{"id":838,"date":"2025-06-20T17:16:43","date_gmt":"2025-06-20T17:16:43","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=838"},"modified":"2025-09-09T19:06:39","modified_gmt":"2025-09-09T19:06:39","slug":"first-order-linear-equations-and-applications-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/first-order-linear-equations-and-applications-fresh-take\/","title":{"raw":"First-Order Linear Equations and Applications: Fresh Take","rendered":"First-Order Linear Equations and Applications: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Write first-order linear differential equations in their standard form<\/li>\r\n \t<li>Find and use integrating factors to solve first-order linear equations<\/li>\r\n \t<li>Understand how carrying capacity affects population growth in the logistic model<\/li>\r\n \t<li>Work with logistic equations and interpret what their solutions mean<\/li>\r\n \t<li>Solve real-world problems using first-order linear differential equations<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>First-Order Differential Equations<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Think of linear differential equations like upgrading from a basic physics problem to the real world\u2014instead of just gravity pulling a ball down, now we add air resistance, friction, and other forces that make the math more interesting but the results more accurate.<\/p>\r\n<p class=\"whitespace-normal break-words\">What Makes an Equation \"Linear\"? The unknown function [latex]y[\/latex] and its derivative [latex]y'[\/latex] appear only to the first power and never get multiplied together. The general form is: [latex]a(x)y' + b(x)y = c(x)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why \"Linear\" Matters:<\/strong><\/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\">Linear equations have systematic solution methods<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Nonlinear equations often require special techniques or numerical approximation<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Linear systems are more predictable and well-behaved<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Standard Form<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Standard form is like having all your tools organized before starting a project. Once your equation looks like [latex]y' + p(x)y = q(x)[\/latex], you can apply systematic solution methods (like integrating factors) that work every time.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>What Are [latex]p(x)[\/latex] and [latex]q(x)[\/latex]?<\/strong><\/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\">[latex]p(x)[\/latex] is the coefficient of [latex]y[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]q(x)[\/latex] is everything on the right-hand side<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Both depend only on [latex]x[\/latex], not on [latex]y[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/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>Rearrange:<\/strong> Move all terms with [latex]y[\/latex] or [latex]y'[\/latex] to the left side<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Factor:<\/strong> Make sure [latex]y'[\/latex] and [latex]y[\/latex] terms are separated<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Normalize:<\/strong> Divide everything by the coefficient of [latex]y'[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Quick Examples:<\/strong><\/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\">[latex]y' = 3x - 4y[\/latex] \u2192 [latex]y' + 4y = 3x[\/latex] (just rearrange)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]3y' - 6y = 9x[\/latex] \u2192 [latex]y' - 2y = 3x[\/latex] (divide by [latex]3[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]xy' + 2y = x^2[\/latex] \u2192 [latex]y' + \\frac{2}{x}y = x[\/latex] (divide by [latex]x[\/latex])<\/li>\r\n<\/ul>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170572331248\" data-type=\"problem\">\r\n\r\nPut the equation [latex]\\frac{\\left(x+3\\right)y^{\\prime} }{2x - 3y - 4}=5[\/latex] into standard form and identify [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1170572241407\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170572484113\">Multiply both sides by the common denominator, then collect all terms involving [latex]y[\/latex] on one side.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1170572452404\" data-type=\"solution\">\r\n<p id=\"fs-id1170572169346\">[latex]y^{\\prime} +\\frac{15}{x+3}y=\\frac{10x - 20}{x+3};p\\left(x\\right)=\\frac{15}{x+3}[\/latex] and [latex]q\\left(x\\right)=\\frac{10x - 20}{x+3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2 data-type=\"title\">Integrating Factors<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Think of an integrating factor as the special function that makes the left side of your equation collapse into something you can actually work with. The integrating factor [latex]\\mu(x) = e^{\\int p(x)dx}[\/latex] is designed to turn the messy left side [latex]y' + p(x)y[\/latex] into the clean derivative [latex]\\frac{d}{dx}[\\mu(x)y][\/latex]. This transforms your differential equation into something you can integrate directly.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/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\"><strong>Step 1:<\/strong> Get your equation into standard form [latex]y' + p(x)y = q(x)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Step 2:<\/strong> Calculate [latex]\\mu(x) = e^{\\int p(x)dx}[\/latex] (no constant needed here!)<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Step 3:<\/strong> Multiply the entire equation by [latex]\\mu(x)[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Step 4:<\/strong> The left side becomes [latex]\\frac{d}{dx}[\\mu(x)y][\/latex], so integrate both sides<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Step 5:<\/strong> Solve for [latex]y[\/latex] and apply initial conditions if given<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">After multiplying by [latex]\\mu(x)[\/latex], the left side always becomes [latex]\\frac{d}{dx}[\\mu(x)y] = \\mu(x)q(x)[\/latex]. This isn't magic\u2014it's the product rule working in reverse.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the general solution to the differential equation [latex]\\left(x - 2\\right)y^{\\prime} +y=3{x}^{2}+2x[\/latex]. Assume [latex]x&gt;2[\/latex].[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1170571690904\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170571679832\">Use the method outlined in the problem-solving strategy for first-order linear differential equations.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1170571611044\" data-type=\"solution\">\r\n<p id=\"fs-id1170571655144\">[latex]y=\\frac{{x}^{3}+{x}^{2}+C}{x - 2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170572505501\" data-type=\"problem\">\r\n<p id=\"fs-id1170572505503\">Solve the initial-value problem [latex]y^{\\prime} -2y=4x+3, y\\left(0\\right)=-2[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1170572449330\" data-type=\"solution\">\r\n<p id=\"fs-id1170572449332\">[latex]y=-2x - 4+2{e}^{2x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h1>Population Growth and Carrying Capacity<\/h1>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Population growth models evolve from simple exponential growth to more realistic patterns that account for environmental limits. The derivative [latex]\\frac{dP}{dt}[\/latex] represents how fast a population is changing at any given moment.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Exponential Growth:<\/strong> The equation [latex]\\frac{dP}{dt} = rP[\/latex] creates unlimited J-shaped growth where bigger populations grow faster. This works well for small populations with abundant resources, but it predicts unrealistic infinite growth over time.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Logistic Growth Model:<\/strong> The equation [latex]\\frac{dP}{dt} = rP\\left(1 - \\frac{P}{K}\\right)[\/latex] creates realistic S-shaped growth curves by incorporating carrying capacity.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>How the Logistic Model Behaves:<\/strong><\/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\">When [latex]P[\/latex] is small: [latex]\\frac{P}{K} \\approx 0[\/latex], so growth looks nearly exponential<\/li>\r\n \t<li class=\"whitespace-normal break-words\">As [latex]P[\/latex] approaches [latex]K[\/latex]: The term [latex]\\left(1 - \\frac{P}{K}\\right)[\/latex] shrinks, slowing growth<\/li>\r\n \t<li class=\"whitespace-normal break-words\">When [latex]P = K[\/latex]: Growth rate becomes zero (equilibrium)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">When [latex]P &gt; K[\/latex]: Growth rate turns negative, population decreases toward [latex]K[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">Real populations face constraints like limited food, space, and resources. This leads to the concept of carrying capacity [latex]K[\/latex]\u2014the maximum population an environment can sustain long-term. Carrying capacity acts like a population magnet\u2014populations below [latex]K[\/latex] grow toward it, populations above [latex]K[\/latex] decline toward it. This creates the characteristic S-curve that starts exponential, slows down, and levels off at the carrying capacity.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit <a href=\"https:\/\/www.otherwise.com\/population\/logistic.html\" target=\"_blank\" rel=\"noopener\">this website for more information on logistic growth<\/a>.<\/section>\r\n<h1>Solving the Logistic Differential Equation<\/h1>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The logistic differential equation [latex]\\frac{dP}{dt} = rP\\left(1 - \\frac{P}{K}\\right)[\/latex] can be solved exactly using separation of variables, giving us the complete formula for realistic population growth.<\/p>\r\n<p class=\"whitespace-normal break-words\">The solution process involves separating variables to get [latex]\\frac{dP}{P(K-P)} = \\frac{r}{K}dt[\/latex], then using partial fraction decomposition on the left side: [latex]\\frac{K}{P(K-P)} = \\frac{1}{P} + \\frac{1}{K-P}[\/latex]. After integration and algebraic manipulation, we arrive at the general solution.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Complete Solution:<\/strong> [latex]P(t) = \\frac{P_0 K e^{rt}}{(K-P_0) + P_0 e^{rt}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Behaviors of This Solution:<\/strong><\/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\">When [latex]t \\to -\\infty[\/latex]: [latex]P(t) \\to 0[\/latex] (population approaches zero in distant past)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">When [latex]t \\to \\infty[\/latex]: [latex]P(t) \\to K[\/latex] (population approaches carrying capacity)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The curve is S-shaped, starting slow, accelerating, then leveling off<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>The Inflection Point:<\/strong> The population growth changes from accelerating to decelerating at [latex]t = \\frac{1}{r}\\ln\\frac{K-P_0}{P_0}[\/latex], which occurs when the population reaches exactly [latex]P = \\frac{K}{2}[\/latex] (half the carrying capacity). This is where the growth rate is fastest.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Equilibrium Solutions:<\/strong> [latex]P = 0[\/latex] (extinction) and [latex]P = K[\/latex] (carrying capacity) are constant solutions. The first is unstable (small populations grow away from zero), while the second is stable (populations approach the carrying capacity).<\/p>\r\n\r\n<\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Write first-order linear differential equations in their standard form<\/li>\n<li>Find and use integrating factors to solve first-order linear equations<\/li>\n<li>Understand how carrying capacity affects population growth in the logistic model<\/li>\n<li>Work with logistic equations and interpret what their solutions mean<\/li>\n<li>Solve real-world problems using first-order linear differential equations<\/li>\n<\/ul>\n<\/section>\n<h2>First-Order Differential Equations<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Think of linear differential equations like upgrading from a basic physics problem to the real world\u2014instead of just gravity pulling a ball down, now we add air resistance, friction, and other forces that make the math more interesting but the results more accurate.<\/p>\n<p class=\"whitespace-normal break-words\">What Makes an Equation &#8220;Linear&#8221;? The unknown function [latex]y[\/latex] and its derivative [latex]y'[\/latex] appear only to the first power and never get multiplied together. The general form is: [latex]a(x)y' + b(x)y = c(x)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why &#8220;Linear&#8221; Matters:<\/strong><\/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\">Linear equations have systematic solution methods<\/li>\n<li class=\"whitespace-normal break-words\">Nonlinear equations often require special techniques or numerical approximation<\/li>\n<li class=\"whitespace-normal break-words\">Linear systems are more predictable and well-behaved<\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Standard Form<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Standard form is like having all your tools organized before starting a project. Once your equation looks like [latex]y' + p(x)y = q(x)[\/latex], you can apply systematic solution methods (like integrating factors) that work every time.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>What Are [latex]p(x)[\/latex] and [latex]q(x)[\/latex]?<\/strong><\/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\">[latex]p(x)[\/latex] is the coefficient of [latex]y[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]q(x)[\/latex] is everything on the right-hand side<\/li>\n<li class=\"whitespace-normal break-words\">Both depend only on [latex]x[\/latex], not on [latex]y[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/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>Rearrange:<\/strong> Move all terms with [latex]y[\/latex] or [latex]y'[\/latex] to the left side<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Factor:<\/strong> Make sure [latex]y'[\/latex] and [latex]y[\/latex] terms are separated<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Normalize:<\/strong> Divide everything by the coefficient of [latex]y'[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Quick Examples:<\/strong><\/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\">[latex]y' = 3x - 4y[\/latex] \u2192 [latex]y' + 4y = 3x[\/latex] (just rearrange)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]3y' - 6y = 9x[\/latex] \u2192 [latex]y' - 2y = 3x[\/latex] (divide by [latex]3[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]xy' + 2y = x^2[\/latex] \u2192 [latex]y' + \\frac{2}{x}y = x[\/latex] (divide by [latex]x[\/latex])<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170572331248\" data-type=\"problem\">\n<p>Put the equation [latex]\\frac{\\left(x+3\\right)y^{\\prime} }{2x - 3y - 4}=5[\/latex] into standard form and identify [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558897\">Hint<\/button><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572241407\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170572484113\">Multiply both sides by the common denominator, then collect all terms involving [latex]y[\/latex] on one side.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572452404\" data-type=\"solution\">\n<p id=\"fs-id1170572169346\">[latex]y^{\\prime} +\\frac{15}{x+3}y=\\frac{10x - 20}{x+3};p\\left(x\\right)=\\frac{15}{x+3}[\/latex] and [latex]q\\left(x\\right)=\\frac{10x - 20}{x+3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Integrating Factors<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Think of an integrating factor as the special function that makes the left side of your equation collapse into something you can actually work with. The integrating factor [latex]\\mu(x) = e^{\\int p(x)dx}[\/latex] is designed to turn the messy left side [latex]y' + p(x)y[\/latex] into the clean derivative [latex]\\frac{d}{dx}[\\mu(x)y][\/latex]. This transforms your differential equation into something you can integrate directly.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy:<\/strong><\/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\"><strong>Step 1:<\/strong> Get your equation into standard form [latex]y' + p(x)y = q(x)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Step 2:<\/strong> Calculate [latex]\\mu(x) = e^{\\int p(x)dx}[\/latex] (no constant needed here!)<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Step 3:<\/strong> Multiply the entire equation by [latex]\\mu(x)[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Step 4:<\/strong> The left side becomes [latex]\\frac{d}{dx}[\\mu(x)y][\/latex], so integrate both sides<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Step 5:<\/strong> Solve for [latex]y[\/latex] and apply initial conditions if given<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">After multiplying by [latex]\\mu(x)[\/latex], the left side always becomes [latex]\\frac{d}{dx}[\\mu(x)y] = \\mu(x)q(x)[\/latex]. This isn&#8217;t magic\u2014it&#8217;s the product rule working in reverse.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find the general solution to the differential equation [latex]\\left(x - 2\\right)y^{\\prime} +y=3{x}^{2}+2x[\/latex]. Assume [latex]x>2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558894\">Hint<\/button><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571690904\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170571679832\">Use the method outlined in the problem-solving strategy for first-order linear differential equations.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558895\">Show Solution<\/button><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571611044\" data-type=\"solution\">\n<p id=\"fs-id1170571655144\">[latex]y=\\frac{{x}^{3}+{x}^{2}+C}{x - 2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170572505501\" data-type=\"problem\">\n<p id=\"fs-id1170572505503\">Solve the initial-value problem [latex]y^{\\prime} -2y=4x+3, y\\left(0\\right)=-2[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558892\">Show Solution<\/button><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572449330\" data-type=\"solution\">\n<p id=\"fs-id1170572449332\">[latex]y=-2x - 4+2{e}^{2x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<h1>Population Growth and Carrying Capacity<\/h1>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Population growth models evolve from simple exponential growth to more realistic patterns that account for environmental limits. The derivative [latex]\\frac{dP}{dt}[\/latex] represents how fast a population is changing at any given moment.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Exponential Growth:<\/strong> The equation [latex]\\frac{dP}{dt} = rP[\/latex] creates unlimited J-shaped growth where bigger populations grow faster. This works well for small populations with abundant resources, but it predicts unrealistic infinite growth over time.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Logistic Growth Model:<\/strong> The equation [latex]\\frac{dP}{dt} = rP\\left(1 - \\frac{P}{K}\\right)[\/latex] creates realistic S-shaped growth curves by incorporating carrying capacity.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>How the Logistic Model Behaves:<\/strong><\/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\">When [latex]P[\/latex] is small: [latex]\\frac{P}{K} \\approx 0[\/latex], so growth looks nearly exponential<\/li>\n<li class=\"whitespace-normal break-words\">As [latex]P[\/latex] approaches [latex]K[\/latex]: The term [latex]\\left(1 - \\frac{P}{K}\\right)[\/latex] shrinks, slowing growth<\/li>\n<li class=\"whitespace-normal break-words\">When [latex]P = K[\/latex]: Growth rate becomes zero (equilibrium)<\/li>\n<li class=\"whitespace-normal break-words\">When [latex]P > K[\/latex]: Growth rate turns negative, population decreases toward [latex]K[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">Real populations face constraints like limited food, space, and resources. This leads to the concept of carrying capacity [latex]K[\/latex]\u2014the maximum population an environment can sustain long-term. Carrying capacity acts like a population magnet\u2014populations below [latex]K[\/latex] grow toward it, populations above [latex]K[\/latex] decline toward it. This creates the characteristic S-curve that starts exponential, slows down, and levels off at the carrying capacity.<\/p>\n<\/div>\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit <a href=\"https:\/\/www.otherwise.com\/population\/logistic.html\" target=\"_blank\" rel=\"noopener\">this website for more information on logistic growth<\/a>.<\/section>\n<h1>Solving the Logistic Differential Equation<\/h1>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The logistic differential equation [latex]\\frac{dP}{dt} = rP\\left(1 - \\frac{P}{K}\\right)[\/latex] can be solved exactly using separation of variables, giving us the complete formula for realistic population growth.<\/p>\n<p class=\"whitespace-normal break-words\">The solution process involves separating variables to get [latex]\\frac{dP}{P(K-P)} = \\frac{r}{K}dt[\/latex], then using partial fraction decomposition on the left side: [latex]\\frac{K}{P(K-P)} = \\frac{1}{P} + \\frac{1}{K-P}[\/latex]. After integration and algebraic manipulation, we arrive at the general solution.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Complete Solution:<\/strong> [latex]P(t) = \\frac{P_0 K e^{rt}}{(K-P_0) + P_0 e^{rt}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key Behaviors of This Solution:<\/strong><\/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\">When [latex]t \\to -\\infty[\/latex]: [latex]P(t) \\to 0[\/latex] (population approaches zero in distant past)<\/li>\n<li class=\"whitespace-normal break-words\">When [latex]t \\to \\infty[\/latex]: [latex]P(t) \\to K[\/latex] (population approaches carrying capacity)<\/li>\n<li class=\"whitespace-normal break-words\">The curve is S-shaped, starting slow, accelerating, then leveling off<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>The Inflection Point:<\/strong> The population growth changes from accelerating to decelerating at [latex]t = \\frac{1}{r}\\ln\\frac{K-P_0}{P_0}[\/latex], which occurs when the population reaches exactly [latex]P = \\frac{K}{2}[\/latex] (half the carrying capacity). This is where the growth rate is fastest.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Equilibrium Solutions:<\/strong> [latex]P = 0[\/latex] (extinction) and [latex]P = K[\/latex] (carrying capacity) are constant solutions. The first is unstable (small populations grow away from zero), while the second is stable (populations approach the carrying capacity).<\/p>\n<\/div>\n","protected":false},"author":15,"menu_order":28,"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\/838"}],"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\/838\/revisions"}],"predecessor-version":[{"id":2249,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/838\/revisions\/2249"}],"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\/838\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=838"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=838"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=838"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=838"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}