{"id":816,"date":"2025-10-09T15:40:53","date_gmt":"2025-10-09T15:40:53","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=816"},"modified":"2025-11-13T20:55:33","modified_gmt":"2025-11-13T20:55:33","slug":"linear-homogeneous-equations-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/linear-homogeneous-equations-fresh-take\/","title":{"raw":"Linear Homogeneous Equations: Fresh Take","rendered":"Linear Homogeneous Equations: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Identify homogeneous linear equations<\/li>\r\n \t<li>Find and use characteristic equations<\/li>\r\n \t<li>Solve homogeneous equations using characteristic equations<\/li>\r\n \t<li>Solve initial and boundary value problems<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Homogeneous Linear Equations<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Second-order differential equations come in different types, and recognizing whether an equation is linear and whether it's homogeneous determines which solution methods you can use. Understanding these classifications is the first step in solving differential equations.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Linear Second-Order Equations<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">A second-order differential equation is <strong>linear<\/strong> if it can be written as:<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]a_2(x)y'' + a_1(x)y' + a_0(x)y = r(x)[\/latex]<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The key requirement: [latex]y[\/latex] and its derivatives appear only to the first power and are never multiplied together. Terms like [latex]y^2[\/latex], [latex]\\sqrt{y'}[\/latex], [latex]\\sin y[\/latex], or [latex]yy'[\/latex] make an equation nonlinear.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Homogeneous vs. Nonhomogeneous<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Once you've confirmed an equation is linear, the next question is whether it's homogeneous:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Homogeneous<\/strong>: Every term contains [latex]y[\/latex] or one of its derivatives. This means [latex]r(x) = 0[\/latex] identically.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Nonhomogeneous<\/strong>: At least one term involves only functions of [latex]x[\/latex] (no [latex]y[\/latex]). This means [latex]r(x) \\neq 0[\/latex] for some [latex]x[\/latex].<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Practical Strategy<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When given a differential equation:<\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Rewrite in standard form<\/strong> if necessary<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Check linearity<\/strong>: Are [latex]y[\/latex] and its derivatives only to the first power? Are they never multiplied together or inside functions?<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Check homogeneity<\/strong>: Does every term contain [latex]y[\/latex] or its derivatives?<\/li>\r\n<\/ol>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For example, [latex]xy'' + 2x^2y' + 5x^3y = 0[\/latex] is linear and homogeneous, while [latex]xy'' + 2x^2y' + 5x^3y = x^2[\/latex] is linear but nonhomogeneous (because of the [latex]x^2[\/latex] term on the right).<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The classification determines your solution approach. Homogeneous linear equations have specific properties and solution methods that don't apply to nonhomogeneous equations.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit this\u00a0<a href=\"https:\/\/www.khanacademy.org\/math\/differential-equations\/second-order-differential-equations\">website<\/a>\u00a0to study more about second-order linear differential equations.<\/section><section class=\"textbox interact\" aria-label=\"Interact\">Visit this\u00a0<a href=\"http:\/\/www.sosmath.com\/diffeq\/diffeq.html\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a>\u00a0that discusses second-order differential equations.<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1170572107308\">Classify each of the following equations as linear or nonlinear. If the equation is linear, determine further whether it is homogeneous or nonhomogeneous.<\/p>\r\n\r\n<ol id=\"fs-id1170572246040\" type=\"a\">\r\n \t<li>[latex](y'')^{2}-y'+8x^{3}y=0[\/latex]<\/li>\r\n \t<li>[latex](\\sin t)y''+\\cos t-3ty'=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"374981011\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"374981011\"]\r\n\r\na. Nonlinear\r\n\r\nb. Linear, nonhomogeneous\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h2>Finding the General Solution<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">When solving second-order linear homogeneous differential equations, finding one solution isn't enough. You need to find the <strong>general solution<\/strong>, which captures all possible solutions to the equation.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Superposition Principle<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">If [latex]y_1(x)[\/latex] and [latex]y_2(x)[\/latex] are solutions to a linear homogeneous differential equation, then any linear combination [latex]y(x) = c_1y_1(x) + c_2y_2(x)[\/latex] is also a solution. The constants [latex]c_1[\/latex] and [latex]c_2[\/latex] can be any real numbers.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why Two Solutions Aren't Always Enough<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Here's the catch: you can't just pick any two solutions and combine them. For example, [latex]e^{-3t}[\/latex] and [latex]2e^{-3t}[\/latex] are both solutions to [latex]x'' + 7x' + 12x = 0[\/latex], but their combination [latex]c_1e^{-3t} + c_2(2e^{-3t})[\/latex] misses other solutions like [latex]e^{-4t}[\/latex]. The problem is that these two solutions are essentially the same function\u2014one is just a constant multiple of the other.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Linear Independence: The Key Requirement<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Two functions are <strong>linearly independent<\/strong> if neither one is a constant multiple of the other (and neither is identically zero). For practical checking:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]x^2[\/latex] and [latex]5x^2[\/latex] are linearly dependent (one is 5 times the other)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\sin x[\/latex] and [latex]\\cos x[\/latex] are linearly independent (no constant [latex]C[\/latex] makes [latex]\\sin x = C\\cos x[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]e^{3x}[\/latex] and [latex]e^{-3x}[\/latex] are linearly independent (different exponents mean they're independent)<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Once you have two linearly independent solutions [latex]y_1(x)[\/latex] and [latex]y_2(x)[\/latex], write the general solution as [latex]y(x) = c_1y_1(x) + c_2y_2(x)[\/latex]. This formula represents every possible solution to the differential equation. This is powerful: finding two linearly independent solutions means you've found all solutions.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1170572563030\">Show that [latex]y=2x^{2}[\/latex] is a solution to the differential equation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac{1}{2}x^{2}y''-xy'+y=0}[\/latex].<\/p>\r\n[reveal-answer q=\"688213714\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"688213714\"]\r\n\r\nLet [latex]y=2x^2[\/latex]. Then [latex]y'=4x[\/latex] and [latex]y''=4[\/latex]. Substituting into the differential equation, we see that\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\frac{1}{2}x^{2}y''-xy'+y&amp;=\\frac{1}{2}x^2(4)-x(4x)+(2x^2) \\\\\r\n&amp;=2x^2-4x^2+2x^2 \\\\\r\n&amp;=0\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1170572474616\">Consider the differential equation<\/p>\r\n[latex]y''+5y'+6y=0[\/latex].\r\n<p id=\"fs-id1170572415337\">Given that [latex]e^{-2x}[\/latex] and [latex]e^{-3x}[\/latex] are solutions to this differential equation, show that [latex]3e^{-2x}+6e^{-3x}[\/latex] is a solution.<\/p>\r\n[reveal-answer q=\"238127390\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"238127390\"]\r\n\r\nWe have [latex]y(x)=3e^{-2x}+6e^{-3x}[\/latex], so [latex]y'(x)=-6e^{-2x}-18e^{-3x}[\/latex] and [latex]y''(x)=12e^{-2x}+54e^{-3x}[\/latex].\r\n\r\nThen\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny^{\\prime\\prime}+5y^\\prime+6y&amp;=\\left(12e^{-2x}+54e^{-3x}\\right)+5\\left(-6e^{-2x}-18e^{-3x}\\right)+6\\left(3e^{-2x}+6e^{-3x}\\right) \\\\\r\n&amp;=12e^{-2x}+54e^{-3x}-30e^{-2x}-90e^{-3x}+18e^{-2x}+36e^{-3x} \\\\\r\n&amp;=0\r\n\\end{aligned}[\/latex].<\/p>\r\nThus, [latex]y(x)=3e^{-2x}+6e^{-3x}[\/latex] is a solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250335&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=D0gjG63CPZk&amp;video_target=tpm-plugin-sozzg70y-D0gjG63CPZk\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.3_transcript.html\">transcript for \u201cCP 7.3\u201d here (opens in new window).<\/a><\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">Determine whether the following pairs of functions are linearly dependent or linearly independent: [latex]f_1(x)=e^{x}[\/latex], [latex]f_2(x)=3e^{3x}[\/latex].[reveal-answer q=\"283742979\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"283742979\"]Linearly independent.[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">If [latex]y_1(x)=e^{3x}[\/latex] and [latex]y_2(x)=xe^{3x}[\/latex] are solutions to [latex]y''-6y'+9y=0[\/latex], what is the general solution?[reveal-answer q=\"342894572\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342894572\"][latex]y(x)=c_1e^{3t}+c_2xe^{3t}[\/latex][\/hidden-answer]<\/section>\r\n<h2>Second-Order Equations with Constant Coefficients<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Second-order differential equations with constant coefficients have the form [latex]ay'' + by' + cy = 0[\/latex], where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are constants. The solution strategy relies on finding the <strong>characteristic equation<\/strong>.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why Exponential Functions?<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">We try solutions of the form [latex]y(x) = e^{\\lambda x}[\/latex] because exponential functions have a special property: their derivatives are constant multiples of themselves. This helps all terms cancel out when we substitute into the differential equation. Substituting [latex]y = e^{\\lambda x}[\/latex] into the equation gives us [latex]e^{\\lambda x}(a\\lambda^2 + b\\lambda + c) = 0[\/latex]. Since [latex]e^{\\lambda x}[\/latex] is never zero, we need [latex]a\\lambda^2 + b\\lambda + c = 0[\/latex]. This is the <strong>characteristic equation<\/strong>.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Three Cases Based on the Roots<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">The characteristic equation is a quadratic, so it has three possible types of roots:<\/p>\r\n\r\n<ul>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Case 1: Distinct Real Roots<\/strong> ([latex]\\lambda_1[\/latex] and [latex]\\lambda_2[\/latex]) General solution: [latex]y(x) = c_1e^{\\lambda_1 x} + c_2e^{\\lambda_2 x}[\/latex]<\/li>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Case 2: Repeated Real Root<\/strong> ([latex]\\lambda[\/latex]) General solution: [latex]y(x) = c_1e^{\\lambda x} + c_2xe^{\\lambda x}[\/latex] Note the [latex]x[\/latex] multiplier in the second term\u2014this ensures linear independence.<\/li>\r\n \t<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Case 3: Complex Conjugate Roots<\/strong> ([latex]\\alpha \\pm \\beta i[\/latex]) General solution: [latex]y(x) = e^{\\alpha x}(c_1\\cos\\beta x + c_2\\sin\\beta x)[\/latex] The real part [latex]\\alpha[\/latex] appears in the exponential, while the imaginary part [latex]\\beta[\/latex] appears in the trigonometric functions.<\/li>\r\n<\/ul>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Write the equation in standard form: [latex]ay'' + by' + cy = 0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Form the characteristic equation: [latex]a\\lambda^2 + b\\lambda + c = 0[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Solve for [latex]\\lambda[\/latex] using factoring or the quadratic formula<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Use the type of roots to determine which case applies and write the general solution<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1170572332764\">Find the general solution to the following differential equations:<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]y''-2y'+10y=0[\/latex]<\/li>\r\n \t<li>[latex]y''+14y'+49y=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"283472981\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"283472981\"]\r\n\r\na. [latex]y(x)=e^x(c_1\\cos3x+c_2\\sin3x)[\/latex]\r\n\r\nb. [latex]y(x)=c_1e^{-7x}+c_2xe^{-7x}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250336&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=f1v_hH_mO5Y&amp;video_target=tpm-plugin-y5rm8nb2-f1v_hH_mO5Y\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP7.6_transcript.html\">transcript for \u201cCP 7.6\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2>Initial-Value Problems and Boundary-Value Problems<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">Finding the general solution to a differential equation gives you a family of solutions with arbitrary constants. To find a specific solution that describes a particular physical system, you need additional information about the system at specific points in time.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Initial-Value Problems<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">An <strong>initial-value problem<\/strong> provides conditions at a single point in time [latex]t_0[\/latex]. For second-order equations, you need two initial conditions: typically [latex]y(t_0)[\/latex] and [latex]y'(t_0)[\/latex]. These two conditions let you solve for the two arbitrary constants [latex]c_1[\/latex] and [latex]c_2[\/latex] in the general solution. Initial-value problems for constant-coefficient second-order equations always have a unique solution when two initial conditions are provided.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Boundary-Value Problems<\/strong><\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">A <strong>boundary-value problem<\/strong> provides conditions at two different points in time: [latex]y(t_0)[\/latex] and [latex]y(t_1)[\/latex]. Unlike initial-value problems, boundary-value problems are less predictable. Even with two boundary conditions, you might find a unique solution, infinitely many solutions, or no solution at all.<\/p>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Find the general solution to the differential equation<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Calculate [latex]y'(t)[\/latex] from the general solution (if working with initial conditions involving [latex]y'(t_0)[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Substitute the given conditions into [latex]y(t)[\/latex] and [latex]y'(t)[\/latex] to create a system of equations<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Solve the system for [latex]c_1[\/latex] and [latex]c_2[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Write the specific solution by substituting the values of [latex]c_1[\/latex] and [latex]c_2[\/latex] back into the general solution<\/li>\r\n<\/ol>\r\n<p class=\"font-claude-response-body whitespace-normal break-words\">For boundary-value problems, watch for inconsistent equations (no solution) or dependent equations (infinitely many solutions).<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Solve the initial-value problem [latex]y''-3y'-10y=0[\/latex], [latex]y(0)=0[\/latex], [latex]y'(0)=7[\/latex].[reveal-answer q=\"442172786\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"442172786\"][latex]y(x)=-e^{-2x}+e^{-5x}[\/latex].[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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><\/section><section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"487721256\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"487721256\"][latex]y(x)=e^{x}(2\\cos3x-\\sin3x)[\/latex][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]<\/section><section class=\"textbox example\" aria-label=\"Example\">\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<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Identify homogeneous linear equations<\/li>\n<li>Find and use characteristic equations<\/li>\n<li>Solve homogeneous equations using characteristic equations<\/li>\n<li>Solve initial and boundary value problems<\/li>\n<\/ul>\n<\/section>\n<h2>Homogeneous Linear Equations<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Second-order differential equations come in different types, and recognizing whether an equation is linear and whether it&#8217;s homogeneous determines which solution methods you can use. Understanding these classifications is the first step in solving differential equations.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Linear Second-Order Equations<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">A second-order differential equation is <strong>linear<\/strong> if it can be written as:<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\" style=\"text-align: center;\">[latex]a_2(x)y'' + a_1(x)y' + a_0(x)y = r(x)[\/latex]<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The key requirement: [latex]y[\/latex] and its derivatives appear only to the first power and are never multiplied together. Terms like [latex]y^2[\/latex], [latex]\\sqrt{y'}[\/latex], [latex]\\sin y[\/latex], or [latex]yy'[\/latex] make an equation nonlinear.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Homogeneous vs. Nonhomogeneous<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Once you&#8217;ve confirmed an equation is linear, the next question is whether it&#8217;s homogeneous:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Homogeneous<\/strong>: Every term contains [latex]y[\/latex] or one of its derivatives. This means [latex]r(x) = 0[\/latex] identically.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Nonhomogeneous<\/strong>: At least one term involves only functions of [latex]x[\/latex] (no [latex]y[\/latex]). This means [latex]r(x) \\neq 0[\/latex] for some [latex]x[\/latex].<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Practical Strategy<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When given a differential equation:<\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Rewrite in standard form<\/strong> if necessary<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Check linearity<\/strong>: Are [latex]y[\/latex] and its derivatives only to the first power? Are they never multiplied together or inside functions?<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Check homogeneity<\/strong>: Does every term contain [latex]y[\/latex] or its derivatives?<\/li>\n<\/ol>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For example, [latex]xy'' + 2x^2y' + 5x^3y = 0[\/latex] is linear and homogeneous, while [latex]xy'' + 2x^2y' + 5x^3y = x^2[\/latex] is linear but nonhomogeneous (because of the [latex]x^2[\/latex] term on the right).<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why This Matters<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The classification determines your solution approach. Homogeneous linear equations have specific properties and solution methods that don&#8217;t apply to nonhomogeneous equations.<\/p>\n<\/div>\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit this\u00a0<a href=\"https:\/\/www.khanacademy.org\/math\/differential-equations\/second-order-differential-equations\">website<\/a>\u00a0to study more about second-order linear differential equations.<\/section>\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit this\u00a0<a href=\"http:\/\/www.sosmath.com\/diffeq\/diffeq.html\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a>\u00a0that discusses second-order differential equations.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1170572107308\">Classify each of the following equations as linear or nonlinear. If the equation is linear, determine further whether it is homogeneous or nonhomogeneous.<\/p>\n<ol id=\"fs-id1170572246040\" type=\"a\">\n<li>[latex](y'')^{2}-y'+8x^{3}y=0[\/latex]<\/li>\n<li>[latex](\\sin t)y''+\\cos t-3ty'=0[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q374981011\">Show Solution<\/button><\/p>\n<div id=\"q374981011\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. Nonlinear<\/p>\n<p>b. Linear, nonhomogeneous<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2>Finding the General Solution<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">When solving second-order linear homogeneous differential equations, finding one solution isn&#8217;t enough. You need to find the <strong>general solution<\/strong>, which captures all possible solutions to the equation.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>The Superposition Principle<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">If [latex]y_1(x)[\/latex] and [latex]y_2(x)[\/latex] are solutions to a linear homogeneous differential equation, then any linear combination [latex]y(x) = c_1y_1(x) + c_2y_2(x)[\/latex] is also a solution. The constants [latex]c_1[\/latex] and [latex]c_2[\/latex] can be any real numbers.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why Two Solutions Aren&#8217;t Always Enough<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Here&#8217;s the catch: you can&#8217;t just pick any two solutions and combine them. For example, [latex]e^{-3t}[\/latex] and [latex]2e^{-3t}[\/latex] are both solutions to [latex]x'' + 7x' + 12x = 0[\/latex], but their combination [latex]c_1e^{-3t} + c_2(2e^{-3t})[\/latex] misses other solutions like [latex]e^{-4t}[\/latex]. The problem is that these two solutions are essentially the same function\u2014one is just a constant multiple of the other.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Linear Independence: The Key Requirement<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Two functions are <strong>linearly independent<\/strong> if neither one is a constant multiple of the other (and neither is identically zero). For practical checking:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]x^2[\/latex] and [latex]5x^2[\/latex] are linearly dependent (one is 5 times the other)<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\sin x[\/latex] and [latex]\\cos x[\/latex] are linearly independent (no constant [latex]C[\/latex] makes [latex]\\sin x = C\\cos x[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]e^{3x}[\/latex] and [latex]e^{-3x}[\/latex] are linearly independent (different exponents mean they&#8217;re independent)<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Once you have two linearly independent solutions [latex]y_1(x)[\/latex] and [latex]y_2(x)[\/latex], write the general solution as [latex]y(x) = c_1y_1(x) + c_2y_2(x)[\/latex]. This formula represents every possible solution to the differential equation. This is powerful: finding two linearly independent solutions means you&#8217;ve found all solutions.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1170572563030\">Show that [latex]y=2x^{2}[\/latex] is a solution to the differential equation<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac{1}{2}x^{2}y''-xy'+y=0}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q688213714\">Show Solution<\/button><\/p>\n<div id=\"q688213714\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let [latex]y=2x^2[\/latex]. Then [latex]y'=4x[\/latex] and [latex]y''=4[\/latex]. Substituting into the differential equation, we see that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\frac{1}{2}x^{2}y''-xy'+y&=\\frac{1}{2}x^2(4)-x(4x)+(2x^2) \\\\  &=2x^2-4x^2+2x^2 \\\\  &=0  \\end{aligned}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1170572474616\">Consider the differential equation<\/p>\n<p>[latex]y''+5y'+6y=0[\/latex].<\/p>\n<p id=\"fs-id1170572415337\">Given that [latex]e^{-2x}[\/latex] and [latex]e^{-3x}[\/latex] are solutions to this differential equation, show that [latex]3e^{-2x}+6e^{-3x}[\/latex] is a solution.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q238127390\">Show Solution<\/button><\/p>\n<div id=\"q238127390\" class=\"hidden-answer\" style=\"display: none\">\n<p>We have [latex]y(x)=3e^{-2x}+6e^{-3x}[\/latex], so [latex]y'(x)=-6e^{-2x}-18e^{-3x}[\/latex] and [latex]y''(x)=12e^{-2x}+54e^{-3x}[\/latex].<\/p>\n<p>Then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y^{\\prime\\prime}+5y^\\prime+6y&=\\left(12e^{-2x}+54e^{-3x}\\right)+5\\left(-6e^{-2x}-18e^{-3x}\\right)+6\\left(3e^{-2x}+6e^{-3x}\\right) \\\\  &=12e^{-2x}+54e^{-3x}-30e^{-2x}-90e^{-3x}+18e^{-2x}+36e^{-3x} \\\\  &=0  \\end{aligned}[\/latex].<\/p>\n<p>Thus, [latex]y(x)=3e^{-2x}+6e^{-3x}[\/latex] is a solution.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250335&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=D0gjG63CPZk&amp;video_target=tpm-plugin-sozzg70y-D0gjG63CPZk\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP7.3_transcript.html\">transcript for \u201cCP 7.3\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Determine whether the following pairs of functions are linearly dependent or linearly independent: [latex]f_1(x)=e^{x}[\/latex], [latex]f_2(x)=3e^{3x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q283742979\">Show Solution<\/button><\/p>\n<div id=\"q283742979\" class=\"hidden-answer\" style=\"display: none\">Linearly independent.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">If [latex]y_1(x)=e^{3x}[\/latex] and [latex]y_2(x)=xe^{3x}[\/latex] are solutions to [latex]y''-6y'+9y=0[\/latex], what is the general solution?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q342894572\">Show Solution<\/button><\/p>\n<div id=\"q342894572\" class=\"hidden-answer\" style=\"display: none\">[latex]y(x)=c_1e^{3t}+c_2xe^{3t}[\/latex]<\/div>\n<\/div>\n<\/section>\n<h2>Second-Order Equations with Constant Coefficients<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Second-order differential equations with constant coefficients have the form [latex]ay'' + by' + cy = 0[\/latex], where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are constants. The solution strategy relies on finding the <strong>characteristic equation<\/strong>.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Why Exponential Functions?<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">We try solutions of the form [latex]y(x) = e^{\\lambda x}[\/latex] because exponential functions have a special property: their derivatives are constant multiples of themselves. This helps all terms cancel out when we substitute into the differential equation. Substituting [latex]y = e^{\\lambda x}[\/latex] into the equation gives us [latex]e^{\\lambda x}(a\\lambda^2 + b\\lambda + c) = 0[\/latex]. Since [latex]e^{\\lambda x}[\/latex] is never zero, we need [latex]a\\lambda^2 + b\\lambda + c = 0[\/latex]. This is the <strong>characteristic equation<\/strong>.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Three Cases Based on the Roots<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">The characteristic equation is a quadratic, so it has three possible types of roots:<\/p>\n<ul>\n<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Case 1: Distinct Real Roots<\/strong> ([latex]\\lambda_1[\/latex] and [latex]\\lambda_2[\/latex]) General solution: [latex]y(x) = c_1e^{\\lambda_1 x} + c_2e^{\\lambda_2 x}[\/latex]<\/li>\n<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Case 2: Repeated Real Root<\/strong> ([latex]\\lambda[\/latex]) General solution: [latex]y(x) = c_1e^{\\lambda x} + c_2xe^{\\lambda x}[\/latex] Note the [latex]x[\/latex] multiplier in the second term\u2014this ensures linear independence.<\/li>\n<li class=\"font-claude-response-body whitespace-normal break-words\"><strong>Case 3: Complex Conjugate Roots<\/strong> ([latex]\\alpha \\pm \\beta i[\/latex]) General solution: [latex]y(x) = e^{\\alpha x}(c_1\\cos\\beta x + c_2\\sin\\beta x)[\/latex] The real part [latex]\\alpha[\/latex] appears in the exponential, while the imaginary part [latex]\\beta[\/latex] appears in the trigonometric functions.<\/li>\n<\/ul>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Write the equation in standard form: [latex]ay'' + by' + cy = 0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Form the characteristic equation: [latex]a\\lambda^2 + b\\lambda + c = 0[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Solve for [latex]\\lambda[\/latex] using factoring or the quadratic formula<\/li>\n<li class=\"whitespace-normal break-words\">Use the type of roots to determine which case applies and write the general solution<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1170572332764\">Find the general solution to the following differential equations:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]y''-2y'+10y=0[\/latex]<\/li>\n<li>[latex]y''+14y'+49y=0[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q283472981\">Show Solution<\/button><\/p>\n<div id=\"q283472981\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. [latex]y(x)=e^x(c_1\\cos3x+c_2\\sin3x)[\/latex]<\/p>\n<p>b. [latex]y(x)=c_1e^{-7x}+c_2xe^{-7x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250336&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=f1v_hH_mO5Y&amp;video_target=tpm-plugin-y5rm8nb2-f1v_hH_mO5Y\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP7.6_transcript.html\">transcript for \u201cCP 7.6\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2>Initial-Value Problems and Boundary-Value Problems<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">Finding the general solution to a differential equation gives you a family of solutions with arbitrary constants. To find a specific solution that describes a particular physical system, you need additional information about the system at specific points in time.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Initial-Value Problems<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">An <strong>initial-value problem<\/strong> provides conditions at a single point in time [latex]t_0[\/latex]. For second-order equations, you need two initial conditions: typically [latex]y(t_0)[\/latex] and [latex]y'(t_0)[\/latex]. These two conditions let you solve for the two arbitrary constants [latex]c_1[\/latex] and [latex]c_2[\/latex] in the general solution. Initial-value problems for constant-coefficient second-order equations always have a unique solution when two initial conditions are provided.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Boundary-Value Problems<\/strong><\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\">A <strong>boundary-value problem<\/strong> provides conditions at two different points in time: [latex]y(t_0)[\/latex] and [latex]y(t_1)[\/latex]. Unlike initial-value problems, boundary-value problems are less predictable. Even with two boundary conditions, you might find a unique solution, infinitely many solutions, or no solution at all.<\/p>\n<p class=\"font-claude-response-body whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Find the general solution to the differential equation<\/li>\n<li class=\"whitespace-normal break-words\">Calculate [latex]y'(t)[\/latex] from the general solution (if working with initial conditions involving [latex]y'(t_0)[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">Substitute the given conditions into [latex]y(t)[\/latex] and [latex]y'(t)[\/latex] to create a system of equations<\/li>\n<li class=\"whitespace-normal break-words\">Solve the system for [latex]c_1[\/latex] and [latex]c_2[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Write the specific solution by substituting the values of [latex]c_1[\/latex] and [latex]c_2[\/latex] back into the general solution<\/li>\n<\/ol>\n<p class=\"font-claude-response-body whitespace-normal break-words\">For boundary-value problems, watch for inconsistent equations (no solution) or dependent equations (infinitely many solutions).<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q442172786\">Show Solution<\/button><\/p>\n<div id=\"q442172786\" class=\"hidden-answer\" style=\"display: none\">[latex]y(x)=-e^{-2x}+e^{-5x}[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q487721256\">Show Solution<\/button><\/p>\n<div id=\"q487721256\" class=\"hidden-answer\" style=\"display: none\">[latex]y(x)=e^{x}(2\\cos3x-\\sin3x)[\/latex]<\/p>\n<figure id=\"attachment_5613\" aria-describedby=\"caption-attachment-5613\" style=\"width: 971px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption id=\"caption-attachment-5613\" class=\"wp-caption-text\">Figure 2. Graph of [latex]y(x)=e^{x}(2\\cos3x-\\sin3x)[\/latex].<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q845982348\">Show Solution<\/button><\/p>\n<div id=\"q845982348\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y(t)=te^{-7t}[\/latex]<\/p>\n<figure id=\"attachment_5617\" aria-describedby=\"caption-attachment-5617\" style=\"width: 721px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption id=\"caption-attachment-5617\" class=\"wp-caption-text\">Figure 4.<\/figcaption><\/figure>\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<\/section>\n","protected":false},"author":15,"menu_order":11,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-816","chapter","type-chapter","status-publish","hentry"],"part":440,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/816","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/816\/revisions"}],"predecessor-version":[{"id":2262,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/816\/revisions\/2262"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/440"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/816\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=816"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=816"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=816"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=816"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}