{"id":286,"date":"2021-03-25T03:02:51","date_gmt":"2021-03-25T03:02:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&p=286"},"modified":"2022-06-27T17:16:15","modified_gmt":"2022-06-27T17:16:15","slug":"putting-it-together-differential-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/putting-it-together-differential-equations\/","title":{"raw":"Putting It Together: Differential Equations","rendered":"Putting It Together: Differential Equations"},"content":{"raw":"
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Chapter Opener: Examining the Carrying Capacity of a Deer Population<\/h3>\r\n
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<\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]\"This (credit: modification of work by Rachel Kramer, Flickr)[\/caption]<\/figure>\r\n

Let\u2019s consider the population of white-tailed deer (Odocoileus virginianus<\/em>) in the state of Kentucky. The Kentucky Department of Fish and Wildlife Resources (KDFWR) sets guidelines for hunting and fishing in the state. Before the hunting season of [latex]2004[\/latex], it estimated a population of [latex]900,000[\/latex] deer. Johnson notes: \"A deer population that has plenty to eat and is not hunted by humans or other predators will double every three years.\" (George Johnson, \"The Problem of Exploding Deer Populations Has No Attractive Solutions,\" January [latex]12,2001[\/latex], accessed April 9, 2015, http:\/\/www.txtwriter.com\/onscience\/Articles\/deerpops.html.) This observation corresponds to a rate of increase [latex]r=\\frac{\\text{ln}\\left(2\\right)}{3}=0.2311[\/latex], so the approximate growth rate is [latex]23.11\\text{%}[\/latex] per year.<\/em> (This assumes that the population grows exponentially, which is reasonable\u2013\u2013at least in the short term\u2013\u2013with plentiful food supply and no predators.) The KDFWR also reports deer population densities for [latex]32[\/latex] counties in Kentucky, the average of which is approximately [latex]27[\/latex] deer per square mile. Suppose this is the deer density for the whole state ([latex]39,732[\/latex] square miles). The carrying capacity [latex]K[\/latex] is [latex]39,732[\/latex] square miles times [latex]27[\/latex] deer per square mile, or [latex]1,072,764[\/latex] deer.<\/em><\/p>\r\n\r\n

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  1. For this application, we have [latex]{P}_{0}=900,000,K=1,072,764[\/latex], and [latex]r=0.2311[\/latex]. Substitute these values into the logistic differential equation\u00a0and form the initial-value problem.<\/li>\r\n \t
  2. Solve the initial-value problem from part a.<\/li>\r\n \t
  3. According to this model, what will be the population in [latex]3[\/latex] years? Recall that the doubling time predicted by Johnson for the deer population was [latex]3[\/latex] years. How do these values compare?<\/li>\r\n \t
  4. Suppose the population managed to reach [latex]1,200,000[\/latex] deer. What does the logistic equation predict will happen to the population in this scenario?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n
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    1. The initial value problem is\r\n<\/span>\r\n[latex]\\frac{dP}{dt}=0.2311P\\left(1-\\frac{P}{1,072,764}\\right),P\\left(0\\right)=900,000[\/latex].<\/li>\r\n \t
    2. The logistic equation is an autonomous differential equation, so we can use the method of separation of variables.\r\n<\/span>\r\nStep 1: Setting the right-hand side equal to zero gives [latex]P=0[\/latex] and [latex]P=1,072,764[\/latex]. This means that if the population starts at zero it will never change, and if it starts at the carrying capacity, it will never change.\r\n<\/span>\r\nStep 2: Rewrite the differential equation and multiply both sides by:\r\n<\/span>\r\n
      [latex]\\begin{array}{ccc}\\hfill \\frac{dP}{dt}& =\\hfill & 0.2311P\\left(\\frac{1,072,764-P}{1,072,764}\\right)\\hfill \\\\ \\hfill dP& =\\hfill & 0.2311P\\left(\\frac{1,072,764-P}{1,072,764}\\right)dt.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nDivide both sides by [latex]P\\left(1,072,764-P\\right)\\text{:}[\/latex] \r\n<\/span>\r\n
      [latex]\\frac{dP}{P\\left(1,072,764-P\\right)}=\\frac{0.2311}{1,072,764}dt[\/latex].<\/div>\r\n\r\n<\/span>\r\nStep 3: Integrate both sides of the equation using partial fraction decomposition:\r\n<\/span>\r\n
      [latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{dP}{P\\left(1,072,764-P\\right)}}& =\\hfill & {\\displaystyle\\int \\frac{0.2311}{1,072,764}dt}\\hfill \\\\ \\hfill \\frac{1}{1,072,764}{\\displaystyle\\int \\left(\\frac{1}{P}+\\frac{1}{1,072,764-P}\\right)dP}& =\\hfill & \\frac{0.2311t}{1,072,764}+C\\hfill \\\\ \\hfill \\frac{1}{1,072,764}\\left(\\text{ln}|P|-\\text{ln}|1,072,764-P|\\right)& =\\hfill & \\frac{0.2311t}{1,072,764}+C.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nStep 4: Multiply both sides by [latex]1,072,764[\/latex] and use the quotient rule for logarithms:\r\n<\/span>\r\n
      [latex]\\text{ln}|\\frac{P}{1,072,764-P}|=0.2311t+{C}_{1}[\/latex].<\/div>\r\n\r\n<\/span>\r\nHere [latex]{C}_{1}=1,072,764C[\/latex]. Next exponentiate both sides and eliminate the absolute value:\r\n<\/span>\r\n
      [latex]\\begin{array}{ccc}\\hfill {e}^{\\text{ln}|\\frac{P}{1,072,764-P}|}& =\\hfill & {e}^{0.2311t+{C}_{1}}\\hfill \\\\ \\hfill |\\frac{P}{1,072,764-P}|& =\\hfill & {C}_{2}{e}^{0.2311t}\\hfill \\\\ \\hfill \\frac{P}{1,072,764-P}& =\\hfill & {C}_{2}{e}^{0.2311t}.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nHere [latex]{C}_{2}={e}^{{C}_{1}}[\/latex] but after eliminating the absolute value, it can be negative as well. Now solve for:\r\n<\/span>\r\n
      [latex]\\begin{array}{ccc}\\hfill P& =\\hfill & {C}_{2}{e}^{0.2311t}\\left(1,072,764-P\\right).\\hfill \\\\ \\hfill P& =\\hfill & 1,072,764{C}_{2}{e}^{0.2311t}-{C}_{2}P{e}^{0.2311t}\\hfill \\\\ \\hfill P+{C}_{2}P{e}^{0.2311t}& =\\hfill & 1,072,764{C}_{2}{e}^{0.2311t}\\hfill \\\\ \\hfill P\\left(1+{C}_{2}{e}^{0.2311t}\\right)& =\\hfill & 1,072,764{C}_{2}{e}^{0.2311t}\\hfill \\\\ \\hfill P\\left(t\\right)& =\\hfill & \\frac{1,072,764{C}_{2}{e}^{0.2311t}}{1+{C}_{2}{e}^{0.2311t}}.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nStep 5: To determine the value of [latex]{C}_{2}[\/latex], it is actually easier to go back a couple of steps to where [latex]{C}_{2}[\/latex] was defined. In particular, use the equation\r\n<\/span>\r\n
      [latex]\\frac{P}{1,072,764-P}={C}_{2}{e}^{0.2311t}[\/latex].<\/div>\r\n\r\n<\/span>\r\nThe initial condition is [latex]P\\left(0\\right)=900,000[\/latex]. Replace [latex]P[\/latex] with [latex]900,000[\/latex] and [latex]t[\/latex] with zero:\r\n<\/span>\r\n
      [latex]\\begin{array}{ccc}\\hfill \\frac{P}{1,072,764-P}& =\\hfill & {C}_{2}{e}^{0.2311t}\\hfill \\\\ \\hfill \\frac{900,000}{1,072,764 - 900,000}& =\\hfill & {C}_{2}{e}^{0.2311\\left(0\\right)}\\hfill \\\\ \\hfill \\frac{900,000}{172,764}& =\\hfill & {C}_{2}\\hfill \\\\ \\hfill {C}_{2}& =\\hfill & \\frac{25,000}{4,799}\\approx 5.209.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nTherefore\r\n<\/span>\r\n
      [latex]\\begin{array}{cc}\\hfill P\\left(t\\right)& =\\frac{1,072,764\\left(\\frac{25000}{4799}\\right){e}^{0.2311t}}{1+\\left(\\frac{25000}{4799}\\right){e}^{0.2311t}}\\hfill \\\\ & =\\frac{1,072,764\\left(25000\\right){e}^{0.2311t}}{4799+25000{e}^{0.2311t}}.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nDividing the numerator and denominator by [latex]25,000[\/latex] gives\r\n<\/span>\r\n
      [latex]P\\left(t\\right)=\\frac{1,072,764{e}^{0.2311t}}{0.19196+{e}^{0.2311t}}[\/latex].<\/div>\r\n\r\n<\/span>\r\nThe graph below depicts this equation.\r\n<\/span>\r\n
      [caption id=\"\" align=\"aligncenter\" width=\"487\"]\"A Logistic curve for the deer population with an initial population of [latex]900,000[\/latex] deer.[\/caption]<\/figure>\r\n<\/li>\r\n \t
    3. Using this model we can predict the population in [latex]3[\/latex] years.\r\n<\/span>\r\n
      [latex]P\\left(3\\right)=\\frac{1,072,764{e}^{0.2311\\left(3\\right)}}{0.19196+{e}^{0.2311\\left(3\\right)}}\\approx 978,830\\text{ deer}[\/latex]<\/div>\r\n\r\n<\/span>\r\nThis is far short of twice the initial population of [latex]900,000[\/latex]. Remember that the doubling time is based on the assumption that the growth rate never changes, but the logistic model takes this possibility into account.<\/li>\r\n \t
    4. If the population reached [latex]1,200,000[\/latex] deer, then the new initial-value problem would be\r\n<\/span>\r\n
      [latex]\\frac{dP}{dt}=0.2311P\\left(1-\\frac{P}{1,072,764}\\right),P\\left(0\\right)=1,200,000[\/latex].<\/div>\r\n\r\n<\/span>\r\nThe general solution to the differential equation would remain the same.\r\n<\/span>\r\n
      [latex]P\\left(t\\right)=\\frac{1,072,764{C}_{2}{e}^{0.2311t}}{1+{C}_{2}{e}^{0.2311t}}[\/latex]<\/div>\r\n\r\n<\/span>\r\nTo determine the value of the constant, return to the equation\r\n<\/span>\r\n
      [latex]\\frac{P}{1,072,764-P}={C}_{2}{e}^{0.2311t}[\/latex].<\/div>\r\n\r\n<\/span>\r\nSubstituting the values [latex]t=0[\/latex] and [latex]P=1,200,000[\/latex], you get\r\n<\/span>\r\n
      [latex]\\begin{array}{ccc}\\hfill {C}_{2}{e}^{0.2311\\left(0\\right)}& =\\hfill & \\frac{1,200,000}{1,072,764 - 1,200,000}\\hfill \\\\ \\hfill {C}_{2}& =\\hfill & -\\frac{100,000}{10,603}\\approx -9.431.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nTherefore\r\n<\/span>\r\n
      [latex]\\begin{array}{cc}\\hfill P\\left(t\\right)& =\\frac{1,072,764{C}_{2}{e}^{0.2311t}}{1+{C}_{2}{e}^{0.2311t}}\\hfill \\\\ & =\\frac{1,072,764\\left(-\\frac{100,000}{10,603}\\right){e}^{0.2311t}}{1+\\left(-\\frac{100,000}{10,603}\\right){e}^{0.2311t}}\\hfill \\\\ & =-\\frac{107,276,400,000{e}^{0.2311t}}{100,000{e}^{0.2311t}-10,603}\\hfill \\\\ & \\approx \\frac{10,117,551{e}^{0.2311t}}{9.43129{e}^{0.2311t}-1}.\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n<\/span>\r\nThis equation is graphed below.\r\n<\/span>\r\n
      [caption id=\"\" align=\"aligncenter\" width=\"325\"]\"A Logistic curve for the deer population with an initial population of [latex]1,200,000[\/latex] deer.[\/caption]<\/figure>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n \r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Chapter Opener: Examining the Carrying Capacity of a Deer Population\r\n\r\n