{"id":801,"date":"2025-06-20T17:13:35","date_gmt":"2025-06-20T17:13:35","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=801"},"modified":"2025-09-10T14:52:39","modified_gmt":"2025-09-10T14:52:39","slug":"basics-of-differential-equations-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/basics-of-differential-equations-learn-it-4\/","title":{"raw":"Basics of Differential Equations: Learn It 4","rendered":"Basics of Differential Equations: Learn It 4"},"content":{"raw":"<h2>Initial-Value Problems<\/h2>\r\n<p class=\"whitespace-normal break-words\">Usually a differential equation has infinitely many solutions, so we need to ask: which solution do we actually want? To choose one specific solution, we need more information.<\/p>\r\n<p class=\"whitespace-normal break-words\">This additional information often comes in the form of an <strong>initial value<\/strong>\u2014a condition that tells us the value of the function (and possibly its derivatives) at a specific point.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>initial-value problem<\/h3>\r\n<p class=\"whitespace-normal break-words\">An <strong>initial-value problem<\/strong> consists of:<\/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\">A differential equation<\/li>\r\n \t<li class=\"whitespace-normal break-words\">One or more initial values (conditions)<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Rule<\/strong>: The number of initial values needed equals the order of the differential equation.<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">These problems are called \"initial-value problems\" because the independent variable often represents time [latex]t[\/latex], and [latex]t = 0[\/latex] represents the starting point or \"initial\" moment of the problem.<\/section>The initial values allow us to pin down one specific curve from the infinite family of solutions, giving us the particular solution that models our specific situation.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170573720384\" data-type=\"problem\">\r\n<p id=\"fs-id1170571021375\">Verify that the function [latex]y=2{e}^{-2t}+{e}^{t}[\/latex] is a solution to the initial-value problem<\/p>\r\n\r\n<div id=\"fs-id1170573368596\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }+2y=3{e}^{t},y\\left(0\\right)=3[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558891\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558891\"]\r\n<div id=\"fs-id1170573400425\" data-type=\"solution\">\r\n<p id=\"fs-id1170573366765\">For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initial condition. To show that [latex]y[\/latex] satisfies the differential equation, we start by calculating [latex]{y}^{\\prime }[\/latex]. This gives [latex]{y}^{\\prime }=-4{e}^{-2t}+{e}^{t}[\/latex]. Next we substitute both [latex]y[\/latex] and [latex]{y}^{\\prime }[\/latex] into the left-hand side of the differential equation and simplify:<\/p>\r\n\r\n<div id=\"fs-id1170571448884\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{y}^{\\prime }+2y\\hfill &amp; =\\left(-4{e}^{-2t}+{e}^{t}\\right)+2\\left(2{e}^{-2t}+{e}^{t}\\right)\\hfill \\\\ &amp; =-4{e}^{-2t}+{e}^{t}+4{e}^{-2t}+2{e}^{t}\\hfill \\\\ &amp; =3{e}^{t}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573381839\">This is equal to the right-hand side of the differential equation, so [latex]y=2{e}^{-2t}+{e}^{t}[\/latex] solves the differential equation. Next we calculate [latex]y\\left(0\\right)\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170573407815\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}y\\left(0\\right)\\hfill &amp; =2{e}^{-2\\left(0\\right)}+{e}^{0}\\hfill \\\\ &amp; =2+1\\hfill \\\\ &amp; =3.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571120266\">This result verifies the initial value. Therefore the given function satisfies the initial-value problem.<\/p>\r\n\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 the example above.<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tS3d3924OQg?controls=0&amp;start=0&amp;end=171&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>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.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.1.5_0to171_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.1.5\" here (opens in new window)<\/a>.<\/section>\r\n<p class=\"whitespace-normal break-words\">The initial-value problem in the previous example consisted of two essential parts:<\/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\">The differential equation: [latex]y' + 2y = 3e^t[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The initial condition: [latex]y(0) = 3[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\">Together, these two pieces formed the complete initial-value problem.<\/p>\r\n<p class=\"whitespace-normal break-words\">For our example, the general solution to the differential equation [latex]y' + 2y = 3e^t[\/latex] is [latex]y = 2e^{-2t} + Ce^t[\/latex]. This represents an entire family of curves, each corresponding to a different value of [latex]C[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The initial condition [latex]y(0) = 3[\/latex] tells us which curve from this family we actually want. When we substitute [latex]t = 0[\/latex] and [latex]y = 3[\/latex] into the general solution, we find that [latex]C = 1[\/latex], giving us the particular solution [latex]y = 2e^{-2t} + e^t[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Figure 2 shows this family of solutions, with our particular solution highlighted. You can see how the initial condition picks out exactly one curve from the infinite family.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_01_002\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233914\/CNX_Calc_Figure_08_01_002.jpg\" alt=\"A graph of a family of solutions to the differential equation y\u2019 + 2 y = 3 e ^ t, which are of the form y = 2 e ^ (-2 t) + C e ^ t. The versions with C = 1, 0.5, and -0.2 are shown, among others not labeled. For all values of C, the function increases rapidly for t &lt; 0 as t goes to negative infinity. For C &gt; 0, the function changes direction and increases in a gentle curve as t goes to infinity. Larger values of C have a tighter curve closer to the y axis and at a higher y value. For C = 0, the function goes to 0 as t goes to infinity. For C &lt; 0, the function continues to decrease as t goes to infinity.\" width=\"325\" height=\"332\" \/> Figure 2: A family of solutions to the differential equation y\u2032+2y=3et. The particular solution y=2e\u22122t+et is labeled.[\/caption]<\/figure>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1170573623483\" data-type=\"problem\">\r\n<p id=\"fs-id1170573449571\">Solve the following initial-value problem:<\/p>\r\n\r\n<div id=\"fs-id1170571227292\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }=3{e}^{x}+{x}^{2}-4,y\\left(0\\right)=5[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558889\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558889\"]\r\n<div id=\"fs-id1170571069707\" data-type=\"solution\">\r\n<p id=\"fs-id1170573412250\">The first step in solving this initial-value problem is to find a general family of solutions. To do this, we find an antiderivative of both sides of the differential equation<\/p>\r\n\r\n<div id=\"fs-id1170573413466\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int {y}^{\\prime }dx=\\displaystyle\\int \\left(3{e}^{x}+{x}^{2}-4\\right)dx[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571506197\">namely,<\/p>\r\n\r\n<div id=\"fs-id1170571122530\" style=\"text-align: center;\" data-type=\"equation\">[latex]y+{C}_{1}=3{e}^{x}+\\frac{1}{3}{x}^{3}-4x+{C}_{2}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573277276\">We are able to integrate both sides because the <em data-effect=\"italics\">y<\/em> term appears by itself. Notice that there are two integration constants: [latex]{C}_{1}[\/latex] and [latex]{C}_{2}[\/latex]. Solving the previous equation for [latex]y[\/latex] gives<\/p>\r\n\r\n<div id=\"fs-id1170571131874\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=3{e}^{x}+\\frac{1}{3}{x}^{3}-4x+{C}_{2}-{C}_{1}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573425777\">Because [latex]{C}_{1}[\/latex] and [latex]{C}_{2}[\/latex] are both constants, [latex]{C}_{2}-{C}_{1}[\/latex] is also a constant. We can therefore define [latex]C={C}_{2}-{C}_{1}[\/latex], which leads to the equation<\/p>\r\n\r\n<div id=\"fs-id1170571120799\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=3{e}^{x}+\\frac{1}{3}{x}^{3}-4x+C[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571098957\">Next we determine the value of [latex]C[\/latex]. To do this, we substitute [latex]x=0[\/latex] and [latex]y=5[\/latex] into our aforementioned equation and solve for [latex]C\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170571417379\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\hfill 5&amp; =\\hfill &amp; 3{e}^{0}+\\frac{1}{3}{0}^{3}-4\\left(0\\right)+C\\hfill \\\\ \\hfill 5&amp; =\\hfill &amp; 3+C\\hfill \\\\ \\hfill C&amp; =\\hfill &amp; 2.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571115031\">Now we substitute the value [latex]C=2[\/latex] into our equation. The solution to the initial-value problem is [latex]y=3{e}^{x}+\\frac{1}{3}{x}^{3}-4x+2[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170571098825\" data-type=\"commentary\">\r\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\r\n<p id=\"fs-id1170571247622\">The difference between a general solution and a particular solution is that a general solution involves a family of functions, either explicitly or implicitly defined, of the independent variable. The initial value or values determine which particular solution in the family of solutions satisfies the desired conditions.<\/p>\r\n\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\/tS3d3924OQg?controls=0&amp;start=173&amp;end=287&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>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.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.1.5_173to287_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.1.5\" here (opens in new window)<\/a>.\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]311302[\/ohm_question]<\/section>","rendered":"<h2>Initial-Value Problems<\/h2>\n<p class=\"whitespace-normal break-words\">Usually a differential equation has infinitely many solutions, so we need to ask: which solution do we actually want? To choose one specific solution, we need more information.<\/p>\n<p class=\"whitespace-normal break-words\">This additional information often comes in the form of an <strong>initial value<\/strong>\u2014a condition that tells us the value of the function (and possibly its derivatives) at a specific point.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>initial-value problem<\/h3>\n<p class=\"whitespace-normal break-words\">An <strong>initial-value problem<\/strong> consists of:<\/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\">A differential equation<\/li>\n<li class=\"whitespace-normal break-words\">One or more initial values (conditions)<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Key Rule<\/strong>: The number of initial values needed equals the order of the differential equation.<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">These problems are called &#8220;initial-value problems&#8221; because the independent variable often represents time [latex]t[\/latex], and [latex]t = 0[\/latex] represents the starting point or &#8220;initial&#8221; moment of the problem.<\/section>\n<p>The initial values allow us to pin down one specific curve from the infinite family of solutions, giving us the particular solution that models our specific situation.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170573720384\" data-type=\"problem\">\n<p id=\"fs-id1170571021375\">Verify that the function [latex]y=2{e}^{-2t}+{e}^{t}[\/latex] is a solution to the initial-value problem<\/p>\n<div id=\"fs-id1170573368596\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }+2y=3{e}^{t},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=\"q44558891\">Show Solution<\/button><\/p>\n<div id=\"q44558891\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170573400425\" data-type=\"solution\">\n<p id=\"fs-id1170573366765\">For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initial condition. To show that [latex]y[\/latex] satisfies the differential equation, we start by calculating [latex]{y}^{\\prime }[\/latex]. This gives [latex]{y}^{\\prime }=-4{e}^{-2t}+{e}^{t}[\/latex]. Next we substitute both [latex]y[\/latex] and [latex]{y}^{\\prime }[\/latex] into the left-hand side of the differential equation and simplify:<\/p>\n<div id=\"fs-id1170571448884\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}{y}^{\\prime }+2y\\hfill & =\\left(-4{e}^{-2t}+{e}^{t}\\right)+2\\left(2{e}^{-2t}+{e}^{t}\\right)\\hfill \\\\ & =-4{e}^{-2t}+{e}^{t}+4{e}^{-2t}+2{e}^{t}\\hfill \\\\ & =3{e}^{t}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573381839\">This is equal to the right-hand side of the differential equation, so [latex]y=2{e}^{-2t}+{e}^{t}[\/latex] solves the differential equation. Next we calculate [latex]y\\left(0\\right)\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1170573407815\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}y\\left(0\\right)\\hfill & =2{e}^{-2\\left(0\\right)}+{e}^{0}\\hfill \\\\ & =2+1\\hfill \\\\ & =3.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571120266\">This result verifies the initial value. Therefore the given function satisfies the initial-value problem.<\/p>\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 the example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tS3d3924OQg?controls=0&amp;start=0&amp;end=171&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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.You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.1.5_0to171_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.1.5&#8221; here (opens in new window)<\/a>.<\/section>\n<p class=\"whitespace-normal break-words\">The initial-value problem in the previous example consisted of two essential parts:<\/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\">The differential equation: [latex]y' + 2y = 3e^t[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">The initial condition: [latex]y(0) = 3[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\">Together, these two pieces formed the complete initial-value problem.<\/p>\n<p class=\"whitespace-normal break-words\">For our example, the general solution to the differential equation [latex]y' + 2y = 3e^t[\/latex] is [latex]y = 2e^{-2t} + Ce^t[\/latex]. This represents an entire family of curves, each corresponding to a different value of [latex]C[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The initial condition [latex]y(0) = 3[\/latex] tells us which curve from this family we actually want. When we substitute [latex]t = 0[\/latex] and [latex]y = 3[\/latex] into the general solution, we find that [latex]C = 1[\/latex], giving us the particular solution [latex]y = 2e^{-2t} + e^t[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Figure 2 shows this family of solutions, with our particular solution highlighted. You can see how the initial condition picks out exactly one curve from the infinite family.<\/p>\n<figure id=\"CNX_Calc_Figure_08_01_002\">\n<figure style=\"width: 325px\" 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\/11233914\/CNX_Calc_Figure_08_01_002.jpg\" alt=\"A graph of a family of solutions to the differential equation y\u2019 + 2 y = 3 e ^ t, which are of the form y = 2 e ^ (-2 t) + C e ^ t. The versions with C = 1, 0.5, and -0.2 are shown, among others not labeled. For all values of C, the function increases rapidly for t &lt; 0 as t goes to negative infinity. For C &gt; 0, the function changes direction and increases in a gentle curve as t goes to infinity. Larger values of C have a tighter curve closer to the y axis and at a higher y value. For C = 0, the function goes to 0 as t goes to infinity. For C &lt; 0, the function continues to decrease as t goes to infinity.\" width=\"325\" height=\"332\" \/><figcaption class=\"wp-caption-text\">Figure 2: A family of solutions to the differential equation y\u2032+2y=3et. The particular solution y=2e\u22122t+et is labeled.<\/figcaption><\/figure>\n<\/figure>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1170573623483\" data-type=\"problem\">\n<p id=\"fs-id1170573449571\">Solve the following initial-value problem:<\/p>\n<div id=\"fs-id1170571227292\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }=3{e}^{x}+{x}^{2}-4,y\\left(0\\right)=5[\/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=\"q44558889\">Show Solution<\/button><\/p>\n<div id=\"q44558889\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571069707\" data-type=\"solution\">\n<p id=\"fs-id1170573412250\">The first step in solving this initial-value problem is to find a general family of solutions. To do this, we find an antiderivative of both sides of the differential equation<\/p>\n<div id=\"fs-id1170573413466\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\displaystyle\\int {y}^{\\prime }dx=\\displaystyle\\int \\left(3{e}^{x}+{x}^{2}-4\\right)dx[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571506197\">namely,<\/p>\n<div id=\"fs-id1170571122530\" style=\"text-align: center;\" data-type=\"equation\">[latex]y+{C}_{1}=3{e}^{x}+\\frac{1}{3}{x}^{3}-4x+{C}_{2}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573277276\">We are able to integrate both sides because the <em data-effect=\"italics\">y<\/em> term appears by itself. Notice that there are two integration constants: [latex]{C}_{1}[\/latex] and [latex]{C}_{2}[\/latex]. Solving the previous equation for [latex]y[\/latex] gives<\/p>\n<div id=\"fs-id1170571131874\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=3{e}^{x}+\\frac{1}{3}{x}^{3}-4x+{C}_{2}-{C}_{1}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573425777\">Because [latex]{C}_{1}[\/latex] and [latex]{C}_{2}[\/latex] are both constants, [latex]{C}_{2}-{C}_{1}[\/latex] is also a constant. We can therefore define [latex]C={C}_{2}-{C}_{1}[\/latex], which leads to the equation<\/p>\n<div id=\"fs-id1170571120799\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=3{e}^{x}+\\frac{1}{3}{x}^{3}-4x+C[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571098957\">Next we determine the value of [latex]C[\/latex]. To do this, we substitute [latex]x=0[\/latex] and [latex]y=5[\/latex] into our aforementioned equation and solve for [latex]C\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1170571417379\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{}\\\\ \\hfill 5& =\\hfill & 3{e}^{0}+\\frac{1}{3}{0}^{3}-4\\left(0\\right)+C\\hfill \\\\ \\hfill 5& =\\hfill & 3+C\\hfill \\\\ \\hfill C& =\\hfill & 2.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571115031\">Now we substitute the value [latex]C=2[\/latex] into our equation. The solution to the initial-value problem is [latex]y=3{e}^{x}+\\frac{1}{3}{x}^{3}-4x+2[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1170571098825\" data-type=\"commentary\">\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\n<p id=\"fs-id1170571247622\">The difference between a general solution and a particular solution is that a general solution involves a family of functions, either explicitly or implicitly defined, of the independent variable. The initial value or values determine which particular solution in the family of solutions satisfies the desired conditions.<\/p>\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\/tS3d3924OQg?controls=0&amp;start=173&amp;end=287&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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. 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