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That is, \ include\ncomments where appropriate, and use the Format menus to distinguish \ text from headings\nfrom actual computation. It might work best just to do \ all the work and then go back and insert comments, format things right, etc. \ Also, remember that when the instructions don't explicitly require the use of \ ", StyleBox["Mathematica,", FontSlant->"Italic"], " you're free to use pencil and paper to do some of the work--whatever gets \ the job done most effectively." }], "Text"], Cell[CellGroupData[{ Cell["Problem 2.", "Subsection"], Cell[CellGroupData[{ Cell["a. (2 points)", "Subsubsection"], Cell[BoxData[ \(soln[x_]\ = \ First[y[x] /. DSolve[{\(y'\)[x]\ + \ y[x]/x == 2, \ y[1] == c}, y[x], x]]\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["b. (2 points)", "Subsubsection"], Cell[BoxData[ \(\(Plot[Evaluate[Table[soln[x], {c, .8, 1.2, .1}]], {x, .75, 1.25}]; \)\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["c. (2 points)", "Subsubsection"], Cell[BoxData[ \(Table[soln[10^n], {n, \(-2\), 1}, {c, .8, 1.2, .1}]\)], "Input"], Cell[BoxData[ \(% // TableForm\)], "Input"], Cell[BoxData[{ \(TextForm\`The\ columns\ correspond\ to\ c\ = \ .8, \ .9, \ 1, \ 1.1, \ 1.2\ going\ from\ left\), \(TextForm \`to\ right . \ \ The\ rows\ correspond\ to\ x\ = \ \(10\^n = \ .01\), \ .1, \ 1, \ 10\ going\), \(TextForm\`from\ top\ to\ \(bottom . \)\)}], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["d. (4 points)", "Subsubsection"], Cell[BoxData[ \(\(Plot[Evaluate[Table[soln[x], {c, .8, 1.2, .1}]], {x, 0, 2.5}]; \)\)], "Input"], Cell["\<\ All solutions are asymptotic to the same line (looks like y=x) as x goes to infinity--regardless of initial condition. Solutions with initial value greater than one tend to infinity as x goes to zero. Those with initial value less than one tend to negative infinity. The solution with initial value one tends to zero as x goes to zero.\ \>", "Text"], Cell[BoxData[ \(\(\n\n\n\n\n\n\n\)\)], "Input"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 4.", "Subsection"], Cell[CellGroupData[{ Cell["a. (2 points) ", "Subsubsection"], Cell[BoxData[{ \(Clear[y]\n\), \(DSolve[ \(y'\)[x]\ == \ \((x\ - \ Exp[\(-x\)])\)/\((y[x] + Exp[y[x]])\), \n \t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y[x], x]\)}], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["b. (2 points) ", "Subsubsection"], Cell["\<\ I'll cut and paste the left hand side and give it a convenient name (it's important to remove the [x]'s from the y[x]'s):\ \>", "Text"], Cell[BoxData[ \(lhs[x_, y_]\ = \ Simplify[\ 1\/2\ E\^\(-x\)\ \((\(-2\) + 2\ E\^\(x + y\) - E\^x\ x\^2 + E\^x\ y\^2)\)]\)], "Input"], Cell[BoxData[ \(\(ContourPlot[lhs[x, y], {x, \(-1\), 3}, {y, \(-2\), 2}, ContourShading -> False]; \)\)], "Input"] }, Open ]], Cell[CellGroupData[{ Cell["c. (3 points) ", "Subsubsection"], Cell[BoxData[ \(\(plot1\ = ContourPlot[lhs[x, y], {x, \(-1\), 3}, {y, \(-2\), 2}, Contours\ -> \ {lhs[1.5, .5]}, \ ContourShading -> False]; \)\)], "Input"], Cell["\<\ The top curve is the graph of the solution since it (evidently) passes through the point (1.5,.5). \t\t\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["d. (3 points) ", "Subsubsection"], Cell["\<\ I'll use the previous picture to take a guess at the correct value of y and then use FindRoot to improve the guess: \ \>", "Text"], Cell[BoxData[{ \(FindRoot[lhs[0, y]\ == \ lhs[1.5, .5], {y, .5}]\), \(FindRoot[lhs[1, y]\ == \ lhs[1.5, .5], {y, .5}]\), \(FindRoot[lhs[1.8, y]\ == \ lhs[1.5, .5], {y, .5}]\), \(FindRoot[lhs[2.1, y]\ == \ lhs[1.5, .5], {y, .5}]\)}], "Input"], Cell["\<\ That is: \t y(0) = .318386... y(1) = .235633... \t y(1.8) = .682187... y(2.1) = .866212...\ \>", "Text"], Cell[BoxData[ \(\(Show[plot1, Graphics[{PointSize[0.03], Point[{0, .318386}], Point[{1, .235633}], \n\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Point[{1.8, .682187}], \ Point[{2.1, .866212}]}]]; \)\)], "Input"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 10.", "Subsection"], Cell[CellGroupData[{ Cell["a. (4 points)", "Subsubsection"], Cell["\<\ Before I plot the vector field, I'd like to check the graph of the right side of the ODE as a function of y. This will give me some idea of what vertical range to use. Since the equation is autonomous, the horizontal range is unimportant.\ \>", "Text"], Cell[BoxData[ \(\(Plot[3 Sin[y] + y - 2, {y, \(-10\), 10}]; \)\)], "Input"], Cell["\<\ looks like I should expect equilibrium solutions at about y=1/2, y=4 and y=5. Therefore, I'll choose [-1,6] as a range for y.\ \>", "Text"], Cell[BoxData[ \(Needs["\"]\)], "Input"], Cell[BoxData[ \(\(PlotVectorField[{1, 3 Sin[y]\ + \ y\ - \ 2}, {t, 0, 3}, {y, \(-1\), 6}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1&)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 15]; \)\)], "Input"], Cell["To find the equilibrium solutions:", "Text"], Cell[BoxData[{ \(FindRoot[3 Sin[y] + y - 2 == 0, {y, .5}]\), \(FindRoot[3 Sin[y] + y - 2 == 0, {y, 4}]\), \(FindRoot[3 Sin[y] + y - 2 == 0, {y, 5}]\n\)}], "Input"], Cell["\<\ From this and the two plots above, it is apparent that the ODE has two unstable equilibrium solutions: y= 4.9295 and y=.517049 and one stable equilibrium solution: y= 3.77452.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["b. (4 points)", "Subsubsection"], Cell["\<\ This ODE is not autonomous, so I cannot proceed as I did in the \ last example. I'll have to get along with just the direction field. Note that the right side vanishes (i.e. the direction field is horizontal) when y=0 and y=x.\ \>", "Text"], Cell[BoxData[ \(\(PlotVectorField[{1, y^2\ - x*y}, {x, \(-4\), 4}, {y, \(-4\), 4}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1&)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 15]; \)\)], "Input"], Cell["\<\ Looks like y=0 is a constant solution of the ODE--an observation that is easily verified by plugging into the ODE. All solutions y[x] starting below y= 0 appear to converge to y=0 as x goes to infinity. Solutions y[x] starting above y= 0 appear to converge to either infinity or 0. Solutions exhibiting these two types of behavior appear to be separated by a curve asymptotic to the line y=x as x goes to infinity and y=0 as x goes to negative infinity. At any rate it is clear from the right side of the differential equation that the dividing curve must lie at or above the line y=x, because the right side is negative when y < x (and y > 0) so that solutions lying between y=0 and y=x will slope downward as x becomes large (and positive). \t \ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["c. (4 points)", "Subsubsection"], Cell[BoxData[ \(DSolve[{\(y'\)[x] == \ y[x]^2\ - \ x\ *y[x], y[0]\ == \ c}, y[x], x] \)], "Input"], Cell["What the *&^&* is this Erf thing?", "Text"], Cell[BoxData[ \(\(Plot[Erf[x/Sqrt[2]], {x, \(-10\), 10}]; \)\)], "Input"], Cell[BoxData[{ FormBox[ \(Doesn' t\ really\ matter, \ I\ guess . \ \ The\ point\ is\ that\ Erf[x/\@2]\), TextForm], FormBox[ RowBox[{ FormBox[ \(increases\ from\ - 1\ to\ 1\ as\ x\ goes\ from\ negative\ infinity\ to\), "TextForm"], " "}], TextForm], FormBox[ RowBox[{ FormBox[\(infinity . \ Therefore, \ if\ c = y[0]\ is\ less\ than\ \@\(2/\[Pi]\)\ then\ it' s\ clear\), "TextForm"], " "}], TextForm], FormBox[ \(that\ y[x]\ is\ defined\ for\ all\ x\ \((that\ is, \ the\ denominator\ never\nvanishes)\) and\ that\ y[x]\ converges\ to\ zero\ as\ x\ goes\ to\ infinity\), TextForm], FormBox[ \(\((because\ the\ numerator\ of\ y[x]\ definitely\ goes\ to\ zero\ while \ the\ndenominator\ doesn' t)\) . \ \ If\ y[0]\ > \@\(2/\[Pi]\), \ then\ at\ some\ finite\), TextForm], FormBox[ \(x\ the\ denominator\ vanishes\ and\ the\ solution\ goes\ to\ infinity\ along\), TextForm], FormBox[ \(a\ vertical\ asymptote\ \((this\ is\ different\ than\ what\ I\ predicted\ from\ the\ \n direction\ field)\) . \ \ It' s\ harder\ to\ tell\ just\ what\ happens\ if\), TextForm], FormBox[ \(y[0]\ = \ \@\(2/\[Pi]\), \ so\ I\ plot\ that\ particular\ \(solution : \)\), TextForm]}], "Text"], Cell[BoxData[ \(\(Plot[ \(-\(\(2\ \@\(2/\[Pi]\)\ E\^\(-\(x\^2\/2\)\)\)\/\(\(-2\) + \(\@\(2/\[Pi]\)\) \@\(2\ \[Pi]\)\ Erf[x\/\@2]\)\)\), {x, 0, 7}]; \)\)], "Input"], Cell["\<\ This looks pretty close to a straight line, so I compare with the \ line y=x that looked important in the direction field:\ \>", "Text"], Cell[BoxData[ \(\(Plot[{ \(-\(\(2\ \@\(2/\[Pi]\)\ E\^\(-\(x\^2\/2\)\)\)\/\(\(-2\) + \(\@\(2/\[Pi]\)\) \@\(2\ \[Pi]\)\ Erf[x\/\@2]\)\)\), x}, {x, 0, 7}]; \)\)], "Input"], Cell["\<\ Sure enough, the \"dividing\" solution is asymptotic to the line \ y=x.\ \>", "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 12.", "Subsection"], Cell[CellGroupData[{ Cell["a. (3 points)", "Subsubsection"], Cell[BoxData[ \(\(PlotVectorField[{1, \(-t\)*x^2}, {t, \(-3\), 3}, {x, \(-3\), 3}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1&)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 19]; \)\)], "Input"], Cell["\<\ Looks like x = 0 is a constant solution.Moreover, it looks like solutions x[t] that start above x=0 will converge to zero as t goes to either positive or negative infinity,whereas solutions x[t] that start below x=0 will converge to negative infinity.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["b. (3 points) ", "Subsubsection"], Cell["\<\ I'm going to get the solution to the ODE in a slightly different form than is asked for in the book.\ \>", "Text"], Cell[BoxData[ \(soln\ = \ First[x[t] /. DSolve[{\(x'\)[t] == \(-t\)*x[t]^2, x[0]\ == \ c}, x[t], t]]\)], "Input"], Cell[BoxData[ \(\(Plot[Evaluate[Table[soln, {c, \(-3\), 3, 1}]], {t, \(-5\), 5}]; \)\)], "Input"], Cell["\<\ So in fact, my predictions from the direction field weren't quite right. It's true that any solution satisfying x[0] >0 will converge to 0 as t goes to positive or negative infinity. However, any negative solution appears to have to vertical asymptotes and to go to negative infinity along both of them--i.e. solutions that start \ below x-axis go to negative infinity at some *finite* positive and negative values of t.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["c. (4 points)", "Subsubsection"], Cell[TextData[ "Five types of solution curves:\n1. x[t] < 0, defined for all t in some \ finite interval symmetric about the \n y-axis, and x[t] tends to negative \ infinity at each end of the interval.\n2. x[t] = 0 for all t\n3. x[t] > 0, \ defined for all t and tending to zero as t goes to \[PlusMinus]\[Infinity].\n\ 4. x[t] > 0 everywhere decreasing, and defined for t in an interval of the \ form [a,\[Infinity])\n where a > 0; asymptote at t = a.\n5. x[t] > 0 \ everywhere, increasing and defined for t in an interval of the form (-\ \[Infinity],-a]\n where a > 0; asymptote at t = -a."], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["d. (2 points)", "Subsubsection"], Cell["\<\ This `Show' command is a little tricky--took some playing around to \ get a good picture.\ \>", "Text"], Cell[BoxData[ \(\(\ \ \(plot1\ = \ PlotVectorField[{1, \(-t\)\ *x^2}, {t, \(-3\), 3}, {x, \(-4\), 4}, \ PlotPoints\ -> \ 17, \n\t\ \ \ ScaleFunction \[Rule] \((1&)\), Axes -> \ True, Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1]; \)\)\)], "Input"], Cell[BoxData[ \(\(\t\ \(plot2\ = \ Plot[Evaluate[Table[soln, {c, \(-3\), 3, 1}]], {t, \(-3\), 3}, \n \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ PlotRange -> {\(-4\), 4}, AspectRatio -> 1]; \)\)\)], "Input"], Cell[BoxData[ \(\(Show[plot1, plot2, PlotRange -> {\(-4\), 4}, Frame -> False, Ticks -> None]; \)\)], "Input"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 16.", "Subsection"], Cell[CellGroupData[{ Cell["a. (3 points)", "Subsubsection"], Cell[BoxData[ \(pop[t_, a_, b_, c_]\ = \ Expand[First[ p[t] /. DSolve[\n \t\t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ { \(p'\)[t]\ == \ a\ p[t] + b, \ p[0]\ == \ c}, \n \t\t\t\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ p[t], t]]] \)], "Input"], Cell["\<\ The parameter c is simply the population of Nevada in 1950--i.e. \ the initial value. The parameter a determines the contribution to population growth coming from births - deaths , and b is the contribution to growth coming from (people moving in) - (people moving out).\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["b (3 points)", "Subsubsection"], Cell[BoxData[ \(Solve[{pop[0, a, b, c] == 285.3, pop[10, a, b, c] == 488.7, \n \t\t\ \ \ \ \ \ \ \ \ \ \ \ pop[20, a, b, c] == 800.5}, {a, b, c}] \)], "Input"], Cell["\<\ Only the first in this list of solutions gives real values for a,b, and c. I'll use that solution.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["c. (4 points)", "Subsubsection"], Cell[BoxData[ \(population[t_]\ = \ pop[t, 0.042718, \ 4.1162, 285.3]\)], "Input"], Cell["\<\ Now Icheck the population in 1950 and 1990 predicted by this model:\ \>", "Text"], Cell[BoxData[ \(population[\(-10\)]\)], "Input"], Cell[BoxData[ \(population[30]\)], "Input"], Cell["\<\ These are at least close to the true values. To predict when the \ population reaches three million, I'll guess that it'll be fifteen years after 1990 (since the \ population nearly tripled in the twenty years between 1970 and 1990).\ \>", "Text"], Cell[BoxData[ \(FindRoot[population[t] == 3000, {t, 45}]\)], "Input"], Cell["\<\ So I'm predicting a population of three million by 2009 Given the real data for 1990 and 1950 this model seems to be projecting too much population growth. I'd probably guess that it'll be a few years after 2009 when the population really does reach three million. Based on the formula above, it looks like I'd have to decrease both a and b to get a more accurate population model.\ \>", "Text"], Cell[BoxData[ \(\(Plot[population[t], {t, \(-10\), 90}]; \)\)], "Input"], Cell["\<\ If the model is right, Nevada will get pretty crowded in time!\ \>", "Text"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Problem 17.", "Subsection"], Cell[CellGroupData[{ Cell["a. (2 points)", "Subsubsection"], Cell["Mathematica is really overkill for this part...", "Text"], Cell[BoxData[ \(Solve[\((\[Alpha] - 1)\) x - \ x^3\ == \ 0, x]\)], "Input"], Cell[TextData[ "That is, we get three real roots when \[Alpha]>1, and only one real\nroot \ when \[Alpha]<1. When \[Alpha]=1, we also get one real root, but with\n\ multiplicity three."], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["b. (2 points)", "Subsubsection"], Cell[BoxData[ \(\(Table[ PlotVectorField[\n \t\t{1, \((\[Alpha] - 1)\) x - \ x^3}, {t, \(-2\), 2}, {x, \(-2\), 2}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1&)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 15], {\[Alpha], \(-2\), 0}]; \)\)], "Input"], Cell["\<\ The picture is more or less the same in all three cases: x=0 is a stable equilibrium solution.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["c. (2 points)", "Subsubsection"], Cell[BoxData[ \(PlotVectorField[\n\t\t{1, \(-x^3\)}, {t, \(-2\), 2}, {x, \(-2\), 2}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1&)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 15]\)], "Input"], Cell["x=0 is still a (barely) stable equilibrium.", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["d. (3 points)", "Subsubsection"], Cell[BoxData[ \(PlotVectorField[\n \t\t{1, .5 x\ - \ x^3}, {t, \(-2\), 2}, {x, \(-2\), 2}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1&)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 17]\)], "Input"], Cell[BoxData[{ \(TextForm\`Now\ x = 0\ is\ an\ unstable\ equilibrium, \ but\ there\ are\ two\ stable\ equilibrium\), \(TextForm\`solutions\ at\ x\ = \ \(\[PlusMinus]\@ .5\)\)}], "Text"], Cell[BoxData[ \(PlotVectorField[\n\t\t{1, x\ - \ x^3}, {t, \(-2\), 2}, {x, \(-2\), 2}, Axes -> \ True, \n\tScaleFunction \[Rule] \((1&)\), Ticks \[Rule] None, Frame \[Rule] True, AspectRatio \[Rule] 1, \ PlotPoints\ -> \ 17]\)], "Input"], Cell["\<\ This is similar to the last case, except that the stable equilibria have migrated to x=\[PlusMinus]1.\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["e. ( 3 points)", "Subsubsection"], Cell[TextData[ "When \[Alpha]<0, we had a single stable equilibrium solution at x=0.\nWhen \ \[Alpha] increased past 0, we suddenly had two stable equilibrium\nthat \ started just above and below x=0 and moved further from\nx=0 as \[Alpha] \ increased."], "Text"] }, Open ]] }, Open ]] }, Open ]] }, FrontEndVersion->"X 3.0", ScreenRectangle->{{0, 1280}, {0, 1024}}, WindowSize->{632, 833}, WindowMargins->{{164, Automatic}, {Automatic, 48}} ] (*********************************************************************** Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. 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