{"id":1620,"date":"2021-07-22T17:01:05","date_gmt":"2021-07-22T17:01:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1620"},"modified":"2022-03-21T22:54:59","modified_gmt":"2022-03-21T22:54:59","slug":"initial-value-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/initial-value-problems\/","title":{"raw":"Initial-Value Problems","rendered":"Initial-Value Problems"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify an initial-value problem<\/li>\r\n \t<li>Identify whether a given function is a solution to a differential equation or an initial-value problem<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170573427887\">Usually a given differential equation has an infinite number of solutions, so it is natural to ask which one we want to use. To choose one solution, more information is needed. Some specific information that can be useful is an <span data-type=\"term\">initial value<\/span>, which is an ordered pair that is used to find a particular solution.<\/p>\r\n<p id=\"fs-id1170573423616\">A differential equation together with one or more initial values is called an <strong>initial-value problem<\/strong>. The general rule is that the number of initial values needed for an initial-value problem is equal to the order of the differential equation. For example, if we have the differential equation [latex]{y}^{\\prime }=2x[\/latex], then [latex]y\\left(3\\right)=7[\/latex] is an initial value, and when taken together, these equations form an initial-value problem. The differential equation [latex]y\\text{''}-3{y}^{\\prime }+2y=4{e}^{x}[\/latex] is second order, so we need two initial values. With initial-value problems of order greater than one, the same value should be used for the independent variable. An example of initial values for this second-order equation would be [latex]y\\left(0\\right)=2[\/latex] and [latex]{y}^{\\prime }\\left(0\\right)=-1[\/latex]. These two initial values together with the differential equation form an initial-value problem. These problems are so named because often the independent variable in the unknown function is [latex]t[\/latex], which represents time. Thus, a value of [latex]t=0[\/latex] represents the beginning of the problem.<\/p>\r\n\r\n<div id=\"fs-id1170573712028\" data-type=\"example\">\r\n<div id=\"fs-id1170573391210\" data-type=\"exercise\">\r\n<div id=\"fs-id1170573720384\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Verifying a Solution to an Initial-Value Problem<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example:\u00a0Verifying a Solution to an Initial-Value Problem[\/caption]\r\n\r\n<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\"><\/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.\r\n\r\nYou 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>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5719[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573246209\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170570975800\" data-type=\"exercise\">\r\n<div id=\"fs-id1170573365823\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170573365823\" data-type=\"problem\">\r\n<p id=\"fs-id1170573429527\">Verify that [latex]y=3{e}^{2t}+4\\sin{t}[\/latex] is a solution to the initial-value problem<\/p>\r\n\r\n<div id=\"fs-id1170573368474\" class=\"unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }-2y=4\\cos{t} - 8\\sin{t},y\\left(0\\right)=3[\/latex].<\/div>\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n<div id=\"fs-id1170571276500\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170573338054\">First verify that [latex]y[\/latex] solves the differential equation. Then check the initial value.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573415467\">In the previous example, the initial-value problem consisted of two parts. The first part was the differential equation [latex]{y}^{\\prime }+2y=3{e}^{x}[\/latex], and the second part was the initial value [latex]y\\left(0\\right)=3[\/latex]. These two equations together formed the initial-value problem.<\/p>\r\n<p id=\"fs-id1170573439347\">The same is true in general. An initial-value problem will consists of two parts: the differential equation and the initial condition. The differential equation has a family of solutions, and the initial condition determines the value of [latex]C[\/latex]. The family of solutions to the differential equation in the example is given by [latex]y=2{e}^{-2t}+C{e}^{t}[\/latex]. This family of solutions is shown in Figure 2, with the particular solution [latex]y=2{e}^{-2t}+{e}^{t}[\/latex] labeled.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_01_002\"><figcaption><\/figcaption>[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 [latex]t[\/latex] &lt; 0 as [latex]t[\/latex] goes to negative infinity. For C &gt; 0, the function changes direction and increases in a gentle curve as [latex]t[\/latex] goes to infinity. Larger values of C have a tighter curve closer to the [latex]y[\/latex]-axis and at a higher y value. For C = 0, the function goes to 0 as [latex]t[\/latex] goes to infinity. For C &lt; 0, the function continues to decrease as [latex]t[\/latex] goes to infinity.\" width=\"325\" height=\"332\" data-media-type=\"image\/jpeg\" \/> Figure 2. A family of solutions to the differential equation [latex]{y}^{\\prime }+2y=3{e}^{t}[\/latex]. The particular solution [latex]y=2{e}^{-2t}+{e}^{t}[\/latex] is labeled.[\/caption]<\/figure>\r\n<div id=\"fs-id1170571123767\" data-type=\"example\">\r\n<div id=\"fs-id1170571169678\" data-type=\"exercise\">\r\n<div id=\"fs-id1170573623483\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Solving an Initial-value Problem<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example:\u00a0Verifying a Solution to an Initial-Value Problem[\/caption]\r\n\r\n<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\"><\/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.\r\n\r\nYou 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<div id=\"fs-id1170571329487\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170571329490\" data-type=\"exercise\">\r\n<div id=\"fs-id1170573362845\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170573362845\" data-type=\"problem\">\r\n<p id=\"fs-id1170573362847\">Solve the initial-value problem<\/p>\r\n\r\n<div id=\"fs-id1170573570394\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }={x}^{2}-4x+3 - 6{e}^{x},y\\left(0\\right)=8[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1170571418500\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170573502068\">First take the antiderivative of both sides of the differential equation. Then substitute [latex]x=0[\/latex] and [latex]y=8[\/latex] into the resulting equation and solve for [latex]C[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558879\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558879\"]\r\n<div id=\"fs-id1170571455482\" data-type=\"solution\">\r\n<p id=\"fs-id1170571203542\">[latex]y=\\frac{1}{3}{x}^{3}-2{x}^{2}+3x - 6{e}^{x}+14[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170573396215\">In physics and engineering applications, we often consider the forces acting upon an object, and use this information to understand the resulting motion that may occur. For example, if we start with an object at Earth\u2019s surface, the primary force acting upon that object is gravity. Physicists and engineers can use this information, along with <span class=\"no-emphasis\" data-type=\"term\">Newton\u2019s second law of motion<\/span> (in equation form [latex]F=ma[\/latex], where [latex]F[\/latex] represents force, [latex]m[\/latex] represents mass, and [latex]a[\/latex] represents acceleration), to derive an equation that can be solved.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_01_003\"><figcaption><\/figcaption>\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\/11233917\/CNX_Calc_Figure_08_01_003.jpg\" alt=\"A picture of a baseball with an arrow underneath it pointing down. The arrow is labeled g = -9.8 m\/sec ^ 2.\" width=\"325\" height=\"212\" data-media-type=\"image\/jpeg\" \/> Figure 3. For a baseball falling in air, the only force acting on it is gravity (neglecting air resistance).[\/caption]<\/figure>\r\n<p id=\"fs-id1170573536932\">In Figure 3. we assume that the only force acting on a baseball is the force of gravity. This assumption ignores air resistance. (The force due to air resistance is considered in a later discussion.) The acceleration due to gravity at Earth\u2019s surface, [latex]g[\/latex], is approximately [latex]9.8{\\text{m\/s}}^{2}[\/latex]. We introduce a frame of reference, where Earth\u2019s surface is at a height of 0 meters. Let [latex]v\\left(t\\right)[\/latex] represent the velocity of the object in meters per second. If [latex]v\\left(t\\right)&gt;0[\/latex], the ball is rising, and if [latex]v\\left(t\\right)&lt;0[\/latex], the ball is falling (Figure 4).<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_08_01_004\"><figcaption><\/figcaption>\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\/11233920\/CNX_Calc_Figure_08_01_004.jpg\" alt=\"A picture of a baseball with an arrow above it pointing up and an arrow below it pointing down. The arrow pointing up is labeled v(t) &gt; 0, and the arrow pointing down is labeled v(t) &lt; 0.\" width=\"325\" height=\"352\" data-media-type=\"image\/jpeg\" \/> Figure 4. Possible velocities for the rising\/falling baseball.[\/caption]<\/figure>\r\n<p id=\"fs-id1170571509335\">Our goal is to solve for the velocity [latex]v\\left(t\\right)[\/latex] at any time [latex]t[\/latex]. To do this, we set up an initial-value problem. Suppose the mass of the ball is [latex]m[\/latex], where [latex]m[\/latex] is measured in kilograms. We use Newton\u2019s second law, which states that the force acting on an object is equal to its mass times its acceleration [latex]\\left(F=ma\\right)[\/latex]. Acceleration is the derivative of velocity, so [latex]a\\left(t\\right)={v}^{\\prime }\\left(t\\right)[\/latex]. Therefore the force acting on the baseball is given by [latex]F=m{v}^{\\prime }\\left(t\\right)[\/latex]. However, this force must be equal to the force of gravity acting on the object, which (again using Newton\u2019s second law) is given by [latex]{F}_{g}=\\text{-}mg[\/latex], since this force acts in a downward direction. Therefore we obtain the equation [latex]F={F}_{g}[\/latex], which becomes [latex]m{v}^{\\prime }\\left(t\\right)=\\text{-}mg[\/latex]. Dividing both sides of the equation by [latex]m[\/latex] gives the equation<\/p>\r\n\r\n<div id=\"fs-id1170573414975\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{v}^{\\prime }\\left(t\\right)=\\text{-}g[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571118060\">Notice that this differential equation remains the same regardless of the mass of the object.<\/p>\r\n<p id=\"fs-id1170571118063\">We now need an initial value. Because we are solving for velocity, it makes sense in the context of the problem to assume that we know the <span data-type=\"term\">initial velocity<\/span>, or the velocity at time [latex]t=0[\/latex]. This is denoted by [latex]v\\left(0\\right)={v}_{0}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1170571084135\" data-type=\"example\">\r\n<div id=\"fs-id1170571084137\" data-type=\"exercise\">\r\n<div id=\"fs-id1170571263710\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Velocity of a Moving Baseball<\/h3>\r\n<div id=\"fs-id1170571263710\" data-type=\"problem\">\r\n<p id=\"fs-id1170571273705\">A baseball is thrown upward from a height of [latex]3[\/latex] meters above Earth\u2019s surface with an initial velocity of [latex]10\\text{m\/s}[\/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15\\text{kg}[\/latex] at Earth\u2019s surface.<\/p>\r\n\r\n<ol id=\"fs-id1170571216089\" type=\"a\">\r\n \t<li>Find the velocity [latex]v\\left(t\\right)[\/latex] of the baseball at time [latex]t[\/latex].<\/li>\r\n \t<li>What is its velocity after [latex]2[\/latex] seconds?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558799\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558799\"]\r\n<div id=\"fs-id1170571270205\" data-type=\"solution\">\r\n<ol id=\"fs-id1170571270207\" type=\"a\">\r\n \t<li>From the preceding discussion, the differential equation that applies in this situation is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573401342\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{v}^{\\prime }\\left(t\\right)=\\text{-}g[\/latex],<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nwhere [latex]g=9.8{\\text{m\/s}}^{2}[\/latex]. The initial condition is [latex]v\\left(0\\right)={v}_{0}[\/latex], where [latex]{v}_{0}=10\\text{m\/s}\\text{.}[\/latex] Therefore the initial-value problem is [latex]{v}^{\\prime }\\left(t\\right)=-9.8{\\text{m\/s}}^{2},v\\left(0\\right)=10\\text{m\/s}\\text{.}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\nThe first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573289010\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int {v}^{\\prime }\\left(t\\right)dt}&amp; =\\hfill &amp; {\\displaystyle\\int -9.8dt}\\hfill \\\\ \\hfill v\\left(t\\right)&amp; =\\hfill &amp; -9.8t+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe next step is to solve for [latex]C[\/latex]. To do this, substitute [latex]t=0[\/latex] and [latex]v\\left(0\\right)=10\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573623436\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill v\\left(t\\right)&amp; =\\hfill &amp; -9.8t+C\\hfill \\\\ \\hfill v\\left(0\\right)&amp; =\\hfill &amp; -9.8\\left(0\\right)+C\\hfill \\\\ \\hfill 10&amp; =\\hfill &amp; C.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore [latex]C=10[\/latex] and the velocity function is given by [latex]v\\left(t\\right)=-9.8t+10[\/latex].<\/li>\r\n \t<li>To find the velocity after [latex]2[\/latex] seconds, substitute [latex]t=2[\/latex] into [latex]v\\left(t\\right)[\/latex]. <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571152980\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill v\\left(t\\right)&amp; =\\hfill &amp; -9.8t+10\\hfill \\\\ \\hfill v\\left(2\\right)&amp; =\\hfill &amp; -9.8\\left(2\\right)+10\\hfill \\\\ \\hfill v\\left(2\\right)&amp; =\\hfill &amp; -9.6.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThe units of velocity are meters per second. Since the answer is negative, the object is falling at a speed of [latex]9.6\\text{m\/s}\\text{.}[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170573419495\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170573419498\" data-type=\"exercise\">\r\n<div id=\"fs-id1170573352098\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170573352098\" data-type=\"problem\">\r\n<p id=\"fs-id1170573352100\">Suppose a rock falls from rest from a height of [latex]100[\/latex] meters and the only force acting on it is gravity. Find an equation for the velocity [latex]v\\left(t\\right)[\/latex] as a function of time, measured in meters per second.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558599\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558599\"]\r\n<div id=\"fs-id1170573318642\" data-type=\"commentary\" data-element-type=\"hint\">\r\n\r\nWhat is the initial velocity of the rock? Use this with the differential equation in the example: Velocity of a Moving Baseball to form an initial-value problem, then solve for [latex]v\\left(t\\right)[\/latex].\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558699\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558699\"]\r\n<div id=\"fs-id1170573493548\" data-type=\"solution\">\r\n<p id=\"fs-id1170573400886\">[latex]v\\left(t\\right)=-9.8t[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571302110\">A natural question to ask after solving this type of problem is how high the object will be above Earth\u2019s surface at a given point in time. Let [latex]s\\left(t\\right)[\/latex] denote the height above Earth\u2019s surface of the object, measured in meters. Because velocity is the derivative of position (in this case height), this assumption gives the equation [latex]{s}^{\\prime }\\left(t\\right)=v\\left(t\\right)[\/latex]. An initial value is necessary; in this case the initial height of the object works well. Let the initial height be given by the equation [latex]s\\left(0\\right)={s}_{0}[\/latex]. Together these assumptions give the initial-value problem<\/p>\r\n\r\n<div id=\"fs-id1170571026947\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{s}^{\\prime }\\left(t\\right)=v\\left(t\\right),s\\left(0\\right)={s}_{0}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573367468\">If the velocity function is known, then it is possible to solve for the position function as well.<\/p>\r\n\r\n<div id=\"fs-id1170571046594\" data-type=\"example\">\r\n<div id=\"fs-id1170571046596\" data-type=\"exercise\">\r\n<div id=\"fs-id1170571046598\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Height of a Moving Baseball<\/h3>\r\n<div id=\"fs-id1170571046598\" data-type=\"problem\">\r\n<p id=\"fs-id1170573207847\">A baseball is thrown upward from a height of [latex]3[\/latex] meters above Earth\u2019s surface with an initial velocity of [latex]10\\text{m\/s}[\/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15[\/latex] kilogram at Earth\u2019s surface.<\/p>\r\n\r\n<ol id=\"fs-id1170573719641\" type=\"a\">\r\n \t<li>Find the position [latex]s\\left(t\\right)[\/latex] of the baseball at time [latex]t[\/latex].<\/li>\r\n \t<li>What is its height after [latex]2[\/latex] seconds?<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44458899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44458899\"]\r\n<div id=\"fs-id1170571232747\" data-type=\"solution\">\r\n<ol id=\"fs-id1170571232749\" type=\"a\">\r\n \t<li>We already know the velocity function for this problem is [latex]v\\left(t\\right)=-9.8t+10[\/latex]. The initial height of the baseball is [latex]3[\/latex] meters, so [latex]{s}_{0}=3[\/latex]. Therefore the initial-value problem for this example is<span data-type=\"newline\">\r\n<\/span>\r\nTo solve the initial-value problem, we first find the antiderivatives:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571207187\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int {s}^{\\prime }\\left(t\\right)dt}&amp; =\\hfill &amp; {\\displaystyle\\int -9.8t+10dt}\\hfill \\\\ \\hfill s\\left(t\\right)&amp; =\\hfill &amp; -4.9{t}^{2}+10t+C.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNext we substitute [latex]t=0[\/latex] and solve for [latex]C\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170573384332\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill s\\left(t\\right)&amp; =\\hfill &amp; -4.9{t}^{2}+10t+C\\hfill \\\\ \\hfill s\\left(0\\right)&amp; =\\hfill &amp; -4.9{\\left(0\\right)}^{2}+10\\left(0\\right)+C\\hfill \\\\ \\hfill 3&amp; =\\hfill &amp; C.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore the position function is [latex]s\\left(t\\right)=-4.9{t}^{2}+10t+3[\/latex].<\/li>\r\n \t<li>The height of the baseball after [latex]2\\text{s}[\/latex] is given by [latex]s\\left(2\\right)\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571064335\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}s\\left(2\\right)\\hfill &amp; =-4.9{\\left(2\\right)}^{2}+10\\left(2\\right)+3\\hfill \\\\ &amp; =-4.9\\left(4\\right)+23\\hfill \\\\ &amp; =3.4.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore the baseball is [latex]3.4[\/latex] meters above Earth\u2019s surface after [latex]2[\/latex] seconds. It is worth noting that the mass of the ball cancelled out completely in the process of solving the problem.<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<section id=\"fs-id1170571025566\" class=\"key-concepts\" data-depth=\"1\"><\/section>\r\n<div data-type=\"glossary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify an initial-value problem<\/li>\n<li>Identify whether a given function is a solution to a differential equation or an initial-value problem<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170573427887\">Usually a given differential equation has an infinite number of solutions, so it is natural to ask which one we want to use. To choose one solution, more information is needed. Some specific information that can be useful is an <span data-type=\"term\">initial value<\/span>, which is an ordered pair that is used to find a particular solution.<\/p>\n<p id=\"fs-id1170573423616\">A differential equation together with one or more initial values is called an <strong>initial-value problem<\/strong>. The general rule is that the number of initial values needed for an initial-value problem is equal to the order of the differential equation. For example, if we have the differential equation [latex]{y}^{\\prime }=2x[\/latex], then [latex]y\\left(3\\right)=7[\/latex] is an initial value, and when taken together, these equations form an initial-value problem. The differential equation [latex]y\\text{''}-3{y}^{\\prime }+2y=4{e}^{x}[\/latex] is second order, so we need two initial values. With initial-value problems of order greater than one, the same value should be used for the independent variable. An example of initial values for this second-order equation would be [latex]y\\left(0\\right)=2[\/latex] and [latex]{y}^{\\prime }\\left(0\\right)=-1[\/latex]. These two initial values together with the differential equation form an initial-value problem. These problems are so named because often the independent variable in the unknown function is [latex]t[\/latex], which represents time. Thus, a value of [latex]t=0[\/latex] represents the beginning of the problem.<\/p>\n<div id=\"fs-id1170573712028\" data-type=\"example\">\n<div id=\"fs-id1170573391210\" data-type=\"exercise\">\n<div id=\"fs-id1170573720384\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Verifying a Solution to an Initial-Value Problem<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558891\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example:\u00a0Verifying a Solution to an Initial-Value Problem<\/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\"><\/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.<\/p>\n<p>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>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5719\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5719&theme=oea&iframe_resize_id=ohm5719&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1170573246209\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170570975800\" data-type=\"exercise\">\n<div id=\"fs-id1170573365823\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170573365823\" data-type=\"problem\">\n<p id=\"fs-id1170573429527\">Verify that [latex]y=3{e}^{2t}+4\\sin{t}[\/latex] is a solution to the initial-value problem<\/p>\n<div id=\"fs-id1170573368474\" class=\"unnumbered\" style=\"text-align: left;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }-2y=4\\cos{t} - 8\\sin{t},y\\left(0\\right)=3[\/latex].<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558890\">Hint<\/span><\/p>\n<div id=\"q44558890\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571276500\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170573338054\">First verify that [latex]y[\/latex] solves the differential equation. Then check the initial value.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573415467\">In the previous example, the initial-value problem consisted of two parts. The first part was the differential equation [latex]{y}^{\\prime }+2y=3{e}^{x}[\/latex], and the second part was the initial value [latex]y\\left(0\\right)=3[\/latex]. These two equations together formed the initial-value problem.<\/p>\n<p id=\"fs-id1170573439347\">The same is true in general. An initial-value problem will consists of two parts: the differential equation and the initial condition. The differential equation has a family of solutions, and the initial condition determines the value of [latex]C[\/latex]. The family of solutions to the differential equation in the example is given by [latex]y=2{e}^{-2t}+C{e}^{t}[\/latex]. This family of solutions is shown in Figure 2, with the particular solution [latex]y=2{e}^{-2t}+{e}^{t}[\/latex] labeled.<\/p>\n<figure id=\"CNX_Calc_Figure_08_01_002\"><figcaption><\/figcaption><div style=\"width: 335px\" 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 [latex]t[\/latex] &lt; 0 as [latex]t[\/latex] goes to negative infinity. For C &gt; 0, the function changes direction and increases in a gentle curve as [latex]t[\/latex] goes to infinity. Larger values of C have a tighter curve closer to the [latex]y[\/latex]-axis and at a higher y value. For C = 0, the function goes to 0 as [latex]t[\/latex] goes to infinity. For C &lt; 0, the function continues to decrease as [latex]t[\/latex] goes to infinity.\" width=\"325\" height=\"332\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. A family of solutions to the differential equation [latex]{y}^{\\prime }+2y=3{e}^{t}[\/latex]. The particular solution [latex]y=2{e}^{-2t}+{e}^{t}[\/latex] is labeled.<\/p>\n<\/div>\n<\/figure>\n<div id=\"fs-id1170571123767\" data-type=\"example\">\n<div id=\"fs-id1170571169678\" data-type=\"exercise\">\n<div id=\"fs-id1170573623483\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Solving an Initial-value Problem<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558889\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example:\u00a0Verifying a Solution to an Initial-Value Problem<\/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\"><\/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.<\/p>\n<p>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 &#8220;4.1.5&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1170571329487\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170571329490\" data-type=\"exercise\">\n<div id=\"fs-id1170573362845\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170573362845\" data-type=\"problem\">\n<p id=\"fs-id1170573362847\">Solve the initial-value problem<\/p>\n<div id=\"fs-id1170573570394\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }={x}^{2}-4x+3 - 6{e}^{x},y\\left(0\\right)=8[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558869\">Hint<\/span><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571418500\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170573502068\">First take the antiderivative of both sides of the differential equation. Then substitute [latex]x=0[\/latex] and [latex]y=8[\/latex] into the resulting equation and solve for [latex]C[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558879\">Show Solution<\/span><\/p>\n<div id=\"q44558879\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571455482\" data-type=\"solution\">\n<p id=\"fs-id1170571203542\">[latex]y=\\frac{1}{3}{x}^{3}-2{x}^{2}+3x - 6{e}^{x}+14[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170573396215\">In physics and engineering applications, we often consider the forces acting upon an object, and use this information to understand the resulting motion that may occur. For example, if we start with an object at Earth\u2019s surface, the primary force acting upon that object is gravity. Physicists and engineers can use this information, along with <span class=\"no-emphasis\" data-type=\"term\">Newton\u2019s second law of motion<\/span> (in equation form [latex]F=ma[\/latex], where [latex]F[\/latex] represents force, [latex]m[\/latex] represents mass, and [latex]a[\/latex] represents acceleration), to derive an equation that can be solved.<\/p>\n<figure id=\"CNX_Calc_Figure_08_01_003\"><figcaption><\/figcaption><div style=\"width: 335px\" 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\/11233917\/CNX_Calc_Figure_08_01_003.jpg\" alt=\"A picture of a baseball with an arrow underneath it pointing down. The arrow is labeled g = -9.8 m\/sec ^ 2.\" width=\"325\" height=\"212\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. For a baseball falling in air, the only force acting on it is gravity (neglecting air resistance).<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1170573536932\">In Figure 3. we assume that the only force acting on a baseball is the force of gravity. This assumption ignores air resistance. (The force due to air resistance is considered in a later discussion.) The acceleration due to gravity at Earth\u2019s surface, [latex]g[\/latex], is approximately [latex]9.8{\\text{m\/s}}^{2}[\/latex]. We introduce a frame of reference, where Earth\u2019s surface is at a height of 0 meters. Let [latex]v\\left(t\\right)[\/latex] represent the velocity of the object in meters per second. If [latex]v\\left(t\\right)>0[\/latex], the ball is rising, and if [latex]v\\left(t\\right)<0[\/latex], the ball is falling (Figure 4).<\/p>\n<figure id=\"CNX_Calc_Figure_08_01_004\"><figcaption><\/figcaption><div style=\"width: 335px\" 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\/11233920\/CNX_Calc_Figure_08_01_004.jpg\" alt=\"A picture of a baseball with an arrow above it pointing up and an arrow below it pointing down. The arrow pointing up is labeled v(t) &gt; 0, and the arrow pointing down is labeled v(t) &lt; 0.\" width=\"325\" height=\"352\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Possible velocities for the rising\/falling baseball.<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1170571509335\">Our goal is to solve for the velocity [latex]v\\left(t\\right)[\/latex] at any time [latex]t[\/latex]. To do this, we set up an initial-value problem. Suppose the mass of the ball is [latex]m[\/latex], where [latex]m[\/latex] is measured in kilograms. We use Newton\u2019s second law, which states that the force acting on an object is equal to its mass times its acceleration [latex]\\left(F=ma\\right)[\/latex]. Acceleration is the derivative of velocity, so [latex]a\\left(t\\right)={v}^{\\prime }\\left(t\\right)[\/latex]. Therefore the force acting on the baseball is given by [latex]F=m{v}^{\\prime }\\left(t\\right)[\/latex]. However, this force must be equal to the force of gravity acting on the object, which (again using Newton\u2019s second law) is given by [latex]{F}_{g}=\\text{-}mg[\/latex], since this force acts in a downward direction. Therefore we obtain the equation [latex]F={F}_{g}[\/latex], which becomes [latex]m{v}^{\\prime }\\left(t\\right)=\\text{-}mg[\/latex]. Dividing both sides of the equation by [latex]m[\/latex] gives the equation<\/p>\n<div id=\"fs-id1170573414975\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{v}^{\\prime }\\left(t\\right)=\\text{-}g[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571118060\">Notice that this differential equation remains the same regardless of the mass of the object.<\/p>\n<p id=\"fs-id1170571118063\">We now need an initial value. Because we are solving for velocity, it makes sense in the context of the problem to assume that we know the <span data-type=\"term\">initial velocity<\/span>, or the velocity at time [latex]t=0[\/latex]. This is denoted by [latex]v\\left(0\\right)={v}_{0}[\/latex].<\/p>\n<div id=\"fs-id1170571084135\" data-type=\"example\">\n<div id=\"fs-id1170571084137\" data-type=\"exercise\">\n<div id=\"fs-id1170571263710\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Velocity of a Moving Baseball<\/h3>\n<div id=\"fs-id1170571263710\" data-type=\"problem\">\n<p id=\"fs-id1170571273705\">A baseball is thrown upward from a height of [latex]3[\/latex] meters above Earth\u2019s surface with an initial velocity of [latex]10\\text{m\/s}[\/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15\\text{kg}[\/latex] at Earth\u2019s surface.<\/p>\n<ol id=\"fs-id1170571216089\" type=\"a\">\n<li>Find the velocity [latex]v\\left(t\\right)[\/latex] of the baseball at time [latex]t[\/latex].<\/li>\n<li>What is its velocity after [latex]2[\/latex] seconds?<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558799\">Show Solution<\/span><\/p>\n<div id=\"q44558799\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571270205\" data-type=\"solution\">\n<ol id=\"fs-id1170571270207\" type=\"a\">\n<li>From the preceding discussion, the differential equation that applies in this situation is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573401342\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{v}^{\\prime }\\left(t\\right)=\\text{-}g[\/latex],<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nwhere [latex]g=9.8{\\text{m\/s}}^{2}[\/latex]. The initial condition is [latex]v\\left(0\\right)={v}_{0}[\/latex], where [latex]{v}_{0}=10\\text{m\/s}\\text{.}[\/latex] Therefore the initial-value problem is [latex]{v}^{\\prime }\\left(t\\right)=-9.8{\\text{m\/s}}^{2},v\\left(0\\right)=10\\text{m\/s}\\text{.}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573289010\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int {v}^{\\prime }\\left(t\\right)dt}& =\\hfill & {\\displaystyle\\int -9.8dt}\\hfill \\\\ \\hfill v\\left(t\\right)& =\\hfill & -9.8t+C.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe next step is to solve for [latex]C[\/latex]. To do this, substitute [latex]t=0[\/latex] and [latex]v\\left(0\\right)=10\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573623436\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill v\\left(t\\right)& =\\hfill & -9.8t+C\\hfill \\\\ \\hfill v\\left(0\\right)& =\\hfill & -9.8\\left(0\\right)+C\\hfill \\\\ \\hfill 10& =\\hfill & C.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore [latex]C=10[\/latex] and the velocity function is given by [latex]v\\left(t\\right)=-9.8t+10[\/latex].<\/li>\n<li>To find the velocity after [latex]2[\/latex] seconds, substitute [latex]t=2[\/latex] into [latex]v\\left(t\\right)[\/latex]. <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571152980\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill v\\left(t\\right)& =\\hfill & -9.8t+10\\hfill \\\\ \\hfill v\\left(2\\right)& =\\hfill & -9.8\\left(2\\right)+10\\hfill \\\\ \\hfill v\\left(2\\right)& =\\hfill & -9.6.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe units of velocity are meters per second. Since the answer is negative, the object is falling at a speed of [latex]9.6\\text{m\/s}\\text{.}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573419495\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170573419498\" data-type=\"exercise\">\n<div id=\"fs-id1170573352098\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170573352098\" data-type=\"problem\">\n<p id=\"fs-id1170573352100\">Suppose a rock falls from rest from a height of [latex]100[\/latex] meters and the only force acting on it is gravity. Find an equation for the velocity [latex]v\\left(t\\right)[\/latex] as a function of time, measured in meters per second.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558599\">Hint<\/span><\/p>\n<div id=\"q44558599\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170573318642\" data-type=\"commentary\" data-element-type=\"hint\">\n<p>What is the initial velocity of the rock? Use this with the differential equation in the example: Velocity of a Moving Baseball to form an initial-value problem, then solve for [latex]v\\left(t\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558699\">Show Solution<\/span><\/p>\n<div id=\"q44558699\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170573493548\" data-type=\"solution\">\n<p id=\"fs-id1170573400886\">[latex]v\\left(t\\right)=-9.8t[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571302110\">A natural question to ask after solving this type of problem is how high the object will be above Earth\u2019s surface at a given point in time. Let [latex]s\\left(t\\right)[\/latex] denote the height above Earth\u2019s surface of the object, measured in meters. Because velocity is the derivative of position (in this case height), this assumption gives the equation [latex]{s}^{\\prime }\\left(t\\right)=v\\left(t\\right)[\/latex]. An initial value is necessary; in this case the initial height of the object works well. Let the initial height be given by the equation [latex]s\\left(0\\right)={s}_{0}[\/latex]. Together these assumptions give the initial-value problem<\/p>\n<div id=\"fs-id1170571026947\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{s}^{\\prime }\\left(t\\right)=v\\left(t\\right),s\\left(0\\right)={s}_{0}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573367468\">If the velocity function is known, then it is possible to solve for the position function as well.<\/p>\n<div id=\"fs-id1170571046594\" data-type=\"example\">\n<div id=\"fs-id1170571046596\" data-type=\"exercise\">\n<div id=\"fs-id1170571046598\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Height of a Moving Baseball<\/h3>\n<div id=\"fs-id1170571046598\" data-type=\"problem\">\n<p id=\"fs-id1170573207847\">A baseball is thrown upward from a height of [latex]3[\/latex] meters above Earth\u2019s surface with an initial velocity of [latex]10\\text{m\/s}[\/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15[\/latex] kilogram at Earth\u2019s surface.<\/p>\n<ol id=\"fs-id1170573719641\" type=\"a\">\n<li>Find the position [latex]s\\left(t\\right)[\/latex] of the baseball at time [latex]t[\/latex].<\/li>\n<li>What is its height after [latex]2[\/latex] seconds?<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44458899\">Show Solution<\/span><\/p>\n<div id=\"q44458899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571232747\" data-type=\"solution\">\n<ol id=\"fs-id1170571232749\" type=\"a\">\n<li>We already know the velocity function for this problem is [latex]v\\left(t\\right)=-9.8t+10[\/latex]. The initial height of the baseball is [latex]3[\/latex] meters, so [latex]{s}_{0}=3[\/latex]. Therefore the initial-value problem for this example is<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTo solve the initial-value problem, we first find the antiderivatives:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571207187\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int {s}^{\\prime }\\left(t\\right)dt}& =\\hfill & {\\displaystyle\\int -9.8t+10dt}\\hfill \\\\ \\hfill s\\left(t\\right)& =\\hfill & -4.9{t}^{2}+10t+C.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNext we substitute [latex]t=0[\/latex] and solve for [latex]C\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170573384332\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill s\\left(t\\right)& =\\hfill & -4.9{t}^{2}+10t+C\\hfill \\\\ \\hfill s\\left(0\\right)& =\\hfill & -4.9{\\left(0\\right)}^{2}+10\\left(0\\right)+C\\hfill \\\\ \\hfill 3& =\\hfill & C.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore the position function is [latex]s\\left(t\\right)=-4.9{t}^{2}+10t+3[\/latex].<\/li>\n<li>The height of the baseball after [latex]2\\text{s}[\/latex] is given by [latex]s\\left(2\\right)\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571064335\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}s\\left(2\\right)\\hfill & =-4.9{\\left(2\\right)}^{2}+10\\left(2\\right)+3\\hfill \\\\ & =-4.9\\left(4\\right)+23\\hfill \\\\ & =3.4.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore the baseball is [latex]3.4[\/latex] meters above Earth\u2019s surface after [latex]2[\/latex] seconds. It is worth noting that the mass of the ball cancelled out completely in the process of solving the problem.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1170571025566\" class=\"key-concepts\" data-depth=\"1\"><\/section>\n<div data-type=\"glossary\"><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1620\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>4.1.5. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.1.5\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1620","chapter","type-chapter","status-publish","hentry"],"part":159,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1620","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1620\/revisions"}],"predecessor-version":[{"id":2202,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1620\/revisions\/2202"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/159"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1620\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1620"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1620"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1620"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1620"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}