{"id":1328,"date":"2021-11-24T06:40:22","date_gmt":"2021-11-24T06:40:22","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1328"},"modified":"2022-11-01T22:40:00","modified_gmt":"2022-11-01T22:40:00","slug":"initial-value-problems-and-boundary-value-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/initial-value-problems-and-boundary-value-problems\/","title":{"raw":"Initial-Value Problems and Boundary-Value Problems","rendered":"Initial-Value Problems and Boundary-Value Problems"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Solve initial-value and boundary-value problems involving linear differential equations.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170571737519\">So far, we have been finding general solutions to differential equations. However, differential equations are often used to describe physical systems, and the person studying that physical system usually knows something about the state of that system at one or more points in time. For example, if a constant-coefficient differential equation is representing how far a motorcycle shock absorber is compressed, we might know that the rider is sitting still on his motorcycle at the start of a race, time [latex]t=t_0[\/latex]. This means the system is at equilibrium, so [latex]y(t_0)=0[\/latex], and the compression of the shock absorber is not changing, so [latex]y'(t_0)=0[\/latex]. With these two initial conditions and the general solution to the differential equation, we can find the\u00a0<em data-effect=\"italics\">specific<\/em>\u00a0solution to the differential equation that satisfies both initial conditions. This process is known as\u00a0<em data-effect=\"italics\">solving an initial-value problem<\/em>. (Recall that we discussed\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term305\" class=\"no-emphasis\" data-type=\"term\">initial-value problems<\/span>\u00a0in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/basics-of-differential-equations\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"1d39a348-071f-4537-85b6-c98912458c3c\" data-page-slug=\"4-introduction\">Introduction to Differential Equations<\/a>.) Note that second-order equations have two arbitrary constants in the general solution, and therefore we require two initial conditions to find the solution to the initial-value problem.<\/p>\r\n<p id=\"fs-id1170572480451\">Sometimes we know the condition of the system at two different times. For example, we might know [latex]y(t_0)=y_0[\/latex] and [latex]y(t_1)=y_1[\/latex]. These conditions are called\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term306\" data-type=\"term\">boundary conditions<\/span>, and finding the solution to the differential equation that satisfies the boundary conditions is called solving a\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term307\" data-type=\"term\">boundary-value problem<\/span>.<\/p>\r\n<p id=\"fs-id1170572624720\">Mathematicians, scientists, and engineers are interested in understanding the conditions under which an initial-value problem or a boundary-value problem has a unique solution. Although a complete treatment of this topic is beyond the scope of this text, it is useful to know that, within the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: solving an initial-value problem<\/h3>\r\nSolve the following initial-value problem: [latex]y''+3y'-4y=0[\/latex],\u00a0[latex]y(0)=1[\/latex],\u00a0[latex]y'(0)=-9[\/latex].\r\n\r\n[reveal-answer q=\"380210266\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"380210266\"]\r\n<p id=\"fs-id1170571547368\">We already solved this differential equation in\u00a0Example \"Solving Second-Order Equations with Constant Coefficients\" part a. and found the general solution to be<\/p>\r\n<p style=\"text-align: center;\">[latex]y(x)=c_1e^{-4x}+c_2e^x[\/latex].<\/p>\r\n<p id=\"fs-id1170571547427\">Then<\/p>\r\n<p style=\"text-align: center;\">[latex]y^\\prime=-4c_1e^{-4x}+c_2e^x[\/latex].<\/p>\r\n<p id=\"fs-id1170571653723\">When [latex]x=0[\/latex], we have [latex]y(0)=c_1+c_2[\/latex] and [latex]y^\\prime(0)=-4c_1+c_2[\/latex]. Applying the initial conditions, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nc_1+c_2&amp;=1 \\\\\r\n-4c_1+c_2&amp;=-9\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170572368489\">Then [latex]c_1=1-c_2[\/latex]. Substituting this expression into the second equation, we see that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n-4(1-c_2)+c_2&amp;=-9 \\\\\r\n-4+4c_2+c_2&amp;=-9 \\\\\r\n5c_2&amp;=-5 \\\\\r\nc_2&amp;=-1\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170571712855\">So, [latex]c_1=2[\/latex] and the solution to the initial-value problem is<\/p>\r\n<p style=\"text-align: center;\">[latex]y(x)=2e^{-4x}-e^x[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nSolve the initial-value problem [latex]y''-3y'-10y=0[\/latex], [latex]y(0)=0[\/latex], [latex]y'(0)=7[\/latex].\r\n\r\n[reveal-answer q=\"442172786\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"442172786\"]\r\n\r\n[latex]y(x)=-e^{-2x}+e^{-5x}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250337&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=_0K0kAUUNd0&amp;video_target=tpm-plugin-7f3wowm0-_0K0kAUUNd0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.7_transcript.html\">transcript for \u201cCP 7.7\u201d here (opens in new window).<\/a><\/center>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: solving an initial-value problem and graphing the solution<\/h3>\r\n<p id=\"fs-id1170572307689\">Solve the following initial-value problem and graph the solution:<\/p>\r\n<p style=\"text-align: center;\">[latex]y^{\\prime\\prime}+6y^\\prime+13y=0, \\ y(0)=0, \\ y^\\prime(0)=2[\/latex].<\/p>\r\n[reveal-answer q=\"835401193\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"835401193\"]\r\n<p id=\"fs-id1170572218518\">We already solved this differential equation in\u00a0Example \"Solving Second-Order Equations with Constant Coefficients\" part b. and found the general solution to be<\/p>\r\n<p style=\"text-align: center;\">[latex]y(x)=e^{-3x}(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\r\nThen\r\n<div id=\"fs-id1170572218526\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\r\n<p style=\"text-align: center;\">[latex]y^\\prime(x)=e^{-3x}(-2c_1\\sin2x+2c_2\\cos2x)-3e^{-3x}(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\r\n<p id=\"fs-id1170571700166\">When [latex]x=0[\/latex], we have [latex]y(0)=c_1[\/latex] and [latex]y^\\prime(0)=2c_2-3c_1[\/latex]. Applying the initial conditions, we obtain<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nc_1&amp;=0 \\\\\r\n-3c_1+2c_2&amp;= 2\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170571746542\">Therefore, [latex]c_1=0[\/latex], [latex]c_2=1[\/latex], and the solution to the initial value problem is shown in the following graph.<\/p>\r\n<p style=\"text-align: center;\">[latex]y=e^{-3x}\\sin2x[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"attachment_3273\" align=\"aligncenter\" width=\"723\"]<img class=\"wp-image-3273 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014029\/7-1-1.jpeg\" alt=\"This figure is a graph of the function y = e^\u22123x sin 2x. The x axis is scaled in increments of tenths. The y axis is scaled in increments of even tenths. The curve passes through the origin and has a horizontal asymptote of the positive x axis.\" width=\"723\" height=\"272\" \/> Figure 1.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nSolve the following initial-value problem and graph the solution: [latex]y''-2y'+10y=10=0[\/latex],\u00a0[latex]y(0)=2[\/latex],\u00a0[latex]y'(0)=-1[\/latex].\r\n\r\n[reveal-answer q=\"487721256\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"487721256\"]\r\n\r\n[latex]y(x)=e^{x}(2\\cos3x-\\sin3x)[\/latex]\r\n\r\n[caption id=\"attachment_5613\" align=\"aligncenter\" width=\"971\"]<img class=\"size-full wp-image-5613\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/03051247\/7.8.jpeg\" alt=\"This figure is the graph of y(x) = e^x(2 cos 3x \u2212 sin 3x) It has the positive x axis scaled in increments of even tenths. The y axis is scaled in increments of twenty. The graph itself starts at the origin. Its amplitude increases as x increases.\" width=\"971\" height=\"572\" \/> Figure 2. Graph of [latex]y(x)=e^{x}(2\\cos3x-\\sin3x)[\/latex].[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: initial-value problem representing a spring-mass system<\/h3>\r\n<p id=\"fs-id1170572306943\">The following initial-value problem models the position of an object with mass attached to a spring. Spring-mass systems are examined in detail in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-applications\/\" data-page-slug=\"7-3-applications\" data-page-uuid=\"1f3e94ba-eee2-44b2-881a-5e5f8fe37136\" data-page-fragment=\"page_1f3e94ba-eee2-44b2-881a-5e5f8fe37136\">Applications<\/a>. The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. (Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.)<\/p>\r\n<p style=\"text-align: center;\">[latex]y^{\\prime\\prime}+2y^\\prime+y=0, \\ y(0)=1, \\ y^\\prime(0)=0[\/latex]<\/p>\r\n<p id=\"fs-id1170572223520\">Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=2[\/latex] sec? How fast is the mass moving at time [latex]t=1[\/latex] sec? In what direction?<\/p>\r\n[reveal-answer q=\"394824182\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"394824182\"]\r\n<p id=\"fs-id1170572223546\">In\u00a0Example\u00a0\"Solving Second-Order Equations with Constant Coefficients\" part c. we found the general solution to this differential equation to be<\/p>\r\n<p style=\"text-align: center;\">[latex]y(t)=c_1e^{-t}+c_2te^{-t}[\/latex].<\/p>\r\n<p id=\"fs-id1170572489109\">Then<\/p>\r\n<p style=\"text-align: center;\">[latex]y^\\prime(t)=-c_1e^{-t}+c_2(-te^{-t}+e^{-t})[\/latex].<\/p>\r\n<p id=\"fs-id1170572489192\">When [latex]t=0[\/latex], we have [latex]y(0)=c_1[\/latex] and [latex]y'(0)=-c_1+c_2[\/latex]. Applying the initial conditions, we obtain<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nc_1&amp;=1 \\\\\r\n-c_1+c_2&amp;= 0\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170572554210\">Thus, [latex]c_1=1[\/latex], [latex]c_2=1[\/latex], and the solution to the initial value problem is<\/p>\r\n<p style=\"text-align: center;\">[latex]y(t)=e^{-t}+te^{-t}[\/latex].<\/p>\r\n<p id=\"fs-id1170572554289\">This solution is represented in the following graph. At time [latex]t=2[\/latex], the mass is at position [latex]y(2)=e^{-2}+2e^{-2}=3e^{-2}\\approx0.406[\/latex] [latex]m[\/latex] below equilibrium.<\/p>\r\n\r\n\r\n[caption id=\"attachment_3276\" align=\"aligncenter\" width=\"702\"]<img class=\"wp-image-3276 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014137\/7-1-2.jpeg\" alt=\"This figure is the graph of y(t) = e^\u2212t + te^\u2212t. The horizontal axis is labeled with t and is scaled in increments of even tenths. The y axis is scaled in increments of 0.5. The graph passes through positive one and decreases with a horizontal asymptote of the positive t axis.\" width=\"702\" height=\"234\" \/> Figure 3.[\/caption]\r\n\r\n&nbsp;\r\n<p id=\"fs-id1170571611981\">To calculate the velocity at time [latex]t=1[\/latex], we need to find the derivative. We have [latex]y(t)=e^{-t}+te^{-t}[\/latex], so<\/p>\r\n<p style=\"text-align: center;\">[latex]y^\\prime(t)=-e^{-t}+e^{-t}-te^{-t}=-te^{-t}[\/latex].<\/p>\r\n<p id=\"fs-id1170572516810\">Then [latex]y^\\prime(1)=e^{-1}\\approx-0.3679[\/latex]. At time [latex]t=1[\/latex], the mass is moving upward at [latex]0.3679[\/latex] m\/sec.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<p id=\"fs-id1170572516875\">Suppose the following initial-value problem models the position (in feet) of a mass in a spring-mass system at any given time. Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=0.3[\/latex] sec? How fast is it moving at time [latex]t=0.1[\/latex] sec? In what direction?<\/p>\r\n<p style=\"text-align: center;\">[latex]y^{\\prime\\prime}+14y^\\prime+49y=0, \\ y(0)=0, \\ y^\\prime(0)=1[\/latex]<\/p>\r\n[reveal-answer q=\"845982348\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"845982348\"]\r\n\r\n[latex]y(t)=te^{-7t}[\/latex]\r\n\r\n[caption id=\"attachment_5617\" align=\"aligncenter\" width=\"721\"]<img class=\"size-full wp-image-5617\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/03051459\/7.9.jpeg\" alt=\"This figure is the graph of y(t) = te^\u22127t. The horizontal axis is labeled with t and is scaled in increments of tenths. The y axis is scaled in increments of 0.5. The graph passes through the origin and has a horizontal asymptote of the positive t axis.\" width=\"721\" height=\"572\" \/> Figure 4.[\/caption]\r\n\r\nAt time [latex]t=0.3[\/latex],\u00a0[latex]y(0.3)=0.3^{(-7*0.3)}=0.3e^{-2.1}\\approx0.0367[\/latex]. The mass is [latex]0.0367[\/latex] ft below equilibrium. At time [latex]t=0.1[\/latex], [latex]y^\\prime(0.1)=0.3e^{-0.7}\\approx0.1490[\/latex]. The mass is moving downward at a speed of [latex]0.1490[\/latex] ft\/sec.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: solving a boundary-value problem<\/h3>\r\n<p id=\"fs-id1170572292389\">In\u00a0Example \"Solving Second-Order Equations with Constant Coefficients\" part f. we solved the differential equation [latex]y''+16y=0[\/latex] and found the general solution to be [latex]y(t)=c_1\\cos4t+c_2\\sin4t[\/latex]. If possible, solve the boundary-value problem if the boundary conditions are the following:<\/p>\r\n\r\n<ol id=\"fs-id1170572373491\" type=\"a\">\r\n \t<li>[latex]y(0)=0[\/latex],\u00a0[latex]y\\left(\\frac{\\pi}4\\right)=0[\/latex]<\/li>\r\n \t<li>[latex]y(0)=1[\/latex], [latex]y\\left(\\frac{\\pi}8\\right)=0[\/latex]<\/li>\r\n \t<li>[latex]y\\left(\\frac{\\pi}8\\right)=0[\/latex],\u00a0[latex]y\\left(\\frac{3\\pi}8\\right)=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"437288237\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"437288237\"]\r\n<p id=\"fs-id1170571781236\">We have<\/p>\r\n\r\n<div id=\"fs-id1170571781239\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]y(x)=c_1\\cos{4t}+c_2\\sin{4t}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<ol id=\"fs-id1170571781288\" type=\"a\">\r\n \t<li style=\"text-align: center;\">Applying the first boundary condition given here, we get [latex]y(0)=c_1=0[\/latex]. So the solution is of the form [latex]y(t)=c_2\\sin4t[\/latex]. When we apply the second boundary condition, though, we get [latex]y\\left(\\frac{\\pi}4\\right)=c_2\\sin\\left(4\\left(\\frac{\\pi}4\\right)\\right)=c_2\\sin\\pi=0[\/latex] for all values of [latex]c_2[\/latex]. The boundary conditions are not sufficient to determine a value for [latex]c_2[\/latex], so this boundary-value problem has infinitely many solutions. Thus, [latex]y(t)=c_2\\sin4t[\/latex] is a solution for any value of [latex]c_2[\/latex].<\/li>\r\n \t<li>Applying the first boundary condition given here, we get [latex]y(0)=c_1=1[\/latex]. Applying the second boundary condition gives[latex]y(\\frac{\\pi}{8})=c_2=0[\/latex], so [latex]c_2=0[\/latex]. In this case, we have a unique solution: [latex]y(t)=\\cos 4t[\/latex].<\/li>\r\n \t<li>Applying the first boundary condition given here, we get [latex]y\\left(\\frac{\\pi}8\\right)=c_2=0[\/latex]. However, applying the second boundary condition gives [latex]y\\left(\\frac{3\\pi}8\\right)=-c_2=2[\/latex], so [latex]c_2=-2[\/latex]. We cannot have [latex]c_2=0=-2[\/latex] so this boundary value problem has no solution.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div data-type=\"note\"><\/div>\r\n<div data-type=\"note\">\r\n\r\n&nbsp;\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Solve initial-value and boundary-value problems involving linear differential equations.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170571737519\">So far, we have been finding general solutions to differential equations. However, differential equations are often used to describe physical systems, and the person studying that physical system usually knows something about the state of that system at one or more points in time. For example, if a constant-coefficient differential equation is representing how far a motorcycle shock absorber is compressed, we might know that the rider is sitting still on his motorcycle at the start of a race, time [latex]t=t_0[\/latex]. This means the system is at equilibrium, so [latex]y(t_0)=0[\/latex], and the compression of the shock absorber is not changing, so [latex]y'(t_0)=0[\/latex]. With these two initial conditions and the general solution to the differential equation, we can find the\u00a0<em data-effect=\"italics\">specific<\/em>\u00a0solution to the differential equation that satisfies both initial conditions. This process is known as\u00a0<em data-effect=\"italics\">solving an initial-value problem<\/em>. (Recall that we discussed\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term305\" class=\"no-emphasis\" data-type=\"term\">initial-value problems<\/span>\u00a0in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/basics-of-differential-equations\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"1d39a348-071f-4537-85b6-c98912458c3c\" data-page-slug=\"4-introduction\">Introduction to Differential Equations<\/a>.) Note that second-order equations have two arbitrary constants in the general solution, and therefore we require two initial conditions to find the solution to the initial-value problem.<\/p>\n<p id=\"fs-id1170572480451\">Sometimes we know the condition of the system at two different times. For example, we might know [latex]y(t_0)=y_0[\/latex] and [latex]y(t_1)=y_1[\/latex]. These conditions are called\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term306\" data-type=\"term\">boundary conditions<\/span>, and finding the solution to the differential equation that satisfies the boundary conditions is called solving a\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term307\" data-type=\"term\">boundary-value problem<\/span>.<\/p>\n<p id=\"fs-id1170572624720\">Mathematicians, scientists, and engineers are interested in understanding the conditions under which an initial-value problem or a boundary-value problem has a unique solution. Although a complete treatment of this topic is beyond the scope of this text, it is useful to know that, within the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: solving an initial-value problem<\/h3>\n<p>Solve the following initial-value problem: [latex]y''+3y'-4y=0[\/latex],\u00a0[latex]y(0)=1[\/latex],\u00a0[latex]y'(0)=-9[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q380210266\">Show Solution<\/span><\/p>\n<div id=\"q380210266\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571547368\">We already solved this differential equation in\u00a0Example &#8220;Solving Second-Order Equations with Constant Coefficients&#8221; part a. and found the general solution to be<\/p>\n<p style=\"text-align: center;\">[latex]y(x)=c_1e^{-4x}+c_2e^x[\/latex].<\/p>\n<p id=\"fs-id1170571547427\">Then<\/p>\n<p style=\"text-align: center;\">[latex]y^\\prime=-4c_1e^{-4x}+c_2e^x[\/latex].<\/p>\n<p id=\"fs-id1170571653723\">When [latex]x=0[\/latex], we have [latex]y(0)=c_1+c_2[\/latex] and [latex]y^\\prime(0)=-4c_1+c_2[\/latex]. Applying the initial conditions, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  c_1+c_2&=1 \\\\  -4c_1+c_2&=-9  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170572368489\">Then [latex]c_1=1-c_2[\/latex]. Substituting this expression into the second equation, we see that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  -4(1-c_2)+c_2&=-9 \\\\  -4+4c_2+c_2&=-9 \\\\  5c_2&=-5 \\\\  c_2&=-1  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170571712855\">So, [latex]c_1=2[\/latex] and the solution to the initial-value problem is<\/p>\n<p style=\"text-align: center;\">[latex]y(x)=2e^{-4x}-e^x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Solve the initial-value problem [latex]y''-3y'-10y=0[\/latex], [latex]y(0)=0[\/latex], [latex]y'(0)=7[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q442172786\">Show Solution<\/span><\/p>\n<div id=\"q442172786\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y(x)=-e^{-2x}+e^{-5x}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250337&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=_0K0kAUUNd0&amp;video_target=tpm-plugin-7f3wowm0-_0K0kAUUNd0\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.7_transcript.html\">transcript for \u201cCP 7.7\u201d here (opens in new window).<\/a><\/div>\n<div class=\"textbox exercises\">\n<h3>Example: solving an initial-value problem and graphing the solution<\/h3>\n<p id=\"fs-id1170572307689\">Solve the following initial-value problem and graph the solution:<\/p>\n<p style=\"text-align: center;\">[latex]y^{\\prime\\prime}+6y^\\prime+13y=0, \\ y(0)=0, \\ y^\\prime(0)=2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q835401193\">Show Solution<\/span><\/p>\n<div id=\"q835401193\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572218518\">We already solved this differential equation in\u00a0Example &#8220;Solving Second-Order Equations with Constant Coefficients&#8221; part b. and found the general solution to be<\/p>\n<p style=\"text-align: center;\">[latex]y(x)=e^{-3x}(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\n<p>Then<\/p>\n<div id=\"fs-id1170572218526\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\n<p style=\"text-align: center;\">[latex]y^\\prime(x)=e^{-3x}(-2c_1\\sin2x+2c_2\\cos2x)-3e^{-3x}(c_1\\cos2x+c_2\\sin2x)[\/latex].<\/p>\n<p id=\"fs-id1170571700166\">When [latex]x=0[\/latex], we have [latex]y(0)=c_1[\/latex] and [latex]y^\\prime(0)=2c_2-3c_1[\/latex]. Applying the initial conditions, we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  c_1&=0 \\\\  -3c_1+2c_2&= 2  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170571746542\">Therefore, [latex]c_1=0[\/latex], [latex]c_2=1[\/latex], and the solution to the initial value problem is shown in the following graph.<\/p>\n<p style=\"text-align: center;\">[latex]y=e^{-3x}\\sin2x[\/latex]<\/p>\n<div id=\"attachment_3273\" style=\"width: 733px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3273\" class=\"wp-image-3273 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014029\/7-1-1.jpeg\" alt=\"This figure is a graph of the function y = e^\u22123x sin 2x. The x axis is scaled in increments of tenths. The y axis is scaled in increments of even tenths. The curve passes through the origin and has a horizontal asymptote of the positive x axis.\" width=\"723\" height=\"272\" \/><\/p>\n<p id=\"caption-attachment-3273\" class=\"wp-caption-text\">Figure 1.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Solve the following initial-value problem and graph the solution: [latex]y''-2y'+10y=10=0[\/latex],\u00a0[latex]y(0)=2[\/latex],\u00a0[latex]y'(0)=-1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q487721256\">Show Solution<\/span><\/p>\n<div id=\"q487721256\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y(x)=e^{x}(2\\cos3x-\\sin3x)[\/latex]<\/p>\n<div id=\"attachment_5613\" style=\"width: 981px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5613\" class=\"size-full wp-image-5613\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/03051247\/7.8.jpeg\" alt=\"This figure is the graph of y(x) = e^x(2 cos 3x \u2212 sin 3x) It has the positive x axis scaled in increments of even tenths. The y axis is scaled in increments of twenty. The graph itself starts at the origin. Its amplitude increases as x increases.\" width=\"971\" height=\"572\" \/><\/p>\n<p id=\"caption-attachment-5613\" class=\"wp-caption-text\">Figure 2. Graph of [latex]y(x)=e^{x}(2\\cos3x-\\sin3x)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: initial-value problem representing a spring-mass system<\/h3>\n<p id=\"fs-id1170572306943\">The following initial-value problem models the position of an object with mass attached to a spring. Spring-mass systems are examined in detail in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-applications\/\" data-page-slug=\"7-3-applications\" data-page-uuid=\"1f3e94ba-eee2-44b2-881a-5e5f8fe37136\" data-page-fragment=\"page_1f3e94ba-eee2-44b2-881a-5e5f8fe37136\">Applications<\/a>. The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. (Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.)<\/p>\n<p style=\"text-align: center;\">[latex]y^{\\prime\\prime}+2y^\\prime+y=0, \\ y(0)=1, \\ y^\\prime(0)=0[\/latex]<\/p>\n<p id=\"fs-id1170572223520\">Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=2[\/latex] sec? How fast is the mass moving at time [latex]t=1[\/latex] sec? In what direction?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q394824182\">Show Solution<\/span><\/p>\n<div id=\"q394824182\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572223546\">In\u00a0Example\u00a0&#8220;Solving Second-Order Equations with Constant Coefficients&#8221; part c. we found the general solution to this differential equation to be<\/p>\n<p style=\"text-align: center;\">[latex]y(t)=c_1e^{-t}+c_2te^{-t}[\/latex].<\/p>\n<p id=\"fs-id1170572489109\">Then<\/p>\n<p style=\"text-align: center;\">[latex]y^\\prime(t)=-c_1e^{-t}+c_2(-te^{-t}+e^{-t})[\/latex].<\/p>\n<p id=\"fs-id1170572489192\">When [latex]t=0[\/latex], we have [latex]y(0)=c_1[\/latex] and [latex]y'(0)=-c_1+c_2[\/latex]. Applying the initial conditions, we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  c_1&=1 \\\\  -c_1+c_2&= 0  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170572554210\">Thus, [latex]c_1=1[\/latex], [latex]c_2=1[\/latex], and the solution to the initial value problem is<\/p>\n<p style=\"text-align: center;\">[latex]y(t)=e^{-t}+te^{-t}[\/latex].<\/p>\n<p id=\"fs-id1170572554289\">This solution is represented in the following graph. At time [latex]t=2[\/latex], the mass is at position [latex]y(2)=e^{-2}+2e^{-2}=3e^{-2}\\approx0.406[\/latex] [latex]m[\/latex] below equilibrium.<\/p>\n<div id=\"attachment_3276\" style=\"width: 712px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3276\" class=\"wp-image-3276 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18014137\/7-1-2.jpeg\" alt=\"This figure is the graph of y(t) = e^\u2212t + te^\u2212t. The horizontal axis is labeled with t and is scaled in increments of even tenths. The y axis is scaled in increments of 0.5. The graph passes through positive one and decreases with a horizontal asymptote of the positive t axis.\" width=\"702\" height=\"234\" \/><\/p>\n<p id=\"caption-attachment-3276\" class=\"wp-caption-text\">Figure 3.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571611981\">To calculate the velocity at time [latex]t=1[\/latex], we need to find the derivative. We have [latex]y(t)=e^{-t}+te^{-t}[\/latex], so<\/p>\n<p style=\"text-align: center;\">[latex]y^\\prime(t)=-e^{-t}+e^{-t}-te^{-t}=-te^{-t}[\/latex].<\/p>\n<p id=\"fs-id1170572516810\">Then [latex]y^\\prime(1)=e^{-1}\\approx-0.3679[\/latex]. At time [latex]t=1[\/latex], the mass is moving upward at [latex]0.3679[\/latex] m\/sec.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-id1170572516875\">Suppose the following initial-value problem models the position (in feet) of a mass in a spring-mass system at any given time. Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=0.3[\/latex] sec? How fast is it moving at time [latex]t=0.1[\/latex] sec? In what direction?<\/p>\n<p style=\"text-align: center;\">[latex]y^{\\prime\\prime}+14y^\\prime+49y=0, \\ y(0)=0, \\ y^\\prime(0)=1[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q845982348\">Show Solution<\/span><\/p>\n<div id=\"q845982348\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y(t)=te^{-7t}[\/latex]<\/p>\n<div id=\"attachment_5617\" style=\"width: 731px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5617\" class=\"size-full wp-image-5617\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/03051459\/7.9.jpeg\" alt=\"This figure is the graph of y(t) = te^\u22127t. The horizontal axis is labeled with t and is scaled in increments of tenths. The y axis is scaled in increments of 0.5. The graph passes through the origin and has a horizontal asymptote of the positive t axis.\" width=\"721\" height=\"572\" \/><\/p>\n<p id=\"caption-attachment-5617\" class=\"wp-caption-text\">Figure 4.<\/p>\n<\/div>\n<p>At time [latex]t=0.3[\/latex],\u00a0[latex]y(0.3)=0.3^{(-7*0.3)}=0.3e^{-2.1}\\approx0.0367[\/latex]. The mass is [latex]0.0367[\/latex] ft below equilibrium. At time [latex]t=0.1[\/latex], [latex]y^\\prime(0.1)=0.3e^{-0.7}\\approx0.1490[\/latex]. The mass is moving downward at a speed of [latex]0.1490[\/latex] ft\/sec.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: solving a boundary-value problem<\/h3>\n<p id=\"fs-id1170572292389\">In\u00a0Example &#8220;Solving Second-Order Equations with Constant Coefficients&#8221; part f. we solved the differential equation [latex]y''+16y=0[\/latex] and found the general solution to be [latex]y(t)=c_1\\cos4t+c_2\\sin4t[\/latex]. If possible, solve the boundary-value problem if the boundary conditions are the following:<\/p>\n<ol id=\"fs-id1170572373491\" type=\"a\">\n<li>[latex]y(0)=0[\/latex],\u00a0[latex]y\\left(\\frac{\\pi}4\\right)=0[\/latex]<\/li>\n<li>[latex]y(0)=1[\/latex], [latex]y\\left(\\frac{\\pi}8\\right)=0[\/latex]<\/li>\n<li>[latex]y\\left(\\frac{\\pi}8\\right)=0[\/latex],\u00a0[latex]y\\left(\\frac{3\\pi}8\\right)=0[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q437288237\">Show Solution<\/span><\/p>\n<div id=\"q437288237\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170571781236\">We have<\/p>\n<div id=\"fs-id1170571781239\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]y(x)=c_1\\cos{4t}+c_2\\sin{4t}[\/latex]<\/p>\n<\/div>\n<ol id=\"fs-id1170571781288\" type=\"a\">\n<li style=\"text-align: center;\">Applying the first boundary condition given here, we get [latex]y(0)=c_1=0[\/latex]. So the solution is of the form [latex]y(t)=c_2\\sin4t[\/latex]. When we apply the second boundary condition, though, we get [latex]y\\left(\\frac{\\pi}4\\right)=c_2\\sin\\left(4\\left(\\frac{\\pi}4\\right)\\right)=c_2\\sin\\pi=0[\/latex] for all values of [latex]c_2[\/latex]. The boundary conditions are not sufficient to determine a value for [latex]c_2[\/latex], so this boundary-value problem has infinitely many solutions. Thus, [latex]y(t)=c_2\\sin4t[\/latex] is a solution for any value of [latex]c_2[\/latex].<\/li>\n<li>Applying the first boundary condition given here, we get [latex]y(0)=c_1=1[\/latex]. Applying the second boundary condition gives[latex]y(\\frac{\\pi}{8})=c_2=0[\/latex], so [latex]c_2=0[\/latex]. In this case, we have a unique solution: [latex]y(t)=\\cos 4t[\/latex].<\/li>\n<li>Applying the first boundary condition given here, we get [latex]y\\left(\\frac{\\pi}8\\right)=c_2=0[\/latex]. However, applying the second boundary condition gives [latex]y\\left(\\frac{3\\pi}8\\right)=-c_2=2[\/latex], so [latex]c_2=-2[\/latex]. We cannot have [latex]c_2=0=-2[\/latex] so this boundary value problem has no solution.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\"><\/div>\n<div data-type=\"note\">\n<p>&nbsp;<\/p>\n<\/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-1328\">\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>CP 7.7. <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 3. <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-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/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-3\/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":428269,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 7.7\",\"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-1328","chapter","type-chapter","status-publish","hentry"],"part":25,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1328","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/428269"}],"version-history":[{"count":45,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1328\/revisions"}],"predecessor-version":[{"id":6462,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1328\/revisions\/6462"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/25"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1328\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1328"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1328"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1328"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1328"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}