3\ t\)\), \(3\ E\^\(2\ t\) - 2\ E\^\(3\ t\)\)} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["Phi[0]//MatrixForm", "Input", AspectRatioFixed->True], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {"1", "0", "0"}, {"0", "1", "0"}, {"0", "0", "1"} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell[TextData[{ "Let's check that ", StyleBox["Phi[t]", FontFamily->"Courier", FontWeight->"Bold"], " satisfies the equation:" }], "Text"], Cell[CellGroupData[{ Cell["Phi'[t]-A.Phi[t]//Simplify", "Input"], Cell[BoxData[ \({{0, 0, 0}, {0, 0, 0}, {0, 0, 0}}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["A.Phi[t]//Simplify//MatrixForm", "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {\(\(-E\^\(-t\)\)\ \((3 + 4\ E\^\(3\ t\))\)\), \(\(-2\)\ E\^\(-t\)\ \((1 + 3\ E\^\(4\ t\))\)\), \(2\ E\^\(2\ t\)\ \((\(-2\) + 3\ E\^t)\)\)}, {\(3\ E\^\(-t\) + 6\ E\^\(2\ t\)\), \(E\^\(-t\)\ \((2 + 9\ E\^\(4\ t\))\)\), \(\(-3\)\ E\^\(2\ t\)\ \((\(-2\) + 3\ E\^t)\)\)}, {\(3\ E\^\(-t\) + 6\ E\^\(2\ t\)\), \(2\ E\^\(-t\) + 6\ E\^\(3\ t\)\), \(\(-6\)\ E\^\(2\ t\)\ \((\(-1\) + E\^t)\)\)} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell["\<\ The solution satisfying the given initial condition is simply:\ \>", "Text"], Cell[CellGroupData[{ Cell[TextData[{ "X[t_] = Phi[t].", StyleBox["{3,-1,1}//Simplify", FontFamily->"Courier", FontWeight->"Bold"] }], "Input"], Cell[BoxData[ \({E\^\(-t\)\ \((7 - 8\ E\^\(3\ t\) + 4\ E\^\(4\ t\))\), \(-7\)\ E\^\(-t\) + 12\ E\^\(2\ t\) - 6\ E\^\(3\ t\), \(-7\)\ E\^\(-t\) + 12\ E\^\(2\ t\) - 4\ E\^\(3\ t\)}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ An alternative method for finding general solution and Special \ solution:\ \>", "Text"], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["A =", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", "Inpute", FontFamily->"Courier", FontWeight->"Bold"], StyleBox["{{-7,-8, 2},\n { 9,11,-3},\n { 9, 8, 0}}", "Inpute"], StyleBox[ ";\nX[t_]:={x1[t],x2[t],x3[t]};\nDSolve[{x1'[t]==Part[A.X[t],1], \ x2'[t]==Part[A.X[t],2], x3'[t]==Part[A.X[t],3]}, X[t],t]", FontFamily->"Courier", FontWeight->"Bold"] }], "Input"], Cell[BoxData[ \({{x1[t] \[Rule] \(-E\^\(-t\)\)\ C[1] - 2\ E\^\(2\ t\)\ C[2] - 2\ E\^\(3\ t\)\ C[3], x2[t] \[Rule] E\^\(-t\)\ C[1] + 3\ E\^\(2\ t\)\ C[2] + 3\ E\^\(3\ t\)\ C[3], x3[t] \[Rule] E\^\(-t\)\ C[1] + 3\ E\^\(2\ t\)\ C[2] + 2\ E\^\(3\ t\)\ C[3]}}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ "\nDSolve[{x1'[t]==Part[A.X[t],1], x2'[t]==Part[A.X[t],2], \ x3'[t]==Part[A.X[t],3],\nx1[0]==3,x2[0]==-1,x3[0]==1}, X[t],t]", FontFamily->"Courier", FontWeight->"Bold"]], "Input"], Cell[BoxData[ \({{x1[t] \[Rule] \(-E\^\(-t\)\)\ \((\(-7\) + 8\ E\^\(3\ t\) - 4\ E\^\(4\ t\))\), x2[t] \[Rule] E\^\(-t\)\ \((\(-7\) + 12\ E\^\(3\ t\) - 6\ E\^\(4\ t\))\), x3[t] \[Rule] E\^\(-t\)\ \((\(-7\) + 12\ E\^\(3\ t\) - 4\ E\^\(4\ t\))\)}}\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["b", FontSlant->"Italic"], ")" }], "Text"], Cell[TextData[{ StyleBox["A =", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", "Inpute", FontFamily->"Courier", FontWeight->"Bold"], StyleBox["{{-4,-2,-1},\n { 2, 6,-3},\n { 2, 4,-1}}", "Inpute"], StyleBox[";\n", FontFamily->"Courier", FontWeight->"Bold"] }], "Input"], Cell[CellGroupData[{ Cell["Eigensystem[A]", "Input", AspectRatioFixed->True], Cell[BoxData[ \({{\(-3\), 2, 2}, {{\(-3\), 1, 1}, {\(-1\), 2, 2}, {0, 0, 0}}}\)], "Output"] }, Open ]], Cell["We get two fundamental solutions from this:", "Text"], Cell[TextData[{ "X1[t_] = {-3,1,1}E^(-3t);\nX2[t_] = ", StyleBox["{-1,2,2}", FontFamily->"Courier", FontWeight->"Bold"], "E^(2t);" }], "Input"], Cell[TextData[{ "The eigenvalue ", StyleBox["2", "Input"], " is repeated and ", StyleBox["Eigensystem[]", FontFamily->"Courier", FontWeight->"Bold"], " finds only one non-trivial eigenvector, ", StyleBox["{-1,2,2}", FontFamily->"Courier", FontWeight->"Bold"], ". A generalized eigenvector is found by solving" }], "Text"], Cell[CellGroupData[{ Cell[TextData[{ "Solve[(A - 2*IdentityMatrix[3]).{w1,w2,w3} == {", StyleBox["-1,2,2", FontFamily->"Courier", FontWeight->"Bold"], "}]" }], "Input", AspectRatioFixed->True], Cell[BoxData[ \(Solve::"svars" \( : \ \) "Equations may not give solutions for all \"solve\" variables."\)], "Message"], Cell[BoxData[ \({{w1 \[Rule] \(-\(w3\/2\)\), w2 \[Rule] 1\/2 + w3}}\)], "Output"] }, Open ]], Cell[TextData[{ "For example, ", StyleBox["{w1,w2,w3} = {-1,5/2,2}", "Input"], ", from which we can derive a third independent solution:" }], "Text"], Cell[TextData[{ "X3[t_] = ", StyleBox["{-1,2,2}", FontFamily->"Courier", FontWeight->"Bold"], " t*E^(2t) + ", StyleBox["{-1,5/2,2}", "Input"], " E^(2t);" }], "Input"], Cell["The general solution is:", "Text"], Cell[CellGroupData[{ Cell["X[t_] = c1 X1[t] + c2 X2[t] + c3 X3[t]", "Input"], Cell[BoxData[ \({\(-3\)\ c1\ E\^\(\(-3\)\ t\) - c2\ E\^\(2\ t\) + c3\ \((\(-E\^\(2\ t\)\) - E\^\(2\ t\)\ t)\), c1\ E\^\(\(-3\)\ t\) + 2\ c2\ E\^\(2\ t\) + c3\ \((\(5\ E\^\(2\ t\)\)\/2 + 2\ E\^\(2\ t\)\ t)\), c1\ E\^\(\(-3\)\ t\) + 2\ c2\ E\^\(2\ t\) + c3\ \((2\ E\^\(2\ t\) + 2\ E\^\(2\ t\)\ t)\)}\)], "Output"] }, Open ]], Cell["Check:", "Text"], Cell[CellGroupData[{ Cell["X'[t]-A.X[t]//Simplify", "Input"], Cell[BoxData[ \({0, 0, 0}\)], "Output"] }, Open ]], Cell["Fundamental matrix:", "Text"], Cell[CellGroupData[{ Cell["\<\ Psi[t_] = Transpose[{X1[t], X2[t], X3[t]}]; Psi[t]//MatrixForm\ \>", "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {\(\(-3\)\ E\^\(\(-3\)\ t\)\), \(-E\^\(2\ t\)\), \(\(-E\^\(2\ t\)\) - E\^\(2\ t\)\ t\)}, {\(E\^\(\(-3\)\ t\)\), \(2\ E\^\(2\ t\)\), \(\(5\ E\^\(2\ t\)\)\/2 + 2\ E\^\(2\ t\)\ t\)}, {\(E\^\(\(-3\)\ t\)\), \(2\ E\^\(2\ t\)\), \(2\ E\^\(2\ t\) + 2\ E\^\(2\ t\)\ t\)} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ Phi[t_] = Psi[t].Inverse[Psi[0]]; Phi[t]//MatrixForm\ \>", "Input"], Cell[BoxData[ TagBox[ RowBox[{"(", GridBox[{ {\(\(6\ E\^\(\(-3\)\ t\)\)\/5 - E\^\(2\ t\)\/5\), \(2\ E\^\(2\ t\) + 2\ \((\(-E\^\(2\ t\)\) - E\^\(2\ t\)\ t)\)\), \(\(3\ E\^\(\(-3\)\ t\)\)\/5 - \(13\ E\^\(2\ t\)\)\/5 - 2\ \((\(-E\^\(2\ t\)\) - E\^\(2\ t\)\ t)\)\)}, {\(\(-\(2\/5\)\)\ E\^\(\(-3\)\ t\) + \(2\ E\^\(2\ t\)\)\/5\), \(\(-4\)\ E\^\(2\ t\) + 2\ \((\(5\ E\^\(2\ t\)\)\/2 + 2\ E\^\(2\ t\)\ t)\)\), \(\(-\(1\/5\)\)\ E\^\(\(-3\)\ t\) + \(26\ E\^\(2\ t\)\)\/5 - 2\ \((\(5\ E\^\(2\ t\)\)\/2 + 2\ E\^\(2\ t\)\ t)\)\)}, {\(\(-\(2\/5\)\)\ E\^\(\(-3\)\ t\) + \(2\ E\^\(2\ t\)\)\/5\), \(\(-4\)\ E\^\(2\ t\) + 2\ \((2\ E\^\(2\ t\) + 2\ E\^\(2\ t\)\ t)\)\), \(\(-\(1\/5\)\)\ E\^\(\(-3\)\ t\) + \(26\ E\^\(2\ t\)\)\/5 - 2\ \((2\ E\^\(2\ t\) + 2\ E\^\(2\ t\)\ t)\)\)} }], ")"}], (MatrixForm[ #]&)]], "Output"] }, Open ]], Cell["The solution satisfying the given initial condition is:", "Text"], Cell[CellGroupData[{ Cell[TextData[{ "X[t_] = Phi[t].", StyleBox["{1,1,-2}//Simplify", FontFamily->"Courier", FontWeight->"Bold"] }], "Input"], Cell[BoxData[ \({\(-E\^\(2\ t\)\)\ \((\(-1\) + 6\ t)\), E\^\(2\ t\)\ \((1 + 12\ t)\), 2\ E\^\(2\ t\)\ \((\(-1\) + 6\ t)\)}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["\<\ An alternative method for finding general solution and Special \ solution:\ \>", "Text"], Cell[CellGroupData[{ Cell[TextData[{ StyleBox["A =", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", "Inpute", FontFamily->"Courier", FontWeight->"Bold"], StyleBox["{{-4,-2,-1},\n { 2, 6,-3},\n { 2, 4,-1}}", "Inpute"], StyleBox[ ";\nX[t_]:={x1[t],x2[t],x3[t]};\nDSolve[{x1'[t]==Part[A.X[t],1], \ x2'[t]==Part[A.X[t],2], x3'[t]==Part[A.X[t],3]}, X[t],t]", FontFamily->"Courier", FontWeight->"Bold"] }], "Input"], Cell[BoxData[ \({{x1[t] \[Rule] \((\(6\ E\^\(\(-3\)\ t\)\)\/5 - E\^\(2\ t\)\/5)\)\ C[1] - 2\ E\^\(2\ t\)\ t\ C[2] + \((\(3\ E\^\(\(-3\)\ t\)\)\/5 - \(3\ E\^\(2\ t\)\)\/5 + 2\ E\^\(2\ t\)\ t)\)\ C[3], x2[t] \[Rule] \((\(-\(2\/5\)\)\ E\^\(\(-3\)\ t\) + \(2\ E\^\(2\ t\)\)\/5)\)\ C[1] + \((E\^\(2\ t\) + 4\ E\^\(2\ t\)\ t)\)\ C[2] + \((\(-\(1\/5\)\)\ E\^\(\(-3\)\ t\) + E\^\(2\ t\)\/5 - 4\ E\^\(2\ t\)\ t)\)\ C[3], x3[t] \[Rule] \((\(-\(2\/5\)\)\ E\^\(\(-3\)\ t\) + \(2\ E\^\(2\ t\)\)\/5)\)\ C[1] + 4\ E\^\(2\ t\)\ t\ C[2] + \((\(-\(1\/5\)\)\ E\^\(\(-3\)\ t\) + \(6\ E\^\(2\ t\)\)\/5 - 4\ E\^\(2\ t\)\ t)\)\ C[3]}}\)], "Output"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell[TextData[StyleBox[ "\nDSolve[{x1'[t]==Part[A.X[t],1], x2'[t]==Part[A.X[t],2], \ x3'[t]==Part[A.X[t],3],\nx1[0]==1,x2[0]==1,x3[0]==-2}, X[t],t]//Simplify", FontFamily->"Courier", FontWeight->"Bold"]], "Input"], Cell[BoxData[ \({{x1[t] \[Rule] \(-E\^\(2\ t\)\)\ \((\(-1\) + 6\ t)\), x2[t] \[Rule] E\^\(2\ t\)\ \((1 + 12\ t)\), x3[t] \[Rule] 2\ E\^\(2\ t\)\ \((\(-1\) + 6\ t)\)}}\)], "Output"] }, Open ]], Cell[TextData[{ StyleBox["c", FontSlant->"Italic"], ")" }], "Text"], Cell[TextData[{ StyleBox["A =", FontFamily->"Courier", FontWeight->"Bold"], StyleBox[" ", "Inputer", FontFamily->"Courier", FontWeight->"Bold"], StyleBox["{{ 5, 3,-1},\n {-16,-6,-1},\n { -4,-3, 2}}", "Inputer"], StyleBox[";", FontFamily->"Courier", FontWeight->"Bold"] }], "Input"], Cell[CellGroupData[{ Cell["Eigensystem[A]", "Input"], Cell[BoxData[ \({{\(-3\)\ I, 3\ I, 1}, {{\(-1\), 2 + I, 1}, {\(-1\), 2 - I, 1}, {\(-1\), 2, 2}}}\)], "Output"] }, Open ]], Cell["\<\ X1[t_] = {-1,2,2}E^t; Z[t_] = {-1,2-I,1}E^(3I*t);\ \>", "Input"], Cell[CellGroupData[{ Cell["X2[t_] = Re[Z[t]]//ComplexExpand", "Input"], Cell[BoxData[ \({\(-Cos[3\ t]\), 2\ Cos[3\ t] + Sin[3\ t], Cos[3\ t]}\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell["X3[t_] = Im[Z[t]]//ComplexExpand", "Input"], Cell[BoxData[ \({\(-Sin[3\ t]\), \(-Cos[3\ t]\) + 2\ Sin[3\ t], Sin[3\ t]}\)], "Output"] }, Open ]], Cell["General solution:", "Text"], Cell[CellGroupData[{ Cell["\<\ X[t_] = c1 X1[t] + c2 X2[t] + c3 X3[t]; X[t]//MatrixForm\ \>", "Input"], Cell[BoxData[ InterpretationBox[ RowBox[{"(", GridBox[{ {\(\(-c1\)\ E\^t - c2\ Cos[3\ t] - c3\ Sin[3\ t]\)}, { \(2\ c1\ E\^t + c2\ \((2\ Cos[3\ t] + Sin[3\ t])\) + c3\ \((\(-Cos[3\ t]\) + 2\ Sin[3\ t])\)\)}, {\(2\ c1\ E\^t + c2\ Cos[3\ t] + c3\ Sin[3\ t]\)} }], ")"}], MatrixForm[ { Plus[ Times[ -1, c1, Power[ E, t]], Times[ -1, c2, Cos[ Times[ 3, t]]], Times[ -1, c3, Sin[ Times[ 3, t]]]], Plus[ Times[ 2, c1, Power[ E, t]], Times[ c2, Plus[ Times[ 2, Cos[ Times[ 3, t]]], Sin[ Times[ 3, t]]]], Times[ c3, Plus[ Times[ -1, Cos[ Times[ 3, t]]], Times[ 2, Sin[ Times[ 3, t]]]]]], Plus[ Times[ 2, c1, Power[ E, t]], Times[ c2, Cos[ Times[ 3, t]]], Times[ c3, Sin[ Times[ 3, t]]]]}]]], "Output"] }, Open ]], Cell["Check:", "Text"], Cell[CellGroupData[{ Cell["X'[t]-A.X[t]//Simplify", "Input"], Cell[BoxData[ \({0, 0, 0}\)], "Output"] }, Open ]], Cell["Fundamental matrix:", "Text"], Cell[CellGroupData[{ Cell["\<\ Psi[t_] = Transpose[{X1[t], X2[t], X3[t]}]; 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