{"id":821,"date":"2025-06-20T17:15:42","date_gmt":"2025-06-20T17:15:42","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=821"},"modified":"2025-09-10T15:03:01","modified_gmt":"2025-09-10T15:03:01","slug":"separation-of-variables-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/separation-of-variables-learn-it-1\/","title":{"raw":"Separation of Variables: Learn It 1","rendered":"Separation of Variables: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Solve differential equations by separating variables<\/li>\r\n \t<li>Apply separation of variables to real-world problems<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Separation of Variables<\/h2>\r\nMany differential equations can be solved using a powerful technique called <strong>separation of variables<\/strong>. This method works when the equation has a special structure that allows us to separate the [latex]x[\/latex] and [latex]y[\/latex] terms completely.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>separable differential equation<\/h3>\r\n<p class=\"whitespace-normal break-words\">A <strong>separable differential equation<\/strong> is any equation that can be written in the form:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]y^{\\prime} = f(x)g(y)[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">The term \"separable\" refers to the fact that the right-hand side can be separated into a function of [latex]x[\/latex] times a function of [latex]y[\/latex].<\/section>Here are examples of separable differential equations.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\"><strong>Example 1<\/strong>: [latex]y^{\\prime} = (x^2 - 4)(3y + 2)[\/latex]<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Here [latex]f(x) = x^2 - 4[\/latex] and [latex]g(y) = 3y + 2[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Example 2<\/strong>: [latex]y^{\\prime} = 6x^2 + 4x[\/latex]<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Here [latex]f(x) = 6x^2 + 4x[\/latex] and [latex]g(y) = 1[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Example 3<\/strong>: [latex]y^{\\prime} = \\sec y + \\tan y[\/latex]<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Here [latex]f(x) = 1[\/latex] and [latex]g(y) = \\sec y + \\tan y[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Example 4<\/strong>: [latex]y^{\\prime} = xy + 3x - 2y - 6[\/latex]<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">This can be factored as [latex](x + 3)(y - 2)[\/latex], so [latex]f(x) = x + 3[\/latex] and [latex]g(y) = y - 2[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>Example 3 above is also called an <strong>autonomous differential equation<\/strong> because the right-hand side depends only on [latex]y[\/latex], not on [latex]x[\/latex].\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">The Method of Separation of Variables<\/h3>\r\n<p class=\"whitespace-normal break-words\">If a differential equation is separable, you can solve it using the method of separation of variables.<\/p>\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy: Separation of Variables<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Check for constant solutions<\/strong>: Find any values of [latex]y[\/latex] that make [latex]g(y) = 0[\/latex]. These correspond to constant solutions.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Separate the variables<\/strong>: Rewrite the differential equation in the form [latex]\\frac{dy}{g(y)} = f(x)dx[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Integrate both sides<\/strong>: [latex]\\int \\frac{dy}{g(y)} = \\int f(x)dx[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Solve for [latex] y[\/latex]<\/strong>: Solve the resulting equation for [latex]y[\/latex] if possible.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Apply initial conditions<\/strong>: If an initial condition exists, substitute the appropriate values for [latex]x[\/latex] and [latex]y[\/latex] into the equation and solve for the constant.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\"><img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"62\" height=\"55\" \/><strong>Important note!<\/strong>\r\n[latex]\\\\[\/latex]\r\nStep 4 states \"solve for [latex]y[\/latex] if possible\" because it's not always possible to obtain [latex]y[\/latex] as an explicit function of [latex]x[\/latex]. Often we must be satisfied with finding [latex]y[\/latex] as an implicit function of [latex]x[\/latex].<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170571303682\" data-type=\"problem\">\r\n<p id=\"fs-id1170573513040\">Find a general solution to the differential equation [latex]y^{\\prime} =\\left({x}^{2}-4\\right)\\left(3y+2\\right)[\/latex] using the method of separation of variables.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1170573368010\" data-type=\"solution\">\r\n<p id=\"fs-id1170573394671\">Follow the five-step method of separation of variables.<\/p>\r\n\r\n<ol id=\"fs-id1170573570811\" type=\"1\">\r\n \t<li>In this example, [latex]f\\left(x\\right)={x}^{2}-4[\/latex] and [latex]g\\left(y\\right)=3y+2[\/latex]. Setting [latex]g\\left(y\\right)=0[\/latex] gives [latex]y=-\\frac{2}{3}[\/latex] as a constant solution.<\/li>\r\n \t<li>Rewrite the differential equation in the form<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573206619\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{dy}{3y+2}=\\left({x}^{2}-4\\right)dx[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Integrate both sides of the equation:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573356123\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{dy}{3y+2}=\\displaystyle\\int \\left({x}^{2}-4\\right)dx[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nLet [latex]u=3y+2[\/latex]. Then [latex]du=3\\frac{dy}{dx}dx[\/latex], so the equation becomes<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571330940\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{3}\\displaystyle\\int \\frac{1}{u}du&amp; =\\hfill &amp; \\frac{1}{3}{x}^{3}-4x+C\\hfill \\\\ \\hfill \\frac{1}{3}\\text{ln}|u|&amp; =\\hfill &amp; \\frac{1}{3}{x}^{3}-4x+C\\hfill \\\\ \\hfill \\frac{1}{3}\\text{ln}|3y+2|&amp; =\\hfill &amp; \\frac{1}{3}{x}^{3}-4x+C.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>To solve this equation for [latex]y[\/latex], first multiply both sides of the equation by [latex]3[\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573425728\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{ln}|3y+2|={x}^{3}-12x+3C[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNow we use some logic in dealing with the constant [latex]C[\/latex]. Since [latex]C[\/latex] represents an arbitrary constant, [latex]3C[\/latex] also represents an arbitrary constant. If we call the second arbitrary constant [latex]{C}_{1}[\/latex], the equation becomes<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571289022\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{ln}|3y+2|={x}^{3}-12x+{C}_{1}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNow exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base [latex]e[\/latex]). <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573240205\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {e}^{\\text{ln}|3y+2|}&amp; =\\hfill &amp; {e}^{{x}^{3}-12x+{C}_{1}}\\hfill \\\\ \\hfill |3y+2|&amp; =\\hfill &amp; {e}^{{C}_{1}}{e}^{{x}^{3}-12x}\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nAgain define a new constant [latex]{C}_{2}={e}^{{c}_{1}}[\/latex] (note that [latex]{C}_{2}&gt;0[\/latex]):<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571058014\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|3y+2|={C}_{2}{e}^{{x}^{3}-12x}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThis corresponds to two separate equations: [latex]3y+2={C}_{2}{e}^{{x}^{3}-12x}[\/latex] and [latex]3y+2=\\text{-}{C}_{2}{e}^{{x}^{3}-12x}[\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\nThe solution to either equation can be written in the form [latex]y=\\frac{-2\\pm {C}_{2}{e}^{{x}^{3}-12x}}{3}[\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\nSince [latex]{C}_{2}&gt;0[\/latex], it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant [latex]C[\/latex] is entirely arbitrary, and can be dropped. Therefore the solution can be written as<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573282406\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\frac{-2+C{e}^{{x}^{3}-12x}}{3}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>No initial condition is imposed, so we are finished.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to example above.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/9-Rhk4KH3Zs?controls=0&amp;start=0&amp;end=262&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.3.1_0to262_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.3.1\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170573574748\" data-type=\"problem\">\r\n<p id=\"fs-id1170571096292\">Using the method of separation of variables, solve the initial-value problem<\/p>\r\n\r\n<div id=\"fs-id1170573423616\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} =\\left(2x+3\\right)\\left({y}^{2}-4\\right),y\\left(0\\right)=-3[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1170573391549\" data-type=\"solution\">\r\n<p id=\"fs-id1170573750521\">Follow the five-step method of separation of variables.<\/p>\r\n\r\n<ol id=\"fs-id1170573593199\" type=\"1\">\r\n \t<li>In this example, [latex]f\\left(x\\right)=2x+3[\/latex] and [latex]g\\left(y\\right)={y}^{2}-4[\/latex]. Setting [latex]g\\left(y\\right)=0[\/latex] gives [latex]y=\\pm 2[\/latex] as constant solutions.<\/li>\r\n \t<li>Divide both sides of the equation by [latex]{y}^{2}-4[\/latex] and multiply by [latex]dx[\/latex]. This gives the equation<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573712029\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\frac{dy}{{y}^{2}-4}=\\left(2x+3\\right)dx[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Next integrate both sides:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573268326\" data-type=\"equation\">[latex]\\displaystyle\\int \\frac{1}{{y}^{2}-4}dy=\\displaystyle\\int \\left(2x+3\\right)dx[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTo evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573336985\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{{y}^{2}-4}=\\frac{1}{4}\\left(\\frac{1}{y - 2}-\\frac{1}{y+2}\\right)[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThen integration becomes<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571123143\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{4}{\\displaystyle\\int \\left(\\frac{1}{y - 2}-\\frac{1}{y+2}\\right)}dy&amp; =\\hfill &amp; {\\displaystyle\\int \\left(2x+3\\right)dx}\\hfill \\\\ \\hfill \\frac{1}{4}\\left(\\text{ln}|y - 2|-\\text{ln}|y+2|\\right)&amp; =\\hfill &amp; {x}^{2}+3x+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nMultiplying both sides of this equation by [latex]4[\/latex] and replacing [latex]4C[\/latex] with [latex]{C}_{1}[\/latex] gives<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573580573\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\text{ln}|y - 2|-\\text{ln}|y+2|&amp; =\\hfill &amp; 4{x}^{2}+12x+{C}_{1}\\hfill \\\\ \\hfill \\text{ln}|\\frac{y - 2}{y+2}|&amp; =\\hfill &amp; 4{x}^{2}+12x+{C}_{1}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>It is possible to solve this equation for <em data-effect=\"italics\">y<\/em>. First exponentiate both sides of the equation and define [latex]{C}_{2}={e}^{{C}_{1}}\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571051107\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|\\frac{y - 2}{y+2}|={C}_{2}{e}^{4{x}^{2}+12x}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNext we can remove the absolute value and let [latex]{C}_{2}[\/latex] be either positive or negative. Then multiply both sides by [latex]y+2[\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573408989\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ y - 2={C}_{2}\\left(y+2\\right){e}^{4{x}^{2}+12x}\\hfill \\\\ y - 2={C}_{2}y{e}^{{}^{4{x}^{2}+12x}}+2{C}_{2}{e}^{{}^{4{x}^{2}+12x}}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNow collect all terms involving <em data-effect=\"italics\">y<\/em> on one side of the equation, and solve for [latex]y\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573627024\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill y-{C}_{2}y{e}^{4{x}^{2}+12x}&amp; =\\hfill &amp; 2+2{C}_{2}{e}^{4{x}^{2}+12x}\\hfill \\\\ \\hfill y\\left(1-{C}_{2}{e}^{4{x}^{2}+12x}\\right)&amp; =\\hfill &amp; 2+2{C}_{2}{e}^{4{x}^{2}+12x}\\hfill \\\\ \\hfill y&amp; =\\hfill &amp; \\frac{2+2{C}_{2}{e}^{4{x}^{2}+12x}}{1-{C}_{2}{e}^{4{x}^{2}+12x}}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>To determine the value of [latex]{C}_{2}[\/latex], substitute [latex]x=0[\/latex] and [latex]y=-1[\/latex] into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation [latex]\\frac{y - 2}{y+2}={C}_{2}{e}^{4{x}^{2}+12}[\/latex]. This is much easier to solve for [latex]{C}_{2}\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571369024\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{y - 2}{y+2}&amp; =\\hfill &amp; {C}_{2}{e}^{4{x}^{2}+12x}\\hfill \\\\ \\hfill \\frac{-1 - 2}{-1+2}&amp; =\\hfill &amp; {C}_{2}{e}^{4{\\left(0\\right)}^{2}+12\\left(0\\right)}\\hfill \\\\ \\hfill {C}_{2}&amp; =\\hfill &amp; -3.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore the solution to the initial-value problem is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573501784\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\frac{2 - 6{e}^{4{x}^{2}+12x}}{1+3{e}^{4{x}^{2}+12x}}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nA graph of this solution appears in Figure 1.<span data-type=\"newline\">\r\n<\/span>\r\n<figure id=\"CNX_Calc_Figure_08_03_001\">[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234116\/CNX_Calc_Figure_08_03_001.jpg\" alt=\"A graph of the solution over [-5, 3] for x and [-3, 2] for y. It begins as a horizontal line at y = -2 from x = -5 to just before -3, almost immediately steps up to y = 2 from just after x = -3 to just before x = 0, and almost immediately steps back down to y = -2 just after x = 0 to x = 3.\" width=\"487\" height=\"275\" data-media-type=\"image\/jpeg\" \/> Figure 1. Graph of the solution to the initial-value problem [latex]y^{\\prime} =\\left(2x+3\\right)\\left({y}^{2}-4\\right),y\\left(0\\right)=-3[\/latex].[\/caption]<\/figure>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311304[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Solve differential equations by separating variables<\/li>\n<li>Apply separation of variables to real-world problems<\/li>\n<\/ul>\n<\/section>\n<h2>Separation of Variables<\/h2>\n<p>Many differential equations can be solved using a powerful technique called <strong>separation of variables<\/strong>. This method works when the equation has a special structure that allows us to separate the [latex]x[\/latex] and [latex]y[\/latex] terms completely.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>separable differential equation<\/h3>\n<p class=\"whitespace-normal break-words\">A <strong>separable differential equation<\/strong> is any equation that can be written in the form:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]y^{\\prime} = f(x)g(y)[\/latex]<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">The term &#8220;separable&#8221; refers to the fact that the right-hand side can be separated into a function of [latex]x[\/latex] times a function of [latex]y[\/latex].<\/section>\n<p>Here are examples of separable differential equations.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\"><strong>Example 1<\/strong>: [latex]y^{\\prime} = (x^2 - 4)(3y + 2)[\/latex]<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Here [latex]f(x) = x^2 - 4[\/latex] and [latex]g(y) = 3y + 2[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Example 2<\/strong>: [latex]y^{\\prime} = 6x^2 + 4x[\/latex]<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Here [latex]f(x) = 6x^2 + 4x[\/latex] and [latex]g(y) = 1[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Example 3<\/strong>: [latex]y^{\\prime} = \\sec y + \\tan y[\/latex]<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Here [latex]f(x) = 1[\/latex] and [latex]g(y) = \\sec y + \\tan y[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Example 4<\/strong>: [latex]y^{\\prime} = xy + 3x - 2y - 6[\/latex]<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">This can be factored as [latex](x + 3)(y - 2)[\/latex], so [latex]f(x) = x + 3[\/latex] and [latex]g(y) = y - 2[\/latex]<\/li>\n<\/ul>\n<\/section>\n<p>Example 3 above is also called an <strong>autonomous differential equation<\/strong> because the right-hand side depends only on [latex]y[\/latex], not on [latex]x[\/latex].<\/p>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">The Method of Separation of Variables<\/h3>\n<p class=\"whitespace-normal break-words\">If a differential equation is separable, you can solve it using the method of separation of variables.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy: Separation of Variables<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Check for constant solutions<\/strong>: Find any values of [latex]y[\/latex] that make [latex]g(y) = 0[\/latex]. These correspond to constant solutions.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Separate the variables<\/strong>: Rewrite the differential equation in the form [latex]\\frac{dy}{g(y)} = f(x)dx[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Integrate both sides<\/strong>: [latex]\\int \\frac{dy}{g(y)} = \\int f(x)dx[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Solve for [latex]y[\/latex]<\/strong>: Solve the resulting equation for [latex]y[\/latex] if possible.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Apply initial conditions<\/strong>: If an initial condition exists, substitute the appropriate values for [latex]x[\/latex] and [latex]y[\/latex] into the equation and solve for the constant.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"62\" height=\"55\" \/><strong>Important note!<\/strong><br \/>\n[latex]\\\\[\/latex]<br \/>\nStep 4 states &#8220;solve for [latex]y[\/latex] if possible&#8221; because it&#8217;s not always possible to obtain [latex]y[\/latex] as an explicit function of [latex]x[\/latex]. Often we must be satisfied with finding [latex]y[\/latex] as an implicit function of [latex]x[\/latex].<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170571303682\" data-type=\"problem\">\n<p id=\"fs-id1170573513040\">Find a general solution to the differential equation [latex]y^{\\prime} =\\left({x}^{2}-4\\right)\\left(3y+2\\right)[\/latex] using the method of separation of variables.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170573368010\" data-type=\"solution\">\n<p id=\"fs-id1170573394671\">Follow the five-step method of separation of variables.<\/p>\n<ol id=\"fs-id1170573570811\" type=\"1\">\n<li>In this example, [latex]f\\left(x\\right)={x}^{2}-4[\/latex] and [latex]g\\left(y\\right)=3y+2[\/latex]. Setting [latex]g\\left(y\\right)=0[\/latex] gives [latex]y=-\\frac{2}{3}[\/latex] as a constant solution.<\/li>\n<li>Rewrite the differential equation in the form<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573206619\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{dy}{3y+2}=\\left({x}^{2}-4\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>Integrate both sides of the equation:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573356123\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int \\frac{dy}{3y+2}=\\displaystyle\\int \\left({x}^{2}-4\\right)dx[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nLet [latex]u=3y+2[\/latex]. Then [latex]du=3\\frac{dy}{dx}dx[\/latex], so the equation becomes<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571330940\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{3}\\displaystyle\\int \\frac{1}{u}du& =\\hfill & \\frac{1}{3}{x}^{3}-4x+C\\hfill \\\\ \\hfill \\frac{1}{3}\\text{ln}|u|& =\\hfill & \\frac{1}{3}{x}^{3}-4x+C\\hfill \\\\ \\hfill \\frac{1}{3}\\text{ln}|3y+2|& =\\hfill & \\frac{1}{3}{x}^{3}-4x+C.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>To solve this equation for [latex]y[\/latex], first multiply both sides of the equation by [latex]3[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573425728\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{ln}|3y+2|={x}^{3}-12x+3C[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNow we use some logic in dealing with the constant [latex]C[\/latex]. Since [latex]C[\/latex] represents an arbitrary constant, [latex]3C[\/latex] also represents an arbitrary constant. If we call the second arbitrary constant [latex]{C}_{1}[\/latex], the equation becomes<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571289022\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\text{ln}|3y+2|={x}^{3}-12x+{C}_{1}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNow exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base [latex]e[\/latex]). <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573240205\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {e}^{\\text{ln}|3y+2|}& =\\hfill & {e}^{{x}^{3}-12x+{C}_{1}}\\hfill \\\\ \\hfill |3y+2|& =\\hfill & {e}^{{C}_{1}}{e}^{{x}^{3}-12x}\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nAgain define a new constant [latex]{C}_{2}={e}^{{c}_{1}}[\/latex] (note that [latex]{C}_{2}>0[\/latex]):<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571058014\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|3y+2|={C}_{2}{e}^{{x}^{3}-12x}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThis corresponds to two separate equations: [latex]3y+2={C}_{2}{e}^{{x}^{3}-12x}[\/latex] and [latex]3y+2=\\text{-}{C}_{2}{e}^{{x}^{3}-12x}[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe solution to either equation can be written in the form [latex]y=\\frac{-2\\pm {C}_{2}{e}^{{x}^{3}-12x}}{3}[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nSince [latex]{C}_{2}>0[\/latex], it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant [latex]C[\/latex] is entirely arbitrary, and can be dropped. Therefore the solution can be written as<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573282406\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\frac{-2+C{e}^{{x}^{3}-12x}}{3}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>No initial condition is imposed, so we are finished.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/9-Rhk4KH3Zs?controls=0&amp;start=0&amp;end=262&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.3.1_0to262_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.3.1&#8221; here (opens in new window)<\/a>.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170573574748\" data-type=\"problem\">\n<p id=\"fs-id1170571096292\">Using the method of separation of variables, solve the initial-value problem<\/p>\n<div id=\"fs-id1170573423616\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} =\\left(2x+3\\right)\\left({y}^{2}-4\\right),y\\left(0\\right)=-3[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558896\">Show Solution<\/button><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170573391549\" data-type=\"solution\">\n<p id=\"fs-id1170573750521\">Follow the five-step method of separation of variables.<\/p>\n<ol id=\"fs-id1170573593199\" type=\"1\">\n<li>In this example, [latex]f\\left(x\\right)=2x+3[\/latex] and [latex]g\\left(y\\right)={y}^{2}-4[\/latex]. Setting [latex]g\\left(y\\right)=0[\/latex] gives [latex]y=\\pm 2[\/latex] as constant solutions.<\/li>\n<li>Divide both sides of the equation by [latex]{y}^{2}-4[\/latex] and multiply by [latex]dx[\/latex]. This gives the equation<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573712029\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]\\frac{dy}{{y}^{2}-4}=\\left(2x+3\\right)dx[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li>Next integrate both sides:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573268326\" data-type=\"equation\">[latex]\\displaystyle\\int \\frac{1}{{y}^{2}-4}dy=\\displaystyle\\int \\left(2x+3\\right)dx[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTo evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573336985\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{1}{{y}^{2}-4}=\\frac{1}{4}\\left(\\frac{1}{y - 2}-\\frac{1}{y+2}\\right)[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThen integration becomes<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571123143\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{1}{4}{\\displaystyle\\int \\left(\\frac{1}{y - 2}-\\frac{1}{y+2}\\right)}dy& =\\hfill & {\\displaystyle\\int \\left(2x+3\\right)dx}\\hfill \\\\ \\hfill \\frac{1}{4}\\left(\\text{ln}|y - 2|-\\text{ln}|y+2|\\right)& =\\hfill & {x}^{2}+3x+C.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nMultiplying both sides of this equation by [latex]4[\/latex] and replacing [latex]4C[\/latex] with [latex]{C}_{1}[\/latex] gives<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573580573\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\text{ln}|y - 2|-\\text{ln}|y+2|& =\\hfill & 4{x}^{2}+12x+{C}_{1}\\hfill \\\\ \\hfill \\text{ln}|\\frac{y - 2}{y+2}|& =\\hfill & 4{x}^{2}+12x+{C}_{1}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>It is possible to solve this equation for <em data-effect=\"italics\">y<\/em>. First exponentiate both sides of the equation and define [latex]{C}_{2}={e}^{{C}_{1}}\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571051107\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]|\\frac{y - 2}{y+2}|={C}_{2}{e}^{4{x}^{2}+12x}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNext we can remove the absolute value and let [latex]{C}_{2}[\/latex] be either positive or negative. Then multiply both sides by [latex]y+2[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573408989\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\\\ y - 2={C}_{2}\\left(y+2\\right){e}^{4{x}^{2}+12x}\\hfill \\\\ y - 2={C}_{2}y{e}^{{}^{4{x}^{2}+12x}}+2{C}_{2}{e}^{{}^{4{x}^{2}+12x}}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNow collect all terms involving <em data-effect=\"italics\">y<\/em> on one side of the equation, and solve for [latex]y\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573627024\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill y-{C}_{2}y{e}^{4{x}^{2}+12x}& =\\hfill & 2+2{C}_{2}{e}^{4{x}^{2}+12x}\\hfill \\\\ \\hfill y\\left(1-{C}_{2}{e}^{4{x}^{2}+12x}\\right)& =\\hfill & 2+2{C}_{2}{e}^{4{x}^{2}+12x}\\hfill \\\\ \\hfill y& =\\hfill & \\frac{2+2{C}_{2}{e}^{4{x}^{2}+12x}}{1-{C}_{2}{e}^{4{x}^{2}+12x}}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>To determine the value of [latex]{C}_{2}[\/latex], substitute [latex]x=0[\/latex] and [latex]y=-1[\/latex] into the general solution. Alternatively, we can put the same values into an earlier equation, namely the equation [latex]\\frac{y - 2}{y+2}={C}_{2}{e}^{4{x}^{2}+12}[\/latex]. This is much easier to solve for [latex]{C}_{2}\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571369024\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{y - 2}{y+2}& =\\hfill & {C}_{2}{e}^{4{x}^{2}+12x}\\hfill \\\\ \\hfill \\frac{-1 - 2}{-1+2}& =\\hfill & {C}_{2}{e}^{4{\\left(0\\right)}^{2}+12\\left(0\\right)}\\hfill \\\\ \\hfill {C}_{2}& =\\hfill & -3.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore the solution to the initial-value problem is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573501784\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\frac{2 - 6{e}^{4{x}^{2}+12x}}{1+3{e}^{4{x}^{2}+12x}}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nA graph of this solution appears in Figure 1.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<figure id=\"CNX_Calc_Figure_08_03_001\">\n<figure style=\"width: 487px\" 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\/11234116\/CNX_Calc_Figure_08_03_001.jpg\" alt=\"A graph of the solution over &#091;-5, 3&#093; for x and &#091;-3, 2&#093; for y. It begins as a horizontal line at y = -2 from x = -5 to just before -3, almost immediately steps up to y = 2 from just after x = -3 to just before x = 0, and almost immediately steps back down to y = -2 just after x = 0 to x = 3.\" width=\"487\" height=\"275\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 1. Graph of the solution to the initial-value problem [latex]y^{\\prime} =\\left(2x+3\\right)\\left({y}^{2}-4\\right),y\\left(0\\right)=-3[\/latex].<\/figcaption><\/figure>\n<\/figure>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm311304\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311304&theme=lumen&iframe_resize_id=ohm311304&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":18,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":669,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/821"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/821\/revisions"}],"predecessor-version":[{"id":2293,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/821\/revisions\/2293"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/669"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/821\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=821"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=821"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=821"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=821"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}