{"id":1621,"date":"2021-07-22T17:08:50","date_gmt":"2021-07-22T17:08:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1621"},"modified":"2022-03-21T22:56:08","modified_gmt":"2022-03-21T22:56:08","slug":"direction-fields","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/direction-fields\/","title":{"raw":"Direction Fields","rendered":"Direction Fields"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Draw the direction field for a given first-order differential equation.<\/li>\r\n \t<li>Use a direction field to draw a solution curve of a first-order differential equation<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1170570991671\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Creating Direction Fields<\/h2>\r\n<p id=\"fs-id1170571047567\">Direction fields (also called slope fields) are useful for investigating first-order differential equations. In particular, we consider a first-order differential equation of the form<\/p>\r\n\r\n<div id=\"fs-id1170573436389\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} =f\\left(x,y\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573368010\">An applied example of this type of differential equation appears in Newton\u2019s law of cooling, which we will solve explicitly later in this chapter. First, though, let us create a direction field for the differential equation<\/p>\r\n\r\n<div id=\"fs-id1170573364732\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{T}^{\\prime }\\left(t\\right)=-0.4\\left(T - 72\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571277998\">Here [latex]T\\left(t\\right)[\/latex] represents the temperature (in degrees Fahrenheit) of an object at time [latex]t[\/latex], and the ambient temperature is [latex]72^\\circ\\text{F}\\text{.}[\/latex] Figure 1 shows the direction field for this equation.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_02_001\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"484\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233925\/CNX_Calc_Figure_08_02_009.jpg\" alt=\"A graph of a direction field for the given differential equation in quadrants one and two. The arrows are pointing directly to the right at y = 72. Below that line, the arrows have increasingly positive slope as y becomes smaller. Above that line, the arrows have increasingly negative slope as y becomes larger. The arrows point to convergence at y = 72. Two solutions are drawn: one for initial temperature less than 72, and one for initial temperatures larger than 72. The upper solution is a decreasing concave up curve, approaching y = 72 as t goes to infinity. The lower solution is an increasing concave down curve, approaching y = 72 as t goes to infinity.\" width=\"484\" height=\"509\" data-media-type=\"image\/jpeg\" \/> Figure 1. Direction field for the differential equation [latex]{T}^{\\prime }\\left(t\\right)=-0.4\\left(T - 72\\right)[\/latex]. Two solutions are plotted: one with initial temperature less than [latex]72^\\circ\\text{F}[\/latex] and the other with initial temperature greater than [latex]72^\\circ\\text{F}\\text{.}[\/latex][\/caption]<\/figure>\r\n<p id=\"fs-id1170573437738\">The idea behind a direction field is the fact that the derivative of a function evaluated at a given point is the slope of the tangent line to the graph of that function at the same point. Other examples of differential equations for which we can create a direction field include<\/p>\r\n\r\n<div id=\"fs-id1170573581634\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}y^{\\prime} =3x+2y - 4\\hfill \\\\ y^{\\prime} ={x}^{2}-{y}^{2}\\hfill \\\\ y^{\\prime} =\\frac{2x+4}{y - 2}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573428280\">To create a direction field, we start with the first equation: [latex]y^{\\prime} =3x+2y - 4[\/latex]. We let [latex]\\left({x}_{0},{y}_{0}\\right)[\/latex] be any ordered pair, and we substitute these numbers into the right-hand side of the differential equation. For example, if we choose [latex]x=1[\/latex] and [latex]y=2[\/latex], substituting into the right-hand side of the differential equation yields<\/p>\r\n\r\n<div id=\"fs-id1170573439082\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{y}^{\\prime }\\hfill &amp; =3x+2y - 4\\hfill \\\\ &amp; =3\\left(1\\right)+2\\left(2\\right)-4=3.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571291390\">This tells us that if a solution to the differential equation [latex]y^{\\prime} =3x+2y - 4[\/latex] passes through the point [latex]\\left(1,2\\right)[\/latex], then the slope of the solution at that point must equal [latex]3[\/latex]. To start creating the direction field, we put a short line segment at the point [latex]\\left(1,2\\right)[\/latex] having slope [latex]3[\/latex]. We can do this for any point in the domain of the function [latex]f\\left(x,y\\right)=3x+2y - 4[\/latex], which consists of all ordered pairs [latex]\\left(x,y\\right)[\/latex] in [latex]{\\mathbb{R} }^{2}[\/latex]. Therefore any point in the Cartesian plane has a slope associated with it, assuming that a solution to the differential equation passes through that point. The direction field for the differential equation [latex]{y}^{\\prime }=3x+2y - 4[\/latex] is shown in Figure 2.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_02_002\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233929\/CNX_Calc_Figure_08_02_001.jpg\" alt=\"A graph of the direction field for the differential equation y\u2019 = 3 x + 2 y \u2013 4 in all four quadrants. In quadrants two and three, the arrows point down and slightly to the right. On a diagonal line, roughly y = -x + 2, the arrows point further and further to the right, curve, and then point up above that line.\" width=\"487\" height=\"467\" data-media-type=\"image\/jpeg\" \/> Figure 2. Direction field for the differential equation [latex]y^{\\prime} =3x+2y - 4[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1170573425994\">We can generate a direction field of this type for any differential equation of the form [latex]y^{\\prime} =f\\left(x,y\\right)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1170570998490\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170573415179\">A <strong>direction field (slope field)<\/strong> is a mathematical object used to graphically represent solutions to a first-order differential equation. At each point in a direction field, a line segment appears whose slope is equal to the slope of a solution to the differential equation passing through that point.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1170571350904\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Using Direction Fields<\/h2>\r\n<p id=\"fs-id1170571078812\">We can use a direction field to predict the behavior of solutions to a differential equation without knowing the actual solution. For example, the direction field in Figure 2 serves as a guide to the behavior of solutions to the differential equation [latex]y^{\\prime} =3x+2y - 4[\/latex]. Before developing this method, it is worthwhile to revisit the notion of a linear approximation.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Linear Approximation OF a Function<\/h3>\r\nThe <strong>linear approximation<\/strong> of a function at a given point, also known as the <strong>linearization<\/strong>, is equivalent to the equation of the tangent line to the graph of the function at that point.\r\n\r\nSince the slope of a function [latex] f(x) [\/latex] at [latex] \\left(x_0,y_0 \\right) [\/latex] is given by\u00a0[latex] f'(x_0) [\/latex], the linear approximation of a function [latex] f(x) [\/latex] at a point [latex] \\left(x_0,y_0 \\right) [\/latex] is:\r\n[latex] L(x) = y_0 + f'(x_0) (x - x_0)[\/latex]\r\n\r\nFor values near the point [latex] \\left(x_0,y_0 \\right) [\/latex], [latex] L(x) \\approx f(x) [\/latex].\u00a0 In other words, [latex] L(x) [\/latex] can be used to predict function values near [latex] \\left(x_0,y_0 \\right) [\/latex].\r\n\r\nPut another way, the actual change in the function output, [latex] \\Delta y [\/latex], can be approximated using the slope at [latex] \\left(x_0,y_0 \\right) [\/latex].\r\n\r\n[latex] \\Delta y \\approx f'(x_0) \\Delta x [\/latex]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170573359408\">To use a direction field, we start by choosing any point in the field. The line segment at that point serves as a signpost telling us what direction to go from there. For example, if a solution to the differential equation passes through the point [latex]\\left(0,1\\right)[\/latex], then the slope of the solution passing through that point is given by [latex]y^{\\prime} =3\\left(0\\right)+2\\left(1\\right)-4=-2[\/latex]. Now let [latex]x[\/latex] increase slightly, say to [latex]x=0.1[\/latex]. Using the method of linear approximations gives a formula for the approximate value of [latex]y[\/latex] for [latex]x=0.1[\/latex]. In particular,<\/p>\r\n\r\n<div id=\"fs-id1170573574070\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}L\\left(x\\right)\\hfill &amp; ={y}_{0}+{f}^{\\prime }\\left({x}_{0}\\right)\\left(x-{x}_{0}\\right)\\hfill \\\\ &amp; =1 - 2\\left({x}_{0}-0\\right)\\hfill \\\\ &amp; =1 - 2{x}_{0}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571509939\">Substituting [latex]{x}_{0}=0.1[\/latex] into [latex]L\\left(x\\right)[\/latex] gives an approximate [latex]y[\/latex] value of [latex]0.8[\/latex].<\/p>\r\n<p id=\"fs-id1170571068254\">At this point the slope of the solution changes (again according to the differential equation). We can keep progressing, recalculating the slope of the solution as we take small steps to the right, and watching the behavior of the solution. Figure 3 shows a graph of the solution passing through the point [latex]\\left(0,1\\right)[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_02_003\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"459\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233932\/CNX_Calc_Figure_08_02_002.jpg\" alt=\"A graph of the direction field for the differential equation y\u2019 = 3 x + 2 y \u2013 4 in all four quadrants. In quadrants two and three, the arrows point down and slightly to the right. On a diagonal line, roughly y = -x + 2, the arrows point further and further to the right, curve, and then point up above that line. The solution passing through the point (0, 1) is shown. It curves down through (-5, 10), (0, 2), (1, 0), and (3, -10).\" width=\"459\" height=\"464\" data-media-type=\"image\/jpeg\" \/> Figure 3. Direction field for the differential equation [latex]y^{\\prime} =3x+2y - 4[\/latex] with the solution passing through the point [latex]\\left(0,1\\right)[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1170573427844\">The curve is the graph of the solution to the initial-value problem<\/p>\r\n\r\n<div id=\"fs-id1170571153236\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} =3x+2y - 4,y\\left(0\\right)=1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571143395\">This curve is called a <span data-type=\"term\">solution curve<\/span> passing through the point [latex]\\left(0,1\\right)[\/latex]. The exact solution to this initial-value problem is<\/p>\r\n\r\n<div id=\"fs-id1170573415442\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=-\\frac{3}{2}x+\\frac{5}{4}-\\frac{1}{4}{e}^{2x}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573318766\">and the graph of this solution is identical to the curve in Figure 3.<\/p>\r\n\r\n<div id=\"fs-id1170571099208\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170573262990\" data-type=\"exercise\">\r\n<div id=\"fs-id1170571119933\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170571119933\" data-type=\"problem\">\r\n<p id=\"fs-id1170573295181\">Create a direction field for the differential equation [latex]y^{\\prime} ={x}^{2}-{y}^{2}[\/latex] and sketch a solution curve passing through the point [latex]\\left(-1,2\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558898\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1170573593135\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170573262665\">Use [latex]x[\/latex] and [latex]y[\/latex] values ranging from [latex]-5[\/latex] to [latex]5[\/latex]. For each coordinate pair, calculate [latex]y^{\\prime} [\/latex] using the right-hand side of the differential equation.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1170573299269\" data-type=\"solution\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"459\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233936\/CNX_Calc_Figure_08_02_003.jpg\" alt=\"A graph of the direction field for the differential equation y\u2019 = x ^ 2 \u2013 y ^ 2. Along y = x and y = -x, the lines are horizontal. On either side of y = x and y = -x, the lines slant and direct solutions along those two functions. The rest of the lines are vertical. The solution going through (-1, 2) is shown. It curves down from about (-2.75, 10), through (-1, 2) and about (0, 1.5), and then up along the diagonal to (10, 10).\" width=\"459\" height=\"464\" data-media-type=\"image\/jpeg\" \/> Figure 4.[\/caption]\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following videos to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6722736&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=NmwwRLHZtkw&amp;video_target=tpm-plugin-zrpsnm0j-NmwwRLHZtkw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.2.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for \"4.2.1\" here (opens in new window)<\/a>.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6722737&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=8HNOR0-GDe8&amp;video_target=tpm-plugin-0tskvtbs-8HNOR0-GDe8\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.2.2_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for \"4.2.2\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1170571224955\" class=\"media-2\" data-type=\"note\">\r\n<div class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<p id=\"fs-id1170573414525\">Visit\u00a0<a href=\"https:\/\/www.mathopenref.com\/calcslopefields.html\" target=\"_blank\" rel=\"noopener\">this Java applet for more practice with slope fields<\/a>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170570993737\">Now consider the direction field for the differential equation [latex]y^{\\prime} =\\left(x - 3\\right)\\left({y}^{2}-4\\right)[\/latex], shown in Figure 5. This direction field has several interesting properties. First of all, at [latex]y=-2[\/latex] and [latex]y=2[\/latex], horizontal dashes appear all the way across the graph. This means that if [latex]y=-2[\/latex], then [latex]y^{\\prime} =0[\/latex]. Substituting this expression into the right-hand side of the differential equation gives<\/p>\r\n\r\n<div id=\"fs-id1170570995991\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc} \\left(x-3\\right)\\left(y^{2}-4\\right)\\hfill &amp; =\\left(x-3\\right)\\left(\\left({-2}\\right)^{2}-4\\right)\\hfill \\\\ &amp; =\\left(x-3\\right)\\left(0\\right)\\hfill \\\\ &amp; =0\\hfill \\\\ &amp; =y^{\\prime} .\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571260861\">Therefore [latex]y=-2[\/latex] is a solution to the differential equation. Similarly, [latex]y=2[\/latex] is a solution to the differential equation. These are the only constant-valued solutions to the differential equation, as we can see from the following argument. Suppose [latex]y=k[\/latex] is a constant solution to the differential equation. Then [latex]{y}^{\\prime }=0[\/latex]. Substituting this expression into the differential equation yields [latex]0=\\left(x - 3\\right)\\left({k}^{2}-4\\right)[\/latex]. This equation must be true for all values of [latex]x[\/latex], so the second factor must equal zero. This result yields the equation [latex]{k}^{2}-4=0[\/latex]. The solutions to this equation are [latex]k=-2[\/latex] and [latex]k=2[\/latex], which are the constant solutions already mentioned. These are called the equilibrium solutions to the differential equation.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_02_005\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233940\/CNX_Calc_Figure_08_02_004.jpg\" alt=\"A direction field for the given differential equation. The arrows are horizontal and pointing to the right at y = -4, y = 4, and x = 6. The closer the arrows are to x = 6, the more horizontal the arrows become. The further away, the more vertical they are. The arrows point down for y &gt; 4 and x &lt; 4, -4 &lt; y &lt; 4 and x &gt; 6, and y &lt; -4 and x &lt; 6. In all other areas, the arrows are pointing up.\" width=\"487\" height=\"445\" data-media-type=\"image\/jpeg\" \/> Figure 5. Direction field for the differential equation [latex]y^{\\prime} =\\left(x - 3\\right)\\left({y}^{2}-4\\right)[\/latex] showing two solutions. These solutions are very close together, but one is barely above the equilibrium solution [latex]x=-2[\/latex] and the other is barely below the same equilibrium solution.[\/caption]<\/figure>\r\n<div id=\"fs-id1170573577673\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170573571376\">Consider the differential equation [latex]y^{\\prime} =f\\left(x,y\\right)[\/latex]. An <strong>equilibrium solution<\/strong> is any solution to the differential equation of the form [latex]y=c[\/latex], where [latex]c[\/latex] is a constant.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170570991022\">To determine the equilibrium solutions to the differential equation [latex]y^{\\prime} =f\\left(x,y\\right)[\/latex], set the right-hand side equal to zero. An equilibrium solution of the differential equation is any function of the form [latex]y=k[\/latex] such that [latex]f\\left(x,k\\right)=0[\/latex] for all values of [latex]x[\/latex] in the domain of [latex]f[\/latex].<\/p>\r\n<p id=\"fs-id1170573425221\">An important characteristic of equilibrium solutions concerns whether or not they approach the line [latex]y=k[\/latex] as an asymptote for large values of [latex]x[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1170571122528\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170571284202\">Consider the differential equation [latex]{y}^{\\prime }=f\\left(x,y\\right)[\/latex], and assume that all solutions to this differential equation are defined for [latex]x\\ge {x}_{0}[\/latex]. Let [latex]y=k[\/latex] be an equilibrium solution to the differential equation.<\/p>\r\n\r\n<ol id=\"fs-id1170570989710\" type=\"1\">\r\n \t<li>[latex]y=k[\/latex] is an<strong> asymptotically stable solution<\/strong> to the differential equation if there exists [latex]\\epsilon &gt;0[\/latex] such that for any value [latex]c\\in \\left(k-\\epsilon ,k+\\epsilon \\right)[\/latex] the solution to the initial-value problem<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571303029\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }=f\\left(x,y\\right),y\\left({x}_{0}\\right)=c[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\napproaches [latex]k[\/latex] as [latex]x[\/latex] approaches infinity.<\/li>\r\n \t<li>[latex]y=k[\/latex] is an <strong>asymptotically unstable solution<\/strong> to the differential equation if there exists [latex]\\epsilon &gt;0[\/latex] such that for any value [latex]c\\in \\left(k-\\epsilon ,k+\\epsilon \\right)[\/latex] the solution to the initial-value problem<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573366618\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }=f\\left(x,y\\right),y\\left({x}_{0}\\right)=c[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nnever approaches [latex]k[\/latex] as [latex]x[\/latex] approaches infinity.<\/li>\r\n \t<li>[latex]y=k[\/latex] is an <strong>asymptotically semi-stable solution<\/strong> to the differential equation if it is neither asymptotically stable nor asymptotically unstable.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573750302\">Now we return to the differential equation [latex]y^{\\prime} =\\left(x - 3\\right)\\left({y}^{2}-4\\right)[\/latex], with the initial condition [latex]y\\left(0\\right)=0.5[\/latex]. The direction field for this initial-value problem, along with the corresponding solution, is shown in Figure 6.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_02_006\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233944\/CNX_Calc_Figure_08_02_005.jpg\" alt=\"A direction field for the given differential equation. The arrows are horizontal and pointing to the right at y = -4, y = 4, and x = 6. The closer the arrows are to x = 6, the more horizontal the arrows become. The further away, the more vertical they are. The arrows point down for y &gt; 4 and x &lt; 4, -4 &lt; y &lt; 4 and x &gt; 6, and y &lt; -4 and x &lt; 6. In all other areas, the arrows are pointing up. A solution is graphed that goes through (0, 0.5). It begins along y = -4 in quadrant three, increases from -4 to 4 between x = -1 and 1, and ends going along y = 4 in quadrant 1.\" width=\"487\" height=\"445\" data-media-type=\"image\/jpeg\" \/> Figure 6. Direction field for the initial-value problem [latex]y^{\\prime} =\\left(x - 3\\right)\\left({y}^{2}-4\\right),y\\left(0\\right)=0.5[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1170573230984\">The values of the solution to this initial-value problem stay between [latex]y=-2[\/latex] and [latex]y=2[\/latex], which are the equilibrium solutions to the differential equation. Furthermore, as [latex]x[\/latex] approaches infinity, [latex]y[\/latex] approaches [latex]2[\/latex]. The behavior of solutions is similar if the initial value is higher than [latex]2[\/latex], for example, [latex]y\\left(0\\right)=2.3[\/latex]. In this case, the solutions decrease and approach [latex]y=2[\/latex] as [latex]x[\/latex] approaches infinity. Therefore [latex]y=2[\/latex] is an asymptotically stable solution to the differential equation.<\/p>\r\n<p id=\"fs-id1170573230988\">What happens when the initial value is below [latex]y=-2?[\/latex] This scenario is illustrated in Figure 7, with the initial value [latex]y\\left(0\\right)=-3[\/latex].<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_02_007\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233948\/CNX_Calc_Figure_08_02_006.jpg\" alt=\"A direction field for the given differential equation. The arrows are horizontal and pointing to the right at y = -4, y = 4, and x = 6. The closer the arrows are to x = 6, the more horizontal the arrows become. The further away, the more vertical they are. The arrows point down for y &gt; 4 and x &lt; 4, -4 &lt; y &lt; 4 and x &gt; 6, and y &lt; -4 and x &lt; 6. In all other areas, the arrows are pointing up. A solution is graphed that goes along y = -4 in quadrant 3 and curves between x = -1 and x = 0 to go to negative infinity along the y axis.\" width=\"487\" height=\"445\" data-media-type=\"image\/jpeg\" \/> Figure 7. Direction field for the initial-value problem [latex]y^{\\prime} =\\left(x - 3\\right)\\left({y}^{2}-4\\right),y\\left(0\\right)=-3[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1170573289098\">The solution decreases rapidly toward negative infinity as [latex]x[\/latex] approaches infinity. Furthermore, if the initial value is slightly higher than [latex]-2[\/latex], then the solution approaches [latex]2[\/latex], which is the other equilibrium solution. Therefore in neither case does the solution approach [latex]y=-2[\/latex], so [latex]y=-2[\/latex] is called an asymptotically unstable, or unstable, equilibrium solution.<\/p>\r\n\r\n<div id=\"fs-id1170573709023\" data-type=\"example\">\r\n<div id=\"fs-id1170573569693\" data-type=\"exercise\">\r\n<div id=\"fs-id1170571120492\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Stability of an Equilibrium Solution<\/h3>\r\n<div id=\"fs-id1170571120492\" data-type=\"problem\">\r\n<p id=\"fs-id1170573333748\">Create a direction field for the differential equation [latex]y^{\\prime} ={\\left(y - 3\\right)}^{2}\\left({y}^{2}+y - 2\\right)[\/latex] and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558897\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1170571038103\" data-type=\"solution\">\r\n<p id=\"fs-id1170571093547\">The direction field is shown in Figure 8.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_02_008\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233951\/CNX_Calc_Figure_08_02_007.jpg\" alt=\"A graph of a direction field with arrows pointing to the right at y = -4, y = 2, and y = 6. For y &lt; -4, the arrows point up. For -4 &lt; y &lt; 2, the arrows point down. For 2 &lt; y &lt;6, the arrows point up, becoming flatter and flatter as they approach y = 6. For y &gt; 6, the arrows point up and become more and more vertical the further they get from y = 6.\" width=\"487\" height=\"445\" data-media-type=\"image\/jpeg\" \/> Figure 8. Direction field for the differential equation [latex]y^{\\prime} ={\\left(y - 3\\right)}^{2}\\left({y}^{2}+y - 2\\right)[\/latex].[\/caption]<\/figure>\r\n<p id=\"fs-id1170573408988\">The equilibrium solutions are [latex]y=-2,y=1[\/latex], and [latex]y=3[\/latex]. To classify each of the solutions, look at an arrow directly above or below each of these values. For example, at [latex]y=-2[\/latex] the arrows directly below this solution point up, and the arrows directly above the solution point down. Therefore all initial conditions close to [latex]y=-2[\/latex] approach [latex]y=-2[\/latex], and the solution is stable. For the solution [latex]y=1[\/latex], all initial conditions above and below [latex]y=1[\/latex] are repelled (pushed away) from [latex]y=1[\/latex], so this solution is unstable. The solution [latex]y=3[\/latex] is semi-stable, because for initial conditions slightly greater than [latex]3[\/latex], the solution approaches infinity, and for initial conditions slightly less than [latex]3[\/latex], the solution approaches [latex]y=3[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573271901\" data-type=\"commentary\">\r\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\r\n<p id=\"fs-id1170571244786\">It is possible to find the equilibrium solutions to the differential equation by setting the right-hand side equal to zero and solving for [latex]y[\/latex]. This approach gives the same equilibrium solutions as those we saw in the direction field.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170571340415\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170571349921\" data-type=\"exercise\">\r\n<div id=\"fs-id1170571290427\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170571290427\" data-type=\"problem\">\r\n<p id=\"fs-id1170573519611\">Create a direction field for the differential equation [latex]y^{\\prime} =\\left(x+5\\right)\\left(y+2\\right)\\left({y}^{2}-4y+4\\right)[\/latex] and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558895\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1170571027317\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170571332593\">First create the direction field and look for horizontal dashes that go all the way across. Then examine the slope lines directly above and below the equilibrium solutions.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1170573255134\" data-type=\"solution\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233955\/CNX_Calc_Figure_08_02_008.jpg\" alt=\"A direction field with arrows pointing to the right at y = -4 and y = 4. The arrows point up for y &gt; -4 and down for y &lt; -4. Close to y = 4, the arrows are more horizontal, but the further away, the more vertical they become.\" width=\"487\" height=\"445\" data-media-type=\"image\/jpeg\" \/> Figure 9.[\/caption]\r\n\r\nThe equilibrium solutions are [latex]y=-2[\/latex] and [latex]y=2[\/latex]. For this equation, [latex]y=-2[\/latex] is an unstable equilibrium solution, and [latex]y=2[\/latex] is a semi-stable equilibrium solution.\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1170571260237\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Draw the direction field for a given first-order differential equation.<\/li>\n<li>Use a direction field to draw a solution curve of a first-order differential equation<\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1170570991671\" data-depth=\"1\">\n<h2 data-type=\"title\">Creating Direction Fields<\/h2>\n<p id=\"fs-id1170571047567\">Direction fields (also called slope fields) are useful for investigating first-order differential equations. In particular, we consider a first-order differential equation of the form<\/p>\n<div id=\"fs-id1170573436389\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} =f\\left(x,y\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573368010\">An applied example of this type of differential equation appears in Newton\u2019s law of cooling, which we will solve explicitly later in this chapter. First, though, let us create a direction field for the differential equation<\/p>\n<div id=\"fs-id1170573364732\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{T}^{\\prime }\\left(t\\right)=-0.4\\left(T - 72\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571277998\">Here [latex]T\\left(t\\right)[\/latex] represents the temperature (in degrees Fahrenheit) of an object at time [latex]t[\/latex], and the ambient temperature is [latex]72^\\circ\\text{F}\\text{.}[\/latex] Figure 1 shows the direction field for this equation.<\/p>\n<figure id=\"CNX_Calc_Figure_08_02_001\"><figcaption><\/figcaption><div style=\"width: 494px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233925\/CNX_Calc_Figure_08_02_009.jpg\" alt=\"A graph of a direction field for the given differential equation in quadrants one and two. The arrows are pointing directly to the right at y = 72. Below that line, the arrows have increasingly positive slope as y becomes smaller. Above that line, the arrows have increasingly negative slope as y becomes larger. The arrows point to convergence at y = 72. Two solutions are drawn: one for initial temperature less than 72, and one for initial temperatures larger than 72. The upper solution is a decreasing concave up curve, approaching y = 72 as t goes to infinity. The lower solution is an increasing concave down curve, approaching y = 72 as t goes to infinity.\" width=\"484\" height=\"509\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. Direction field for the differential equation [latex]{T}^{\\prime }\\left(t\\right)=-0.4\\left(T - 72\\right)[\/latex]. Two solutions are plotted: one with initial temperature less than [latex]72^\\circ\\text{F}[\/latex] and the other with initial temperature greater than [latex]72^\\circ\\text{F}\\text{.}[\/latex]<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1170573437738\">The idea behind a direction field is the fact that the derivative of a function evaluated at a given point is the slope of the tangent line to the graph of that function at the same point. Other examples of differential equations for which we can create a direction field include<\/p>\n<div id=\"fs-id1170573581634\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{c}y^{\\prime} =3x+2y - 4\\hfill \\\\ y^{\\prime} ={x}^{2}-{y}^{2}\\hfill \\\\ y^{\\prime} =\\frac{2x+4}{y - 2}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573428280\">To create a direction field, we start with the first equation: [latex]y^{\\prime} =3x+2y - 4[\/latex]. We let [latex]\\left({x}_{0},{y}_{0}\\right)[\/latex] be any ordered pair, and we substitute these numbers into the right-hand side of the differential equation. For example, if we choose [latex]x=1[\/latex] and [latex]y=2[\/latex], substituting into the right-hand side of the differential equation yields<\/p>\n<div id=\"fs-id1170573439082\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{y}^{\\prime }\\hfill & =3x+2y - 4\\hfill \\\\ & =3\\left(1\\right)+2\\left(2\\right)-4=3.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571291390\">This tells us that if a solution to the differential equation [latex]y^{\\prime} =3x+2y - 4[\/latex] passes through the point [latex]\\left(1,2\\right)[\/latex], then the slope of the solution at that point must equal [latex]3[\/latex]. To start creating the direction field, we put a short line segment at the point [latex]\\left(1,2\\right)[\/latex] having slope [latex]3[\/latex]. We can do this for any point in the domain of the function [latex]f\\left(x,y\\right)=3x+2y - 4[\/latex], which consists of all ordered pairs [latex]\\left(x,y\\right)[\/latex] in [latex]{\\mathbb{R} }^{2}[\/latex]. Therefore any point in the Cartesian plane has a slope associated with it, assuming that a solution to the differential equation passes through that point. The direction field for the differential equation [latex]{y}^{\\prime }=3x+2y - 4[\/latex] is shown in Figure 2.<\/p>\n<figure id=\"CNX_Calc_Figure_08_02_002\"><figcaption><\/figcaption><div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233929\/CNX_Calc_Figure_08_02_001.jpg\" alt=\"A graph of the direction field for the differential equation y\u2019 = 3 x + 2 y \u2013 4 in all four quadrants. In quadrants two and three, the arrows point down and slightly to the right. On a diagonal line, roughly y = -x + 2, the arrows point further and further to the right, curve, and then point up above that line.\" width=\"487\" height=\"467\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. Direction field for the differential equation [latex]y^{\\prime} =3x+2y - 4[\/latex].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1170573425994\">We can generate a direction field of this type for any differential equation of the form [latex]y^{\\prime} =f\\left(x,y\\right)[\/latex].<\/p>\n<div id=\"fs-id1170570998490\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170573415179\">A <strong>direction field (slope field)<\/strong> is a mathematical object used to graphically represent solutions to a first-order differential equation. At each point in a direction field, a line segment appears whose slope is equal to the slope of a solution to the differential equation passing through that point.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1170571350904\" data-depth=\"1\">\n<h2 data-type=\"title\">Using Direction Fields<\/h2>\n<p id=\"fs-id1170571078812\">We can use a direction field to predict the behavior of solutions to a differential equation without knowing the actual solution. For example, the direction field in Figure 2 serves as a guide to the behavior of solutions to the differential equation [latex]y^{\\prime} =3x+2y - 4[\/latex]. Before developing this method, it is worthwhile to revisit the notion of a linear approximation.<\/p>\n<div class=\"textbox examples\">\n<h3>Recall: Linear Approximation OF a Function<\/h3>\n<p>The <strong>linear approximation<\/strong> of a function at a given point, also known as the <strong>linearization<\/strong>, is equivalent to the equation of the tangent line to the graph of the function at that point.<\/p>\n<p>Since the slope of a function [latex]f(x)[\/latex] at [latex]\\left(x_0,y_0 \\right)[\/latex] is given by\u00a0[latex]f'(x_0)[\/latex], the linear approximation of a function [latex]f(x)[\/latex] at a point [latex]\\left(x_0,y_0 \\right)[\/latex] is:<br \/>\n[latex]L(x) = y_0 + f'(x_0) (x - x_0)[\/latex]<\/p>\n<p>For values near the point [latex]\\left(x_0,y_0 \\right)[\/latex], [latex]L(x) \\approx f(x)[\/latex].\u00a0 In other words, [latex]L(x)[\/latex] can be used to predict function values near [latex]\\left(x_0,y_0 \\right)[\/latex].<\/p>\n<p>Put another way, the actual change in the function output, [latex]\\Delta y[\/latex], can be approximated using the slope at [latex]\\left(x_0,y_0 \\right)[\/latex].<\/p>\n<p>[latex]\\Delta y \\approx f'(x_0) \\Delta x[\/latex]<\/p>\n<\/div>\n<p id=\"fs-id1170573359408\">To use a direction field, we start by choosing any point in the field. The line segment at that point serves as a signpost telling us what direction to go from there. For example, if a solution to the differential equation passes through the point [latex]\\left(0,1\\right)[\/latex], then the slope of the solution passing through that point is given by [latex]y^{\\prime} =3\\left(0\\right)+2\\left(1\\right)-4=-2[\/latex]. Now let [latex]x[\/latex] increase slightly, say to [latex]x=0.1[\/latex]. Using the method of linear approximations gives a formula for the approximate value of [latex]y[\/latex] for [latex]x=0.1[\/latex]. In particular,<\/p>\n<div id=\"fs-id1170573574070\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}L\\left(x\\right)\\hfill & ={y}_{0}+{f}^{\\prime }\\left({x}_{0}\\right)\\left(x-{x}_{0}\\right)\\hfill \\\\ & =1 - 2\\left({x}_{0}-0\\right)\\hfill \\\\ & =1 - 2{x}_{0}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571509939\">Substituting [latex]{x}_{0}=0.1[\/latex] into [latex]L\\left(x\\right)[\/latex] gives an approximate [latex]y[\/latex] value of [latex]0.8[\/latex].<\/p>\n<p id=\"fs-id1170571068254\">At this point the slope of the solution changes (again according to the differential equation). We can keep progressing, recalculating the slope of the solution as we take small steps to the right, and watching the behavior of the solution. Figure 3 shows a graph of the solution passing through the point [latex]\\left(0,1\\right)[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_08_02_003\"><figcaption><\/figcaption><div style=\"width: 469px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233932\/CNX_Calc_Figure_08_02_002.jpg\" alt=\"A graph of the direction field for the differential equation y\u2019 = 3 x + 2 y \u2013 4 in all four quadrants. In quadrants two and three, the arrows point down and slightly to the right. On a diagonal line, roughly y = -x + 2, the arrows point further and further to the right, curve, and then point up above that line. The solution passing through the point (0, 1) is shown. It curves down through (-5, 10), (0, 2), (1, 0), and (3, -10).\" width=\"459\" height=\"464\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Direction field for the differential equation [latex]y^{\\prime} =3x+2y - 4[\/latex] with the solution passing through the point [latex]\\left(0,1\\right)[\/latex].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1170573427844\">The curve is the graph of the solution to the initial-value problem<\/p>\n<div id=\"fs-id1170571153236\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} =3x+2y - 4,y\\left(0\\right)=1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571143395\">This curve is called a <span data-type=\"term\">solution curve<\/span> passing through the point [latex]\\left(0,1\\right)[\/latex]. The exact solution to this initial-value problem is<\/p>\n<div id=\"fs-id1170573415442\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=-\\frac{3}{2}x+\\frac{5}{4}-\\frac{1}{4}{e}^{2x}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573318766\">and the graph of this solution is identical to the curve in Figure 3.<\/p>\n<div id=\"fs-id1170571099208\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170573262990\" data-type=\"exercise\">\n<div id=\"fs-id1170571119933\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170571119933\" data-type=\"problem\">\n<p id=\"fs-id1170573295181\">Create a direction field for the differential equation [latex]y^{\\prime} ={x}^{2}-{y}^{2}[\/latex] and sketch a solution curve passing through the point [latex]\\left(-1,2\\right)[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Hint<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170573593135\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170573262665\">Use [latex]x[\/latex] and [latex]y[\/latex] values ranging from [latex]-5[\/latex] to [latex]5[\/latex]. For each coordinate pair, calculate [latex]y^{\\prime}[\/latex] using the right-hand side of the differential equation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170573299269\" data-type=\"solution\">\n<div style=\"width: 469px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233936\/CNX_Calc_Figure_08_02_003.jpg\" alt=\"A graph of the direction field for the differential equation y\u2019 = x ^ 2 \u2013 y ^ 2. Along y = x and y = -x, the lines are horizontal. On either side of y = x and y = -x, the lines slant and direct solutions along those two functions. The rest of the lines are vertical. The solution going through (-1, 2) is shown. It curves down from about (-2.75, 10), through (-1, 2) and about (0, 1.5), and then up along the diagonal to (10, 10).\" width=\"459\" height=\"464\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following videos to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6722736&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=NmwwRLHZtkw&amp;video_target=tpm-plugin-zrpsnm0j-NmwwRLHZtkw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.2.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;4.2.1&#8221; here (opens in new window)<\/a>.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6722737&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=8HNOR0-GDe8&amp;video_target=tpm-plugin-0tskvtbs-8HNOR0-GDe8\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.2.2_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;4.2.2&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1170571224955\" class=\"media-2\" data-type=\"note\">\n<div class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p id=\"fs-id1170573414525\">Visit\u00a0<a href=\"https:\/\/www.mathopenref.com\/calcslopefields.html\" target=\"_blank\" rel=\"noopener\">this Java applet for more practice with slope fields<\/a>.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1170570993737\">Now consider the direction field for the differential equation [latex]y^{\\prime} =\\left(x - 3\\right)\\left({y}^{2}-4\\right)[\/latex], shown in Figure 5. This direction field has several interesting properties. First of all, at [latex]y=-2[\/latex] and [latex]y=2[\/latex], horizontal dashes appear all the way across the graph. This means that if [latex]y=-2[\/latex], then [latex]y^{\\prime} =0[\/latex]. Substituting this expression into the right-hand side of the differential equation gives<\/p>\n<div id=\"fs-id1170570995991\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc} \\left(x-3\\right)\\left(y^{2}-4\\right)\\hfill & =\\left(x-3\\right)\\left(\\left({-2}\\right)^{2}-4\\right)\\hfill \\\\ & =\\left(x-3\\right)\\left(0\\right)\\hfill \\\\ & =0\\hfill \\\\ & =y^{\\prime} .\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571260861\">Therefore [latex]y=-2[\/latex] is a solution to the differential equation. Similarly, [latex]y=2[\/latex] is a solution to the differential equation. These are the only constant-valued solutions to the differential equation, as we can see from the following argument. Suppose [latex]y=k[\/latex] is a constant solution to the differential equation. Then [latex]{y}^{\\prime }=0[\/latex]. Substituting this expression into the differential equation yields [latex]0=\\left(x - 3\\right)\\left({k}^{2}-4\\right)[\/latex]. This equation must be true for all values of [latex]x[\/latex], so the second factor must equal zero. This result yields the equation [latex]{k}^{2}-4=0[\/latex]. The solutions to this equation are [latex]k=-2[\/latex] and [latex]k=2[\/latex], which are the constant solutions already mentioned. These are called the equilibrium solutions to the differential equation.<\/p>\n<figure id=\"CNX_Calc_Figure_08_02_005\"><figcaption><\/figcaption><div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233940\/CNX_Calc_Figure_08_02_004.jpg\" alt=\"A direction field for the given differential equation. The arrows are horizontal and pointing to the right at y = -4, y = 4, and x = 6. The closer the arrows are to x = 6, the more horizontal the arrows become. The further away, the more vertical they are. The arrows point down for y &gt; 4 and x &lt; 4, -4 &lt; y &lt; 4 and x &gt; 6, and y &lt; -4 and x &lt; 6. In all other areas, the arrows are pointing up.\" width=\"487\" height=\"445\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 5. Direction field for the differential equation [latex]y^{\\prime} =\\left(x - 3\\right)\\left({y}^{2}-4\\right)[\/latex] showing two solutions. These solutions are very close together, but one is barely above the equilibrium solution [latex]x=-2[\/latex] and the other is barely below the same equilibrium solution.<\/p>\n<\/div>\n<\/figure>\n<div id=\"fs-id1170573577673\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170573571376\">Consider the differential equation [latex]y^{\\prime} =f\\left(x,y\\right)[\/latex]. An <strong>equilibrium solution<\/strong> is any solution to the differential equation of the form [latex]y=c[\/latex], where [latex]c[\/latex] is a constant.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170570991022\">To determine the equilibrium solutions to the differential equation [latex]y^{\\prime} =f\\left(x,y\\right)[\/latex], set the right-hand side equal to zero. An equilibrium solution of the differential equation is any function of the form [latex]y=k[\/latex] such that [latex]f\\left(x,k\\right)=0[\/latex] for all values of [latex]x[\/latex] in the domain of [latex]f[\/latex].<\/p>\n<p id=\"fs-id1170573425221\">An important characteristic of equilibrium solutions concerns whether or not they approach the line [latex]y=k[\/latex] as an asymptote for large values of [latex]x[\/latex].<\/p>\n<div id=\"fs-id1170571122528\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170571284202\">Consider the differential equation [latex]{y}^{\\prime }=f\\left(x,y\\right)[\/latex], and assume that all solutions to this differential equation are defined for [latex]x\\ge {x}_{0}[\/latex]. Let [latex]y=k[\/latex] be an equilibrium solution to the differential equation.<\/p>\n<ol id=\"fs-id1170570989710\" type=\"1\">\n<li>[latex]y=k[\/latex] is an<strong> asymptotically stable solution<\/strong> to the differential equation if there exists [latex]\\epsilon >0[\/latex] such that for any value [latex]c\\in \\left(k-\\epsilon ,k+\\epsilon \\right)[\/latex] the solution to the initial-value problem<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571303029\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }=f\\left(x,y\\right),y\\left({x}_{0}\\right)=c[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\napproaches [latex]k[\/latex] as [latex]x[\/latex] approaches infinity.<\/li>\n<li>[latex]y=k[\/latex] is an <strong>asymptotically unstable solution<\/strong> to the differential equation if there exists [latex]\\epsilon >0[\/latex] such that for any value [latex]c\\in \\left(k-\\epsilon ,k+\\epsilon \\right)[\/latex] the solution to the initial-value problem<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573366618\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }=f\\left(x,y\\right),y\\left({x}_{0}\\right)=c[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nnever approaches [latex]k[\/latex] as [latex]x[\/latex] approaches infinity.<\/li>\n<li>[latex]y=k[\/latex] is an <strong>asymptotically semi-stable solution<\/strong> to the differential equation if it is neither asymptotically stable nor asymptotically unstable.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573750302\">Now we return to the differential equation [latex]y^{\\prime} =\\left(x - 3\\right)\\left({y}^{2}-4\\right)[\/latex], with the initial condition [latex]y\\left(0\\right)=0.5[\/latex]. The direction field for this initial-value problem, along with the corresponding solution, is shown in Figure 6.<\/p>\n<figure id=\"CNX_Calc_Figure_08_02_006\"><figcaption><\/figcaption><div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233944\/CNX_Calc_Figure_08_02_005.jpg\" alt=\"A direction field for the given differential equation. The arrows are horizontal and pointing to the right at y = -4, y = 4, and x = 6. The closer the arrows are to x = 6, the more horizontal the arrows become. The further away, the more vertical they are. The arrows point down for y &gt; 4 and x &lt; 4, -4 &lt; y &lt; 4 and x &gt; 6, and y &lt; -4 and x &lt; 6. In all other areas, the arrows are pointing up. A solution is graphed that goes through (0, 0.5). It begins along y = -4 in quadrant three, increases from -4 to 4 between x = -1 and 1, and ends going along y = 4 in quadrant 1.\" width=\"487\" height=\"445\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 6. Direction field for the initial-value problem [latex]y^{\\prime} =\\left(x - 3\\right)\\left({y}^{2}-4\\right),y\\left(0\\right)=0.5[\/latex].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1170573230984\">The values of the solution to this initial-value problem stay between [latex]y=-2[\/latex] and [latex]y=2[\/latex], which are the equilibrium solutions to the differential equation. Furthermore, as [latex]x[\/latex] approaches infinity, [latex]y[\/latex] approaches [latex]2[\/latex]. The behavior of solutions is similar if the initial value is higher than [latex]2[\/latex], for example, [latex]y\\left(0\\right)=2.3[\/latex]. In this case, the solutions decrease and approach [latex]y=2[\/latex] as [latex]x[\/latex] approaches infinity. Therefore [latex]y=2[\/latex] is an asymptotically stable solution to the differential equation.<\/p>\n<p id=\"fs-id1170573230988\">What happens when the initial value is below [latex]y=-2?[\/latex] This scenario is illustrated in Figure 7, with the initial value [latex]y\\left(0\\right)=-3[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_08_02_007\"><figcaption><\/figcaption><div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233948\/CNX_Calc_Figure_08_02_006.jpg\" alt=\"A direction field for the given differential equation. The arrows are horizontal and pointing to the right at y = -4, y = 4, and x = 6. The closer the arrows are to x = 6, the more horizontal the arrows become. The further away, the more vertical they are. The arrows point down for y &gt; 4 and x &lt; 4, -4 &lt; y &lt; 4 and x &gt; 6, and y &lt; -4 and x &lt; 6. In all other areas, the arrows are pointing up. A solution is graphed that goes along y = -4 in quadrant 3 and curves between x = -1 and x = 0 to go to negative infinity along the y axis.\" width=\"487\" height=\"445\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 7. Direction field for the initial-value problem [latex]y^{\\prime} =\\left(x - 3\\right)\\left({y}^{2}-4\\right),y\\left(0\\right)=-3[\/latex].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1170573289098\">The solution decreases rapidly toward negative infinity as [latex]x[\/latex] approaches infinity. Furthermore, if the initial value is slightly higher than [latex]-2[\/latex], then the solution approaches [latex]2[\/latex], which is the other equilibrium solution. Therefore in neither case does the solution approach [latex]y=-2[\/latex], so [latex]y=-2[\/latex] is called an asymptotically unstable, or unstable, equilibrium solution.<\/p>\n<div id=\"fs-id1170573709023\" data-type=\"example\">\n<div id=\"fs-id1170573569693\" data-type=\"exercise\">\n<div id=\"fs-id1170571120492\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Stability of an Equilibrium Solution<\/h3>\n<div id=\"fs-id1170571120492\" data-type=\"problem\">\n<p id=\"fs-id1170573333748\">Create a direction field for the differential equation [latex]y^{\\prime} ={\\left(y - 3\\right)}^{2}\\left({y}^{2}+y - 2\\right)[\/latex] and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Show Solution<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571038103\" data-type=\"solution\">\n<p id=\"fs-id1170571093547\">The direction field is shown in Figure 8.<\/p>\n<figure id=\"CNX_Calc_Figure_08_02_008\"><figcaption><\/figcaption><div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233951\/CNX_Calc_Figure_08_02_007.jpg\" alt=\"A graph of a direction field with arrows pointing to the right at y = -4, y = 2, and y = 6. For y &lt; -4, the arrows point up. For -4 &lt; y &lt; 2, the arrows point down. For 2 &lt; y &lt;6, the arrows point up, becoming flatter and flatter as they approach y = 6. For y &gt; 6, the arrows point up and become more and more vertical the further they get from y = 6.\" width=\"487\" height=\"445\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 8. Direction field for the differential equation [latex]y^{\\prime} ={\\left(y - 3\\right)}^{2}\\left({y}^{2}+y - 2\\right)[\/latex].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1170573408988\">The equilibrium solutions are [latex]y=-2,y=1[\/latex], and [latex]y=3[\/latex]. To classify each of the solutions, look at an arrow directly above or below each of these values. For example, at [latex]y=-2[\/latex] the arrows directly below this solution point up, and the arrows directly above the solution point down. Therefore all initial conditions close to [latex]y=-2[\/latex] approach [latex]y=-2[\/latex], and the solution is stable. For the solution [latex]y=1[\/latex], all initial conditions above and below [latex]y=1[\/latex] are repelled (pushed away) from [latex]y=1[\/latex], so this solution is unstable. The solution [latex]y=3[\/latex] is semi-stable, because for initial conditions slightly greater than [latex]3[\/latex], the solution approaches infinity, and for initial conditions slightly less than [latex]3[\/latex], the solution approaches [latex]y=3[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1170573271901\" data-type=\"commentary\">\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\n<p id=\"fs-id1170571244786\">It is possible to find the equilibrium solutions to the differential equation by setting the right-hand side equal to zero and solving for [latex]y[\/latex]. This approach gives the same equilibrium solutions as those we saw in the direction field.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170571340415\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170571349921\" data-type=\"exercise\">\n<div id=\"fs-id1170571290427\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170571290427\" data-type=\"problem\">\n<p id=\"fs-id1170573519611\">Create a direction field for the differential equation [latex]y^{\\prime} =\\left(x+5\\right)\\left(y+2\\right)\\left({y}^{2}-4y+4\\right)[\/latex] and identify any equilibrium solutions. Classify each of the equilibrium solutions as stable, unstable, or semi-stable.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Hint<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571027317\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170571332593\">First create the direction field and look for horizontal dashes that go all the way across. Then examine the slope lines directly above and below the equilibrium solutions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170573255134\" data-type=\"solution\">\n<div style=\"width: 497px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233955\/CNX_Calc_Figure_08_02_008.jpg\" alt=\"A direction field with arrows pointing to the right at y = -4 and y = 4. The arrows point up for y &gt; -4 and down for y &lt; -4. Close to y = 4, the arrows are more horizontal, but the further away, the more vertical they become.\" width=\"487\" height=\"445\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 9.<\/p>\n<\/div>\n<p>The equilibrium solutions are [latex]y=-2[\/latex] and [latex]y=2[\/latex]. For this equation, [latex]y=-2[\/latex] is an unstable equilibrium solution, and [latex]y=2[\/latex] is a semi-stable equilibrium solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1170571260237\" data-depth=\"1\"><\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1621\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>4.2.1. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>4.2.2. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.2.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"4.2.2\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1621","chapter","type-chapter","status-publish","hentry"],"part":159,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1621","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1621\/revisions"}],"predecessor-version":[{"id":2658,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1621\/revisions\/2658"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/159"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1621\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1621"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1621"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1621"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1621"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}