{"id":1835,"date":"2025-08-15T16:51:19","date_gmt":"2025-08-15T16:51:19","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1835"},"modified":"2025-09-10T17:32:26","modified_gmt":"2025-09-10T17:32:26","slug":"first-order-linear-equations-and-applications-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/first-order-linear-equations-and-applications-learn-it-5\/","title":{"raw":"First-Order Linear Equations and Applications: Learn It 5","rendered":"First-Order Linear Equations and Applications: Learn It 5"},"content":{"raw":"<h1>Solving the Logistic Differential Equation<\/h1>\r\n<p id=\"fs-id1170571618234\">The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\"><strong>Step 1: Find the equilibrium solutions<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Setting the right-hand side equal to zero gives us [latex]P = 0[\/latex] and [latex]P = K[\/latex] as constant solutions.<\/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 = 0[\/latex]: If no organisms are present, the population will never grow<\/li>\r\n \t<li class=\"whitespace-normal break-words\">When [latex]P = K[\/latex]: If the population starts at carrying capacity, it remains constant<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 2: Separate the variables<\/strong><\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Rewrite the differential equation as:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\frac{dP}{dt} = \\frac{rP(K-P)}{K}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Multiply both sides by [latex]dt[\/latex] and divide by [latex]P(K-P)[\/latex]:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\frac{dP}{P(K-P)} = \\frac{r}{K}dt[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 3: Prepare for integration<\/strong><\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Multiply both sides by [latex]K[\/latex]:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\int \\frac{K}{P(K-P)}dP = \\int r dt[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 4: Use partial fraction decomposition<\/strong><\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">The left side can be decomposed using partial fractions:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\frac{K}{P(K-P)} = \\frac{1}{P} + \\frac{1}{K-P}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><em>You can verify this decomposition by finding a common denominator on the right side and checking that it equals the left side.<\/em><\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 5: Integrate both sides<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\int \\left(\\frac{1}{P} + \\frac{1}{K-P}\\right)dP = \\int r dt[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\ln|P| - \\ln|K-P| = rt + C[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\ln\\left|\\frac{P}{K-P}\\right| = rt + C[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 6: Solve for [latex]P[\/latex]<\/strong><\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Exponentiate both sides to eliminate the natural logarithm:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]e^{\\ln\\left|\\frac{P}{K-P}\\right|} = e^{rt+C}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\left|\\frac{P}{K-P}\\right| = e^C \\cdot e^{rt}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Defining [latex]C_1 = e^C[\/latex], we get: [latex]\\frac{P}{K-P} = C_1 e^{rt}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 7: Solve for [latex]P(t)[\/latex]<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Starting from [latex]\\frac{P}{K-P} = C_1 e^{rt}[\/latex], multiply both sides by [latex]K-P[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P = C_1 e^{rt}(K-P)[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P = C_1 K e^{rt} - C_1 P e^{rt}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Collect all terms containing [latex]P[\/latex] on the left side:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]P + C_1 P e^{rt} = C_1 K e^{rt}[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Factor out [latex]P[\/latex] and solve:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]P(1 + C_1 e^{rt}) = C_1 K e^{rt}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P(t) = \\frac{C_1 K e^{rt}}{1 + C_1 e^{rt}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 8: Find the constant [latex]C_1[\/latex]<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To determine [latex]C_1[\/latex], use the initial condition. When [latex]t = 0[\/latex], [latex]P(0) = P_0[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Substitute into [latex]\\frac{P}{K-P} = C_1 e^{rt}[\/latex]:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\frac{P_0}{K-P_0} = C_1 e^{r(0)} = C_1[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Therefore: [latex]C_1 = \\frac{P_0}{K-P_0}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Step 9: Complete the solution<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Substitute [latex]C_1 = \\frac{P_0}{K-P_0}[\/latex] into our equation:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P(t) = \\frac{C_1 K e^{rt}}{1 + C_1 e^{rt}} = \\frac{\\frac{P_0}{K-P_0} \\cdot K e^{rt}}{1 + \\frac{P_0}{K-P_0} e^{rt}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Multiply numerator and denominator by [latex]\\left(K-{P}_{0}\\right)[\/latex] to simplify:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill P\\left(t\\right)&amp; =\\frac{\\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}\\hfill \\\\ &amp; =\\frac{\\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}\\cdot \\frac{K-{P}_{0}}{K-{P}_{0}}\\hfill \\\\ &amp; =\\frac{{P}_{0}K{e}^{rt}}{\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}}.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/section>\r\n<p id=\"fs-id1170572111539\">We state this result as a theorem.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>logistic growth solution<\/h3>\r\n<p id=\"fs-id1170571728769\">Consider the logistic differential equation subject to an initial population of [latex]{P}_{0}[\/latex] with carrying capacity [latex]K[\/latex] and growth rate [latex]r[\/latex]. The solution to the corresponding initial-value problem is:<\/p>\r\n\r\n<div id=\"fs-id1170571597871\" style=\"text-align: center;\" data-type=\"equation\">[latex]P\\left(t\\right)=\\dfrac{{P}_{0}K{e}^{rt}}{\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}}[\/latex].<\/div>\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Finding the Point of Inflection<\/h3>\r\n<p class=\"whitespace-normal break-words\">The logistic growth curve has a point of inflection where the growth rate changes from increasing to decreasing. To find this point, we set the second derivative equal to zero.<\/p>\r\n<p class=\"whitespace-normal break-words\">Starting with our solution, we can find the derivatives:<\/p>\r\n\r\n<div id=\"fs-id1170572307551\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill P\\left(t\\right)&amp; =\\hfill &amp; \\frac{{P}_{0}K{e}^{rt}}{\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}}\\hfill \\\\ \\hfill {P}^{\\prime }\\left(t\\right)&amp; =\\hfill &amp; \\frac{r{P}_{0}K\\left(K-{P}_{0}\\right){e}^{rt}}{{\\left(\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}\\right)}^{2}}\\hfill \\\\ \\hfill P^{\\prime\\prime}\\left(t\\right)&amp; =\\hfill &amp; \\frac{{r}^{2}{P}_{0}K{\\left(K-{P}_{0}\\right)}^{2}{e}^{rt}-{r}^{2}{P}_{0}{}^{2}K\\left(K-{P}_{0}\\right){e}^{2rt}}{{\\left(\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}\\right)}^{3}}\\hfill \\\\ &amp; =\\hfill &amp; \\frac{{r}^{2}{P}_{0}K\\left(K-{P}_{0}\\right){e}^{rt}\\left(\\left(K-{P}_{0}\\right)-{P}_{0}{e}^{rt}\\right)}{{\\left(\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}\\right)}^{3}}.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1170571712032\">Setting the numerator equal to zero,<\/p>\r\n\r\n<div id=\"fs-id1170572509015\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{r}^{2}{P}_{0}K\\left(K-{P}_{0}\\right){e}^{rt}\\left(\\left(K-{P}_{0}\\right)-{P}_{0}{e}^{rt}\\right)=0[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572344215\">As long as [latex]{P}_{0}\\ne K[\/latex], the entire quantity before and including [latex]{e}^{rt}[\/latex] is nonzero, so we can divide it out:<\/p>\r\n\r\n<div id=\"fs-id1170572147551\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left(K-{P}_{0}\\right)-{P}_{0}{e}^{rt}=0[\/latex].<\/div>\r\n<p id=\"fs-id1170572236506\">Solving for [latex]t[\/latex],<\/p>\r\n\r\n<div id=\"fs-id1170572404260\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {P}_{0}{e}^{rt}&amp; =\\hfill &amp; K-{P}_{0}\\hfill \\\\ \\hfill {e}^{rt}&amp; =\\hfill &amp; \\frac{K-{P}_{0}}{{P}_{0}}\\hfill \\\\ \\hfill \\text{ln}{e}^{rt}&amp; =\\hfill &amp; \\text{ln}\\frac{K-{P}_{0}}{{P}_{0}}\\hfill \\\\ \\hfill rt&amp; =\\hfill &amp; \\text{ln}\\frac{K-{P}_{0}}{{P}_{0}}\\hfill \\\\ \\hfill t&amp; =\\hfill &amp; \\frac{1}{r}\\text{ln}\\frac{K-{P}_{0}}{{P}_{0}}.\\hfill \\end{array}[\/latex]<\/div>\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">If [latex]P_0 &gt; K[\/latex] (initial population exceeds carrying capacity), this expression becomes undefined because we'd be taking the natural logarithm of a negative number. In this case, the population decreases monotonically toward [latex]K[\/latex] with no inflection point.<\/section>The inflection point marks where the population growth changes from accelerating to decelerating. This is where the \"leveling off\" begins as the population approaches the carrying capacity and growth becomes limited by available resources.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170572604605\" data-type=\"problem\">\r\n<p id=\"fs-id1170572604607\">A population of rabbits in a meadow is observed to be [latex]200[\/latex] rabbits at time [latex]t=0[\/latex]. After a month, the rabbit population is observed to have increased by [latex]4\\text{%}[\/latex]. Using an initial population of [latex]200[\/latex] and a growth rate of [latex]0.04[\/latex], with a carrying capacity of [latex]750[\/latex] rabbits,<\/p>\r\n\r\n<ol id=\"fs-id1170571638238\" type=\"a\">\r\n \t<li>Write the logistic differential equation and initial condition for this model.<\/li>\r\n \t<li>Draw a slope field for this logistic differential equation, and sketch the solution corresponding to an initial population of [latex]200[\/latex] rabbits.<\/li>\r\n \t<li>Solve the initial-value problem for [latex]P\\left(t\\right)[\/latex].<\/li>\r\n \t<li>Use the solution to predict the population after [latex]1[\/latex] year.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558898\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1170571649916\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170572455116\">First determine the values of [latex]r,K[\/latex], and [latex]{P}_{0}[\/latex]. Then create the initial-value problem, draw the direction field, and solve the problem.<\/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-id1170572596902\" data-type=\"solution\">\r\n<ol id=\"fs-id1170572596904\" type=\"a\">\r\n \t<li>\r\n<p id=\"fs-id1167024121600\">[latex]\\frac{dP}{dt}=0.04\\left(1-\\frac{P}{750}\\right),P\\left(0\\right)=200[\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p id=\"fs-id1167023779503\"><span data-type=\"newline\">\u00a0<\/span><\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234151\/CNX_Calc_Figure_08_04_005.jpg\" alt=\"A direction field with horizontal lines on the x-axis and y = 15. The other lines are vertical, except for those curving into the x-axis and y = 15. A solution is drawn that crosses the y-axis at about (0, 4) and asymptotically approaches y = 15.\" width=\"325\" height=\"333\" data-media-type=\"image\/jpeg\" \/> Figure 4.[\/caption]<\/li>\r\n \t<li>\r\n<p id=\"fs-id1167023779524\">[latex]P\\left(t\\right)=\\frac{3000{e}^{.04t}}{11+4{e}^{.04t}}[\/latex]<\/p>\r\n<\/li>\r\n \t<li>\r\n<p id=\"fs-id1167023916424\">After [latex]12[\/latex] months, the population will be [latex]P\\left(12\\right)\\approx 278[\/latex] rabbits.<\/p>\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]311314[\/ohm_question]<\/section>","rendered":"<h1>Solving the Logistic Differential Equation<\/h1>\n<p id=\"fs-id1170571618234\">The logistic differential equation is an autonomous differential equation, so we can use separation of variables to find the general solution.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\"><strong>Step 1: Find the equilibrium solutions<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Setting the right-hand side equal to zero gives us [latex]P = 0[\/latex] and [latex]P = K[\/latex] as constant solutions.<\/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 = 0[\/latex]: If no organisms are present, the population will never grow<\/li>\n<li class=\"whitespace-normal break-words\">When [latex]P = K[\/latex]: If the population starts at carrying capacity, it remains constant<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Step 2: Separate the variables<\/strong><\/p>\n<p class=\"whitespace-pre-wrap break-words\">Rewrite the differential equation as:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\frac{dP}{dt} = \\frac{rP(K-P)}{K}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Multiply both sides by [latex]dt[\/latex] and divide by [latex]P(K-P)[\/latex]:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\frac{dP}{P(K-P)} = \\frac{r}{K}dt[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 3: Prepare for integration<\/strong><\/p>\n<p class=\"whitespace-pre-wrap break-words\">Multiply both sides by [latex]K[\/latex]:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\int \\frac{K}{P(K-P)}dP = \\int r dt[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 4: Use partial fraction decomposition<\/strong><\/p>\n<p class=\"whitespace-pre-wrap break-words\">The left side can be decomposed using partial fractions:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\frac{K}{P(K-P)} = \\frac{1}{P} + \\frac{1}{K-P}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><em>You can verify this decomposition by finding a common denominator on the right side and checking that it equals the left side.<\/em><\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 5: Integrate both sides<\/strong><\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\int \\left(\\frac{1}{P} + \\frac{1}{K-P}\\right)dP = \\int r dt[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\ln|P| - \\ln|K-P| = rt + C[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\ln\\left|\\frac{P}{K-P}\\right| = rt + C[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 6: Solve for [latex]P[\/latex]<\/strong><\/p>\n<p class=\"whitespace-pre-wrap break-words\">Exponentiate both sides to eliminate the natural logarithm:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]e^{\\ln\\left|\\frac{P}{K-P}\\right|} = e^{rt+C}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\left|\\frac{P}{K-P}\\right| = e^C \\cdot e^{rt}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Defining [latex]C_1 = e^C[\/latex], we get: [latex]\\frac{P}{K-P} = C_1 e^{rt}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 7: Solve for [latex]P(t)[\/latex]<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Starting from [latex]\\frac{P}{K-P} = C_1 e^{rt}[\/latex], multiply both sides by [latex]K-P[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P = C_1 e^{rt}(K-P)[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P = C_1 K e^{rt} - C_1 P e^{rt}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Collect all terms containing [latex]P[\/latex] on the left side:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]P + C_1 P e^{rt} = C_1 K e^{rt}[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Factor out [latex]P[\/latex] and solve:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]P(1 + C_1 e^{rt}) = C_1 K e^{rt}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P(t) = \\frac{C_1 K e^{rt}}{1 + C_1 e^{rt}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 8: Find the constant [latex]C_1[\/latex]<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To determine [latex]C_1[\/latex], use the initial condition. When [latex]t = 0[\/latex], [latex]P(0) = P_0[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Substitute into [latex]\\frac{P}{K-P} = C_1 e^{rt}[\/latex]:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\frac{P_0}{K-P_0} = C_1 e^{r(0)} = C_1[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Therefore: [latex]C_1 = \\frac{P_0}{K-P_0}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Step 9: Complete the solution<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Substitute [latex]C_1 = \\frac{P_0}{K-P_0}[\/latex] into our equation:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]P(t) = \\frac{C_1 K e^{rt}}{1 + C_1 e^{rt}} = \\frac{\\frac{P_0}{K-P_0} \\cdot K e^{rt}}{1 + \\frac{P_0}{K-P_0} e^{rt}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Multiply numerator and denominator by [latex]\\left(K-{P}_{0}\\right)[\/latex] to simplify:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill P\\left(t\\right)& =\\frac{\\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}\\hfill \\\\ & =\\frac{\\frac{{P}_{0}}{K-{P}_{0}}K{e}^{rt}}{1+\\frac{{P}_{0}}{K-{P}_{0}}{e}^{rt}}\\cdot \\frac{K-{P}_{0}}{K-{P}_{0}}\\hfill \\\\ & =\\frac{{P}_{0}K{e}^{rt}}{\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}}.\\hfill \\end{array}[\/latex]<\/p>\n<\/section>\n<p id=\"fs-id1170572111539\">We state this result as a theorem.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>logistic growth solution<\/h3>\n<p id=\"fs-id1170571728769\">Consider the logistic differential equation subject to an initial population of [latex]{P}_{0}[\/latex] with carrying capacity [latex]K[\/latex] and growth rate [latex]r[\/latex]. The solution to the corresponding initial-value problem is:<\/p>\n<div id=\"fs-id1170571597871\" style=\"text-align: center;\" data-type=\"equation\">[latex]P\\left(t\\right)=\\dfrac{{P}_{0}K{e}^{rt}}{\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}}[\/latex].<\/div>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">Finding the Point of Inflection<\/h3>\n<p class=\"whitespace-normal break-words\">The logistic growth curve has a point of inflection where the growth rate changes from increasing to decreasing. To find this point, we set the second derivative equal to zero.<\/p>\n<p class=\"whitespace-normal break-words\">Starting with our solution, we can find the derivatives:<\/p>\n<div id=\"fs-id1170572307551\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill P\\left(t\\right)& =\\hfill & \\frac{{P}_{0}K{e}^{rt}}{\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}}\\hfill \\\\ \\hfill {P}^{\\prime }\\left(t\\right)& =\\hfill & \\frac{r{P}_{0}K\\left(K-{P}_{0}\\right){e}^{rt}}{{\\left(\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}\\right)}^{2}}\\hfill \\\\ \\hfill P^{\\prime\\prime}\\left(t\\right)& =\\hfill & \\frac{{r}^{2}{P}_{0}K{\\left(K-{P}_{0}\\right)}^{2}{e}^{rt}-{r}^{2}{P}_{0}{}^{2}K\\left(K-{P}_{0}\\right){e}^{2rt}}{{\\left(\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}\\right)}^{3}}\\hfill \\\\ & =\\hfill & \\frac{{r}^{2}{P}_{0}K\\left(K-{P}_{0}\\right){e}^{rt}\\left(\\left(K-{P}_{0}\\right)-{P}_{0}{e}^{rt}\\right)}{{\\left(\\left(K-{P}_{0}\\right)+{P}_{0}{e}^{rt}\\right)}^{3}}.\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1170571712032\">Setting the numerator equal to zero,<\/p>\n<div id=\"fs-id1170572509015\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{r}^{2}{P}_{0}K\\left(K-{P}_{0}\\right){e}^{rt}\\left(\\left(K-{P}_{0}\\right)-{P}_{0}{e}^{rt}\\right)=0[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572344215\">As long as [latex]{P}_{0}\\ne K[\/latex], the entire quantity before and including [latex]{e}^{rt}[\/latex] is nonzero, so we can divide it out:<\/p>\n<div id=\"fs-id1170572147551\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left(K-{P}_{0}\\right)-{P}_{0}{e}^{rt}=0[\/latex].<\/div>\n<p id=\"fs-id1170572236506\">Solving for [latex]t[\/latex],<\/p>\n<div id=\"fs-id1170572404260\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {P}_{0}{e}^{rt}& =\\hfill & K-{P}_{0}\\hfill \\\\ \\hfill {e}^{rt}& =\\hfill & \\frac{K-{P}_{0}}{{P}_{0}}\\hfill \\\\ \\hfill \\text{ln}{e}^{rt}& =\\hfill & \\text{ln}\\frac{K-{P}_{0}}{{P}_{0}}\\hfill \\\\ \\hfill rt& =\\hfill & \\text{ln}\\frac{K-{P}_{0}}{{P}_{0}}\\hfill \\\\ \\hfill t& =\\hfill & \\frac{1}{r}\\text{ln}\\frac{K-{P}_{0}}{{P}_{0}}.\\hfill \\end{array}[\/latex]<\/div>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">If [latex]P_0 > K[\/latex] (initial population exceeds carrying capacity), this expression becomes undefined because we&#8217;d be taking the natural logarithm of a negative number. In this case, the population decreases monotonically toward [latex]K[\/latex] with no inflection point.<\/section>\n<p>The inflection point marks where the population growth changes from accelerating to decelerating. This is where the &#8220;leveling off&#8221; begins as the population approaches the carrying capacity and growth becomes limited by available resources.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170572604605\" data-type=\"problem\">\n<p id=\"fs-id1170572604607\">A population of rabbits in a meadow is observed to be [latex]200[\/latex] rabbits at time [latex]t=0[\/latex]. After a month, the rabbit population is observed to have increased by [latex]4\\text{%}[\/latex]. Using an initial population of [latex]200[\/latex] and a growth rate of [latex]0.04[\/latex], with a carrying capacity of [latex]750[\/latex] rabbits,<\/p>\n<ol id=\"fs-id1170571638238\" type=\"a\">\n<li>Write the logistic differential equation and initial condition for this model.<\/li>\n<li>Draw a slope field for this logistic differential equation, and sketch the solution corresponding to an initial population of [latex]200[\/latex] rabbits.<\/li>\n<li>Solve the initial-value problem for [latex]P\\left(t\\right)[\/latex].<\/li>\n<li>Use the solution to predict the population after [latex]1[\/latex] year.<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Hint<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571649916\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170572455116\">First determine the values of [latex]r,K[\/latex], and [latex]{P}_{0}[\/latex]. Then create the initial-value problem, draw the direction field, and solve the problem.<\/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=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572596902\" data-type=\"solution\">\n<ol id=\"fs-id1170572596904\" type=\"a\">\n<li>\n<p id=\"fs-id1167024121600\">[latex]\\frac{dP}{dt}=0.04\\left(1-\\frac{P}{750}\\right),P\\left(0\\right)=200[\/latex]<\/p>\n<\/li>\n<li>\n<p id=\"fs-id1167023779503\"><span data-type=\"newline\">\u00a0<\/span><\/p>\n<figure style=\"width: 325px\" 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\/11234151\/CNX_Calc_Figure_08_04_005.jpg\" alt=\"A direction field with horizontal lines on the x-axis and y = 15. The other lines are vertical, except for those curving into the x-axis and y = 15. A solution is drawn that crosses the y-axis at about (0, 4) and asymptotically approaches y = 15.\" width=\"325\" height=\"333\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 4.<\/figcaption><\/figure>\n<\/li>\n<li>\n<p id=\"fs-id1167023779524\">[latex]P\\left(t\\right)=\\frac{3000{e}^{.04t}}{11+4{e}^{.04t}}[\/latex]<\/p>\n<\/li>\n<li>\n<p id=\"fs-id1167023916424\">After [latex]12[\/latex] months, the population will be [latex]P\\left(12\\right)\\approx 278[\/latex] rabbits.<\/p>\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=\"ohm311314\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=311314&theme=lumen&iframe_resize_id=ohm311314&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":26,"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\/1835"}],"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":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1835\/revisions"}],"predecessor-version":[{"id":2297,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1835\/revisions\/2297"}],"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\/1835\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1835"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1835"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1835"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1835"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}