{"id":5519,"date":"2022-06-02T19:04:57","date_gmt":"2022-06-02T19:04:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5519"},"modified":"2022-11-01T22:37:48","modified_gmt":"2022-11-01T22:37:48","slug":"second-order-equations-with-constant-coefficients","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/second-order-equations-with-constant-coefficients\/","title":{"raw":"Second-Order Equations with Constant Coefficients","rendered":"Second-Order Equations with Constant Coefficients"},"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\">Use the roots of the characteristic equation to find the solution to a homogeneous linear equation.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170572185069\">Now that we have a better feel for linear differential equations, we are going to concentrate on solving second-order equations of the form<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{ay^{\\prime\\prime}+by^\\prime+cy=0}[\/latex]<\/p>\r\nwhere [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are constants.\r\n<p id=\"fs-id1170572185130\">Since all the coefficients are constants, the solutions are probably going to be functions with derivatives that are constant multiples of themselves. We need all the terms to cancel out, and if taking a derivative introduces a term that is not a constant multiple of the original function, it is difficult to see how that term cancels out. Exponential functions have derivatives that are constant multiples of the original function, so let\u2019s see what happens when we try a solution of the form [latex]y(x)=e^{\\lambda x}[\/latex], where [latex]\\lambda[\/latex] (the lowercase Greek letter lambda) is some constant.<\/p>\r\n<p id=\"fs-id1170571571130\">If [latex]y(x)=e^{\\lambda x}[\/latex], then [latex]y^\\prime(x)=\\lambda e^{\\lambda x}[\/latex] and [latex]y^{\\prime\\prime}(x)=\\lambda^2e^{\\lambda x}[\/latex]. Substituting these expressions into a linear second-order differential equation, we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nay^{\\prime\\prime}+by^\\prime+cy&amp;=a(\\lambda^2e^{\\lambda x})+b(\\lambda e^{\\lambda x})+ce^{\\lambda x} \\\\\r\n&amp;=e^{\\lambda x}(a\\lambda^2+b\\lambda+c)\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170572386166\">Since [latex]e^{\\lambda x}[\/latex] is never zero, this expression can be equal to zero for all [latex]x[\/latex] only if<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{a\\lambda^1+b\\lambda+c=0}[\/latex].<\/p>\r\n<p id=\"fs-id1170572168698\">We call this the characteristic equation of the differential equation.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nThe\u00a0<strong><span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term301\" data-type=\"term\">characteristic equation<\/span><\/strong>\u00a0of the differential equation [latex]ay^{\\prime\\prime}+by^\\prime+cy=0[\/latex] is [latex]a\\lambda^2+b\\lambda+c+0[\/latex].\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572168770\">The characteristic equation is very important in finding solutions to differential equations of this form. We can solve the characteristic equation either by factoring or by using the quadratic formula<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\lambda=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}}[\/latex].<\/p>\r\n<p id=\"fs-id1170572439943\">This gives three cases. The characteristic equation has (1) distinct real roots; (2) a single, repeated real root; or (3) complex conjugate roots. We consider each of these cases separately.<\/p>\r\n\r\n<section id=\"fs-id1170572439948\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Distinct Real Roots<\/h3>\r\n<p id=\"fs-id1170572439954\">If the characteristic equation has distinct real roots [latex]\\lambda_1[\/latex] and [latex]\\lambda_2[\/latex], then [latex]e^{\\lambda_1 x}[\/latex] and [latex]e^{\\lambda_2 x}[\/latex] are linearly independent solutions to\u00a0Example \"Classifying Second-Order Equations\", and the general solution is given by<\/p>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1em;\">[latex]\\large{y(x)=c_1e^{\\lambda_1 x}+c_2e^{\\lambda_2 x}}[\/latex],<\/span><\/p>\r\n<p id=\"fs-id1170572218644\">where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\r\n<p id=\"fs-id1170572218664\">For example, the differential equation [latex]y^{\\prime\\prime}9y^\\prime+14y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2+9\\lambda+14=0[\/latex]. This factors into [latex](\\lambda+2)(\\lambda+7)=0[\/latex], which has roots [latex]\\lambda_1=-2[\/latex] and [latex]\\lambda_2=-7[\/latex]. Therefore, the general solution to this differential equation is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1e^{-2}+c_2e^{-7x}}[\/latex].<\/p>\r\n\r\n<\/section><section id=\"fs-id1170572185027\" data-depth=\"2\">\r\n<h3 data-type=\"title\">Single Repeated Real Root<\/h3>\r\n<p id=\"fs-id1170572185033\">Things are a little more complicated if the characteristic equation has a repeated real root, [latex]\\lambda[\/latex]. In this case, we know [latex]e^{\\lambda x}[\/latex] is a solution to\u00a0Example \"Classifying Second-Order Equations\", but it is only one solution and we need two linearly independent solutions to determine the general solution. We might be tempted to try a function of the form [latex]ke^{\\lambda x}[\/latex], where [latex]k[\/latex] is some constant, but it would not be linearly independent of [latex]e^{\\lambda x}[\/latex]. Therefore, let\u2019s try [latex]xe^{\\lambda x}[\/latex] as the second solution. First, note that by the quadratic formula,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\lambda=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}}[\/latex].<\/p>\r\n<p id=\"fs-id1170571579633\">But, [latex]\\lambda[\/latex] is a repeated root, so [latex]b^2-4ac=0[\/latex] and [latex]\\lambda=\\frac{-b}{2a}[\/latex]. Thus, if [latex]y=xe^{\\lambda x}[\/latex], we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y^\\prime=e^{\\lambda x}+\\lambda xe^{\\lambda x}\\text{ and }y^{\\prime\\prime}=2\\lambda e^{\\lambda x}+\\lambda^2xe^{\\lambda x}}[\/latex].<\/p>\r\n<p id=\"fs-id1170571629723\">Substituting these expressions into a linear second-order differential equation, we see that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nay^{\\prime\\prime}+by^\\prime+cy&amp;=a(2\\lambda e^{\\lambda x}+\\lambda^2e^{\\lambda x})+b(e^{\\lambda x}+\\lambda xe^{\\lambda x})+cxe^{\\lambda x} \\\\\r\n&amp;=xe^{\\lambda x}(a\\lambda^2+b\\lambda+c)+e^{\\lambda x}(2a\\lambda+b) \\\\\r\n&amp;=xe^{\\lambda x}(0)+e^{\\lambda x}\\left(2\\left(\\frac{-b}{2a}\\right)+b\\right) \\\\\r\n&amp;=0+e^{\\lambda x}(0) \\\\\r\n=0\r\n\\end{aligned}[\/latex].<\/p>\r\nThis shows that [latex]xe^{\\lambda x}[\/latex] is a solution to a linear second-order differential equation. Since [latex]e^{\\lambda x}[\/latex] and [latex]xe^{\\lambda x}[\/latex] are linearly independent, when the characteristic equation has a repeated root [latex]\\lambda[\/latex], the general solution to a linear second-order differential equation is given by\r\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1e^{\\lambda x}+c_2xe^{\\lambda x}}[\/latex],<\/p>\r\n<p id=\"fs-id1170571695755\">where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\r\n<p id=\"fs-id1170572569934\">For example, the differential equation [latex]y^{\\prime\\prime}+12y^\\prime+36y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2+12\\lambda+36=0[\/latex]. This factors into [latex](\\lambda+6)^2[\/latex], which has a repeated root [latex]\\lambda=-6[\/latex]. Therefore, the general solution to this differential equation is<\/p>\r\n\r\n<div style=\"text-align: center;\" data-type=\"note\">[latex]\\large{y(x)=c_1e^{-6x}+c_2xe^{-6x}}[\/latex].<\/div>\r\n<div data-type=\"note\"><\/div>\r\n<h3 data-type=\"title\">Complex Conjugate Roots<\/h3>\r\n<p id=\"fs-id1170571592522\">The third case we must consider is when [latex]b^2-4ac&lt;0[\/latex]. In this case, when we apply the quadratic formula, we are taking the square root of a negative number. We must use the imaginary number [latex]i=\\sqrt{-1}[\/latex] to find the roots, which take the form [latex]\\lambda_1=\\alpha+\\beta i[\/latex] and [latex]\\lambda_2=\\alpha-\\beta i[\/latex]. The\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term302\" class=\"no-emphasis\" data-type=\"term\">complex number [latex]\\alpha+\\beta i[\/latex]\u00a0<\/span>is called the\u00a0<em data-effect=\"italics\">conjugate<\/em>\u00a0of [latex]\\alpha-\\beta i[\/latex]. Thus, we see that when [latex]b^2-4ac&lt;0[\/latex], the roots of our characteristic equation are always\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term303\" class=\"no-emphasis\" data-type=\"term\">complex conjugates<\/span>.<\/p>\r\n<p id=\"fs-id1170572456336\">This creates a little bit of a problem for us. If we follow the same process we used for distinct real roots\u2014using the roots of the characteristic equation as the coefficients in the exponents of exponential functions\u2014we get the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] as our solutions. However, there are problems with this approach. First, these functions take on complex (imaginary) values, and a complete discussion of such functions is beyond the scope of this text. Second, even if we were comfortable with complex-value functions, in this course we do not address the idea of a derivative for such functions. So, if possible, we\u2019d like to find two linearly independent\u00a0<em data-effect=\"italics\">real-value<\/em>\u00a0solutions to the differential equation. For purposes of this development, we are going to manipulate and differentiate the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] as if they were real-value functions. For these particular functions, this approach is valid mathematically, but be aware that there are other instances when complex-value functions do not follow the same rules as real-value functions. Those of you interested in a more in-depth discussion of complex-value functions should consult a complex analysis text.<\/p>\r\n<p id=\"fs-id1170572548197\">Based on the roots [latex]\\alpha\\pm\\beta i[\/latex] of the characteristic equation, the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] are linearly independent solutions to the differential equation. and the general solution is given by<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1e^{(\\alpha+\\beta i)x}+c_2e^{(\\alpha-\\beta i)x}}[\/latex].<\/p>\r\n<p id=\"fs-id1170572331344\">Using some smart choices for [latex]c_1[\/latex] and [latex]c_2[\/latex], and a little bit of algebraic manipulation, we can find two linearly independent, real-value solutions to\u00a0a linear second-order differential equation\u00a0and express our general solution in those terms.<\/p>\r\n<p id=\"fs-id1170571700943\">We encountered exponential functions with complex exponents earlier. One of the key tools we used to express these exponential functions in terms of sines and cosines was\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term304\" class=\"no-emphasis\" data-type=\"term\">Euler\u2019s formula<\/span>, which tells us that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{e^{i\\theta}=\\cos\\theta+i\\sin\\theta}[\/latex]<\/p>\r\n<p id=\"fs-id1170571700984\">for all real numbers [latex]\\theta[\/latex].<\/p>\r\n<p id=\"fs-id1170571700993\">Going back to the general solution, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}\r\ny(x)&amp;=c_1e^{(\\alpha+\\beta i)x}+c_2e^{(\\alpha-\\beta i)x} \\\\\r\n&amp;=c_1e^{\\alpha x}e^{\\beta ix}+c_2e^{\\alpha x}e^{-\\beta ix} \\\\\r\n&amp;=e^{\\alpha x}\\left(c_1e^{\\beta ix}+c_2e^{-\\beta ix}\\right)\r\n\\end{aligned}}[\/latex].<\/p>\r\n<p id=\"fs-id1170572453504\">Applying Euler\u2019s formula together with the identities [latex]\\cos(-x)=\\cos x[\/latex] and [latex]\\sin(-x)=-\\sin x[\/latex], we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}\r\ny(x)&amp;=e^{\\alpha x}[c_1(\\cos\\beta x+i\\sin\\beta x)+c_2(\\cos(-\\beta x)+i\\sin(-\\beta x))] \\\\\r\n&amp;=e^{\\alpha x}[(c_1+c_2)\\cos\\beta x+(c_1-c_2)i\\sin\\beta x]\r\n\\end{aligned}}[\/latex].<\/p>\r\n<p id=\"fs-id1170571736591\">Now, if we choose [latex]c_1=c_2=\\frac{1}{2}[\/latex], the second term is zero and we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y(x)=e^{\\alpha x}\\cos\\beta x}[\/latex]<\/p>\r\n<p id=\"fs-id1170571736656\">as a real-value solution to\u00a0a linear second-order differential equation. Similarly, if we choose [latex]c_1=-\\frac{i}2[\/latex] and [latex]c_2=\\frac{i}2[\/latex], the first term is zero and we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y(x)=e^{\\alpha x}\\sin\\beta x}[\/latex]<\/p>\r\n<p id=\"fs-id1170572088637\">as a second, linearly independent, real-value solution to\u00a0a linear second-order differential equation.<\/p>\r\n<p id=\"fs-id1170572088645\">Based on this, we see that if the characteristic equation has complex conjugate roots [latex]\\alpha\\pm\\beta i[\/latex], then the general solution to\u00a0a linear second-order differential equation\u00a0is given by<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}\r\ny(x)&amp;=c_1e^{\\alpha x}\\cos\\beta x+c_2e^{\\alpha x}\\cos\\sin\\beta x \\\\\r\n&amp;=e^{\\alpha x}(c_1\\cos\\beta x+c_2\\sin\\beta x)\r\n\\end{aligned}}[\/latex].<\/p>\r\n<p id=\"fs-id1170572330194\">where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\r\n<p id=\"fs-id1170572330214\">For example, the differential equation [latex]y^{\\prime\\prime}-2y^\\prime+5y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2-2\\lambda+5=0[\/latex]. By the quadratic formula, the roots of the characteristic equation are [latex]1\\pm2i[\/latex]. Therefore, the general solution to this differential equation is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y(x)=e^x(c_1\\cos2x+c_2\\sin2x)}[\/latex].<\/p>\r\n\r\n<div data-type=\"note\">\r\n<h3 data-type=\"title\">Summary of Results<\/h3>\r\n<p id=\"fs-id1170571769731\">We can solve second-order, linear, homogeneous differential equations with constant coefficients by finding the roots of the associated characteristic equation. The form of the general solution varies, depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots. The three cases are summarized in\u00a0Table 7.1 Summary of Characteristic Equation Cases below.<\/p>\r\n\r\n<div id=\"fs-id1170571769749\" class=\"os-table\">\r\n<table style=\"width: 620px;\" data-id=\"fs-id1170571769749\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th style=\"width: 196.391px;\" scope=\"col\" data-align=\"left\">Characteristic Equation Roots<\/th>\r\n<th style=\"width: 397.609px;\" scope=\"col\" data-align=\"left\">General Solution to the Differential Equation<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td style=\"width: 196.391px;\" data-align=\"left\">Distinct real roots,\u00a0[latex]\\lambda_1[\/latex] and\u00a0[latex]\\lambda_2[\/latex]<\/td>\r\n<td style=\"width: 397.609px;\" data-align=\"left\">[latex]y(x)=c_1e^{\\lambda_1x}+c_2e^{\\lambda_2x}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"width: 196.391px;\" data-align=\"left\">A repeated real root, [latex]\\lambda[\/latex]<\/td>\r\n<td style=\"width: 397.609px;\" data-align=\"left\">[latex]y(x)=c_1e^{\\lambda x}+c_2xe^{\\lambda x}[\/latex]<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"width: 196.391px;\" data-align=\"left\">Complex conjugate roots [latex]\\alpha\\pm\\beta i[\/latex]<\/td>\r\n<td style=\"width: 397.609px;\" data-align=\"left\">[latex]y(x)=e^{\\lambda x}(c_1\\cos\\beta x+c_2\\sin\\beta x)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"os-caption-container\"><span class=\"os-title-label\">Table<\/span>\u00a0<span class=\"os-number\">7.1<\/span>\u00a0<span class=\"os-title\" data-type=\"title\">Summary of Characteristic Equation Cases<\/span><\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1167794333153\" class=\"problem-solving\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3>problem-solving strategy using the characteristic equation to solve second order differential equations with constant coefficients<\/h3>\r\n<ol id=\"fs-id1170572292216\" type=\"1\">\r\n \t<li>Write the differential equation in the form [latex]ay''+by'+cy=0[\/latex].<\/li>\r\n \t<li>Find the corresponding characteristic equation [latex]a\\lambda^2+b\\lambda+c=0[\/latex].<\/li>\r\n \t<li>Either factor the characteristic equation or use the quadratic formula to find the roots.<\/li>\r\n \t<li>Determine the form of the general solution based on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: solving second-order equations with constant coefficients<\/h3>\r\n<p id=\"fs-id1170571735188\">Find the general solution to the following differential equations. Give your answers as functions of [latex]x[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1170571735197\" type=\"a\">\r\n \t<li>[latex]y''+3y'-4y=0[\/latex]<\/li>\r\n \t<li>[latex]y''+6y'+13y=0[\/latex]<\/li>\r\n \t<li>[latex]y''+2y'+y=0[\/latex]<\/li>\r\n \t<li>[latex]y''-5y'=0[\/latex]<\/li>\r\n \t<li>[latex]y''-16y=0[\/latex]<\/li>\r\n \t<li>[latex]y''+16y=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"372498512\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"372498512\"]\r\n<p id=\"fs-id1170572268258\">Note that all these equations are already given in standard form (step 1).<\/p>\r\n\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol type=\"a\">\r\n \t<li>The characteristic equation is [latex]\\lambda^2+3\\lambda-4=0[\/latex] (step 2). This factors into [latex](\\lambda+4)(\\lambda-1)=0[\/latex], so the roots of the characteristic equation are [latex]\\lambda_1=-4[\/latex] and [latex]\\lambda_2=1[\/latex] (step 3). Then the general solution to the differential equation is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572589239\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]y(x)=c_1e^{-4x}+c_2e^x[\/latex] (step 4).<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li>The characteristic equation is [latex]\\lambda^2+6\\lambda+13=0[\/latex] (step 2). Applying the quadratic formula, we see this equation has complex conjugate roots [latex](-3\\pm2i[\/latex] (step 3). Then the general solution to the differential equation is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572296619\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]y(t)=e^{-3t}(c_1\\cos2t+c_2\\sin2t)[\/latex] (step 4).<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li>The characteristic equation is [latex]\\lambda^2+2\\lambda+1=0[\/latex] (step 2). This factors into [latex](\\lambda+1)^2=0[\/latex], so the characteristic equation has a repeated real root [latex]\\lambda=-1[\/latex] (step 3). Then the general solution to the differential equation is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572468885\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]y(t)=c_1e^{-t}+c_2te^{-t}[\/latex] (step 4).<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li>The characteristic equation is [latex]\\lambda^2-5\\lambda=0[\/latex] (step 2). This factors into [latex]\\lambda(\\lambda-5)=0[\/latex], so the roots of the characteristic equation are [latex]\\lambda_1=0[\/latex] and [latex]\\lambda_2=5[\/latex] (step 3). Note that [latex]e^{0x}=e^0=1[\/latex], so our first solution is just a constant. Then the general solution to the differential equation is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571531346\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]y(x)=c_1+c_2e^{5x}[\/latex] (step 4).<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li>The characteristic equation is [latex]\\lambda^2-16=0[\/latex] (step 2). This factors into [latex](\\lambda+4)(\\lambda-4)=0[\/latex], so the roots of the characteristic equation are [latex]\\lambda_1=4[\/latex] and [latex]\\lambda_2=-4[\/latex] (step 3). Then the general solution to the differential equation is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572306377\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]y(x)=c_1e^{4x}+c_2e^{-4x}[\/latex] (step 4).<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li>The characteristic equation is [latex]\\lambda^2+16=0[\/latex] (step 2). This has complex conjugate roots [latex]\\pm4i[\/latex] (step 3). Note that [latex]e^{0x}=e^0=1[\/latex], so the exponential term in our solution is just a constant. Then the general solution to the differential equation is<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">[latex]y(t)=c_1\\cos4t+c_2\\sin4t[\/latex] (step 4).<\/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-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<\/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=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\"><\/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><\/div>\r\n<\/div>\r\n<\/section>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Use the roots of the characteristic equation to find the solution to a homogeneous linear equation.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170572185069\">Now that we have a better feel for linear differential equations, we are going to concentrate on solving second-order equations of the form<\/p>\n<p style=\"text-align: center;\">[latex]\\large{ay^{\\prime\\prime}+by^\\prime+cy=0}[\/latex]<\/p>\n<p>where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex] are constants.<\/p>\n<p id=\"fs-id1170572185130\">Since all the coefficients are constants, the solutions are probably going to be functions with derivatives that are constant multiples of themselves. We need all the terms to cancel out, and if taking a derivative introduces a term that is not a constant multiple of the original function, it is difficult to see how that term cancels out. Exponential functions have derivatives that are constant multiples of the original function, so let\u2019s see what happens when we try a solution of the form [latex]y(x)=e^{\\lambda x}[\/latex], where [latex]\\lambda[\/latex] (the lowercase Greek letter lambda) is some constant.<\/p>\n<p id=\"fs-id1170571571130\">If [latex]y(x)=e^{\\lambda x}[\/latex], then [latex]y^\\prime(x)=\\lambda e^{\\lambda x}[\/latex] and [latex]y^{\\prime\\prime}(x)=\\lambda^2e^{\\lambda x}[\/latex]. Substituting these expressions into a linear second-order differential equation, we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  ay^{\\prime\\prime}+by^\\prime+cy&=a(\\lambda^2e^{\\lambda x})+b(\\lambda e^{\\lambda x})+ce^{\\lambda x} \\\\  &=e^{\\lambda x}(a\\lambda^2+b\\lambda+c)  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170572386166\">Since [latex]e^{\\lambda x}[\/latex] is never zero, this expression can be equal to zero for all [latex]x[\/latex] only if<\/p>\n<p style=\"text-align: center;\">[latex]\\large{a\\lambda^1+b\\lambda+c=0}[\/latex].<\/p>\n<p id=\"fs-id1170572168698\">We call this the characteristic equation of the differential equation.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p>The\u00a0<strong><span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term301\" data-type=\"term\">characteristic equation<\/span><\/strong>\u00a0of the differential equation [latex]ay^{\\prime\\prime}+by^\\prime+cy=0[\/latex] is [latex]a\\lambda^2+b\\lambda+c+0[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1170572168770\">The characteristic equation is very important in finding solutions to differential equations of this form. We can solve the characteristic equation either by factoring or by using the quadratic formula<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\lambda=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}}[\/latex].<\/p>\n<p id=\"fs-id1170572439943\">This gives three cases. The characteristic equation has (1) distinct real roots; (2) a single, repeated real root; or (3) complex conjugate roots. We consider each of these cases separately.<\/p>\n<section id=\"fs-id1170572439948\" data-depth=\"2\">\n<h3 data-type=\"title\">Distinct Real Roots<\/h3>\n<p id=\"fs-id1170572439954\">If the characteristic equation has distinct real roots [latex]\\lambda_1[\/latex] and [latex]\\lambda_2[\/latex], then [latex]e^{\\lambda_1 x}[\/latex] and [latex]e^{\\lambda_2 x}[\/latex] are linearly independent solutions to\u00a0Example &#8220;Classifying Second-Order Equations&#8221;, and the general solution is given by<\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1em;\">[latex]\\large{y(x)=c_1e^{\\lambda_1 x}+c_2e^{\\lambda_2 x}}[\/latex],<\/span><\/p>\n<p id=\"fs-id1170572218644\">where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\n<p id=\"fs-id1170572218664\">For example, the differential equation [latex]y^{\\prime\\prime}9y^\\prime+14y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2+9\\lambda+14=0[\/latex]. This factors into [latex](\\lambda+2)(\\lambda+7)=0[\/latex], which has roots [latex]\\lambda_1=-2[\/latex] and [latex]\\lambda_2=-7[\/latex]. Therefore, the general solution to this differential equation is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1e^{-2}+c_2e^{-7x}}[\/latex].<\/p>\n<\/section>\n<section id=\"fs-id1170572185027\" data-depth=\"2\">\n<h3 data-type=\"title\">Single Repeated Real Root<\/h3>\n<p id=\"fs-id1170572185033\">Things are a little more complicated if the characteristic equation has a repeated real root, [latex]\\lambda[\/latex]. In this case, we know [latex]e^{\\lambda x}[\/latex] is a solution to\u00a0Example &#8220;Classifying Second-Order Equations&#8221;, but it is only one solution and we need two linearly independent solutions to determine the general solution. We might be tempted to try a function of the form [latex]ke^{\\lambda x}[\/latex], where [latex]k[\/latex] is some constant, but it would not be linearly independent of [latex]e^{\\lambda x}[\/latex]. Therefore, let\u2019s try [latex]xe^{\\lambda x}[\/latex] as the second solution. First, note that by the quadratic formula,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\lambda=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}}[\/latex].<\/p>\n<p id=\"fs-id1170571579633\">But, [latex]\\lambda[\/latex] is a repeated root, so [latex]b^2-4ac=0[\/latex] and [latex]\\lambda=\\frac{-b}{2a}[\/latex]. Thus, if [latex]y=xe^{\\lambda x}[\/latex], we have<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y^\\prime=e^{\\lambda x}+\\lambda xe^{\\lambda x}\\text{ and }y^{\\prime\\prime}=2\\lambda e^{\\lambda x}+\\lambda^2xe^{\\lambda x}}[\/latex].<\/p>\n<p id=\"fs-id1170571629723\">Substituting these expressions into a linear second-order differential equation, we see that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  ay^{\\prime\\prime}+by^\\prime+cy&=a(2\\lambda e^{\\lambda x}+\\lambda^2e^{\\lambda x})+b(e^{\\lambda x}+\\lambda xe^{\\lambda x})+cxe^{\\lambda x} \\\\  &=xe^{\\lambda x}(a\\lambda^2+b\\lambda+c)+e^{\\lambda x}(2a\\lambda+b) \\\\  &=xe^{\\lambda x}(0)+e^{\\lambda x}\\left(2\\left(\\frac{-b}{2a}\\right)+b\\right) \\\\  &=0+e^{\\lambda x}(0) \\\\  =0  \\end{aligned}[\/latex].<\/p>\n<p>This shows that [latex]xe^{\\lambda x}[\/latex] is a solution to a linear second-order differential equation. Since [latex]e^{\\lambda x}[\/latex] and [latex]xe^{\\lambda x}[\/latex] are linearly independent, when the characteristic equation has a repeated root [latex]\\lambda[\/latex], the general solution to a linear second-order differential equation is given by<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1e^{\\lambda x}+c_2xe^{\\lambda x}}[\/latex],<\/p>\n<p id=\"fs-id1170571695755\">where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\n<p id=\"fs-id1170572569934\">For example, the differential equation [latex]y^{\\prime\\prime}+12y^\\prime+36y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2+12\\lambda+36=0[\/latex]. This factors into [latex](\\lambda+6)^2[\/latex], which has a repeated root [latex]\\lambda=-6[\/latex]. Therefore, the general solution to this differential equation is<\/p>\n<div style=\"text-align: center;\" data-type=\"note\">[latex]\\large{y(x)=c_1e^{-6x}+c_2xe^{-6x}}[\/latex].<\/div>\n<div data-type=\"note\"><\/div>\n<h3 data-type=\"title\">Complex Conjugate Roots<\/h3>\n<p id=\"fs-id1170571592522\">The third case we must consider is when [latex]b^2-4ac<0[\/latex]. In this case, when we apply the quadratic formula, we are taking the square root of a negative number. We must use the imaginary number [latex]i=\\sqrt{-1}[\/latex] to find the roots, which take the form [latex]\\lambda_1=\\alpha+\\beta i[\/latex] and [latex]\\lambda_2=\\alpha-\\beta i[\/latex]. The\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term302\" class=\"no-emphasis\" data-type=\"term\">complex number [latex]\\alpha+\\beta i[\/latex]\u00a0<\/span>is called the\u00a0<em data-effect=\"italics\">conjugate<\/em>\u00a0of [latex]\\alpha-\\beta i[\/latex]. Thus, we see that when [latex]b^2-4ac<0[\/latex], the roots of our characteristic equation are always\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term303\" class=\"no-emphasis\" data-type=\"term\">complex conjugates<\/span>.<\/p>\n<p id=\"fs-id1170572456336\">This creates a little bit of a problem for us. If we follow the same process we used for distinct real roots\u2014using the roots of the characteristic equation as the coefficients in the exponents of exponential functions\u2014we get the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] as our solutions. However, there are problems with this approach. First, these functions take on complex (imaginary) values, and a complete discussion of such functions is beyond the scope of this text. Second, even if we were comfortable with complex-value functions, in this course we do not address the idea of a derivative for such functions. So, if possible, we\u2019d like to find two linearly independent\u00a0<em data-effect=\"italics\">real-value<\/em>\u00a0solutions to the differential equation. For purposes of this development, we are going to manipulate and differentiate the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] as if they were real-value functions. For these particular functions, this approach is valid mathematically, but be aware that there are other instances when complex-value functions do not follow the same rules as real-value functions. Those of you interested in a more in-depth discussion of complex-value functions should consult a complex analysis text.<\/p>\n<p id=\"fs-id1170572548197\">Based on the roots [latex]\\alpha\\pm\\beta i[\/latex] of the characteristic equation, the functions [latex]e^{(\\alpha+\\beta i)x}[\/latex] and [latex]e^{(\\alpha-\\beta i)x}[\/latex] are linearly independent solutions to the differential equation. and the general solution is given by<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1e^{(\\alpha+\\beta i)x}+c_2e^{(\\alpha-\\beta i)x}}[\/latex].<\/p>\n<p id=\"fs-id1170572331344\">Using some smart choices for [latex]c_1[\/latex] and [latex]c_2[\/latex], and a little bit of algebraic manipulation, we can find two linearly independent, real-value solutions to\u00a0a linear second-order differential equation\u00a0and express our general solution in those terms.<\/p>\n<p id=\"fs-id1170571700943\">We encountered exponential functions with complex exponents earlier. One of the key tools we used to express these exponential functions in terms of sines and cosines was\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term304\" class=\"no-emphasis\" data-type=\"term\">Euler\u2019s formula<\/span>, which tells us that<\/p>\n<p style=\"text-align: center;\">[latex]\\large{e^{i\\theta}=\\cos\\theta+i\\sin\\theta}[\/latex]<\/p>\n<p id=\"fs-id1170571700984\">for all real numbers [latex]\\theta[\/latex].<\/p>\n<p id=\"fs-id1170571700993\">Going back to the general solution, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}  y(x)&=c_1e^{(\\alpha+\\beta i)x}+c_2e^{(\\alpha-\\beta i)x} \\\\  &=c_1e^{\\alpha x}e^{\\beta ix}+c_2e^{\\alpha x}e^{-\\beta ix} \\\\  &=e^{\\alpha x}\\left(c_1e^{\\beta ix}+c_2e^{-\\beta ix}\\right)  \\end{aligned}}[\/latex].<\/p>\n<p id=\"fs-id1170572453504\">Applying Euler\u2019s formula together with the identities [latex]\\cos(-x)=\\cos x[\/latex] and [latex]\\sin(-x)=-\\sin x[\/latex], we get<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}  y(x)&=e^{\\alpha x}[c_1(\\cos\\beta x+i\\sin\\beta x)+c_2(\\cos(-\\beta x)+i\\sin(-\\beta x))] \\\\  &=e^{\\alpha x}[(c_1+c_2)\\cos\\beta x+(c_1-c_2)i\\sin\\beta x]  \\end{aligned}}[\/latex].<\/p>\n<p id=\"fs-id1170571736591\">Now, if we choose [latex]c_1=c_2=\\frac{1}{2}[\/latex], the second term is zero and we get<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y(x)=e^{\\alpha x}\\cos\\beta x}[\/latex]<\/p>\n<p id=\"fs-id1170571736656\">as a real-value solution to\u00a0a linear second-order differential equation. Similarly, if we choose [latex]c_1=-\\frac{i}2[\/latex] and [latex]c_2=\\frac{i}2[\/latex], the first term is zero and we get<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y(x)=e^{\\alpha x}\\sin\\beta x}[\/latex]<\/p>\n<p id=\"fs-id1170572088637\">as a second, linearly independent, real-value solution to\u00a0a linear second-order differential equation.<\/p>\n<p id=\"fs-id1170572088645\">Based on this, we see that if the characteristic equation has complex conjugate roots [latex]\\alpha\\pm\\beta i[\/latex], then the general solution to\u00a0a linear second-order differential equation\u00a0is given by<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\begin{aligned}  y(x)&=c_1e^{\\alpha x}\\cos\\beta x+c_2e^{\\alpha x}\\cos\\sin\\beta x \\\\  &=e^{\\alpha x}(c_1\\cos\\beta x+c_2\\sin\\beta x)  \\end{aligned}}[\/latex].<\/p>\n<p id=\"fs-id1170572330194\">where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\n<p id=\"fs-id1170572330214\">For example, the differential equation [latex]y^{\\prime\\prime}-2y^\\prime+5y=0[\/latex] has the associated characteristic equation [latex]\\lambda^2-2\\lambda+5=0[\/latex]. By the quadratic formula, the roots of the characteristic equation are [latex]1\\pm2i[\/latex]. Therefore, the general solution to this differential equation is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y(x)=e^x(c_1\\cos2x+c_2\\sin2x)}[\/latex].<\/p>\n<div data-type=\"note\">\n<h3 data-type=\"title\">Summary of Results<\/h3>\n<p id=\"fs-id1170571769731\">We can solve second-order, linear, homogeneous differential equations with constant coefficients by finding the roots of the associated characteristic equation. The form of the general solution varies, depending on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots. The three cases are summarized in\u00a0Table 7.1 Summary of Characteristic Equation Cases below.<\/p>\n<div id=\"fs-id1170571769749\" class=\"os-table\">\n<table style=\"width: 620px;\" data-id=\"fs-id1170571769749\">\n<thead>\n<tr valign=\"top\">\n<th style=\"width: 196.391px;\" scope=\"col\" data-align=\"left\">Characteristic Equation Roots<\/th>\n<th style=\"width: 397.609px;\" scope=\"col\" data-align=\"left\">General Solution to the Differential Equation<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td style=\"width: 196.391px;\" data-align=\"left\">Distinct real roots,\u00a0[latex]\\lambda_1[\/latex] and\u00a0[latex]\\lambda_2[\/latex]<\/td>\n<td style=\"width: 397.609px;\" data-align=\"left\">[latex]y(x)=c_1e^{\\lambda_1x}+c_2e^{\\lambda_2x}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"width: 196.391px;\" data-align=\"left\">A repeated real root, [latex]\\lambda[\/latex]<\/td>\n<td style=\"width: 397.609px;\" data-align=\"left\">[latex]y(x)=c_1e^{\\lambda x}+c_2xe^{\\lambda x}[\/latex]<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"width: 196.391px;\" data-align=\"left\">Complex conjugate roots [latex]\\alpha\\pm\\beta i[\/latex]<\/td>\n<td style=\"width: 397.609px;\" data-align=\"left\">[latex]y(x)=e^{\\lambda x}(c_1\\cos\\beta x+c_2\\sin\\beta x)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"os-caption-container\"><span class=\"os-title-label\">Table<\/span>\u00a0<span class=\"os-number\">7.1<\/span>\u00a0<span class=\"os-title\" data-type=\"title\">Summary of Characteristic Equation Cases<\/span><\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1167794333153\" class=\"problem-solving\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3>problem-solving strategy using the characteristic equation to solve second order differential equations with constant coefficients<\/h3>\n<ol id=\"fs-id1170572292216\" type=\"1\">\n<li>Write the differential equation in the form [latex]ay''+by'+cy=0[\/latex].<\/li>\n<li>Find the corresponding characteristic equation [latex]a\\lambda^2+b\\lambda+c=0[\/latex].<\/li>\n<li>Either factor the characteristic equation or use the quadratic formula to find the roots.<\/li>\n<li>Determine the form of the general solution based on whether the characteristic equation has distinct, real roots; a single, repeated real root; or complex conjugate roots.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: solving second-order equations with constant coefficients<\/h3>\n<p id=\"fs-id1170571735188\">Find the general solution to the following differential equations. Give your answers as functions of [latex]x[\/latex].<\/p>\n<ol id=\"fs-id1170571735197\" type=\"a\">\n<li>[latex]y''+3y'-4y=0[\/latex]<\/li>\n<li>[latex]y''+6y'+13y=0[\/latex]<\/li>\n<li>[latex]y''+2y'+y=0[\/latex]<\/li>\n<li>[latex]y''-5y'=0[\/latex]<\/li>\n<li>[latex]y''-16y=0[\/latex]<\/li>\n<li>[latex]y''+16y=0[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q372498512\">Show Solution<\/span><\/p>\n<div id=\"q372498512\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572268258\">Note that all these equations are already given in standard form (step 1).<\/p>\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ol type=\"a\">\n<li>The characteristic equation is [latex]\\lambda^2+3\\lambda-4=0[\/latex] (step 2). This factors into [latex](\\lambda+4)(\\lambda-1)=0[\/latex], so the roots of the characteristic equation are [latex]\\lambda_1=-4[\/latex] and [latex]\\lambda_2=1[\/latex] (step 3). Then the general solution to the differential equation is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572589239\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]y(x)=c_1e^{-4x}+c_2e^x[\/latex] (step 4).<\/p>\n<\/div>\n<\/li>\n<li>The characteristic equation is [latex]\\lambda^2+6\\lambda+13=0[\/latex] (step 2). Applying the quadratic formula, we see this equation has complex conjugate roots [latex](-3\\pm2i[\/latex] (step 3). Then the general solution to the differential equation is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572296619\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]y(t)=e^{-3t}(c_1\\cos2t+c_2\\sin2t)[\/latex] (step 4).<\/p>\n<\/div>\n<\/li>\n<li>The characteristic equation is [latex]\\lambda^2+2\\lambda+1=0[\/latex] (step 2). This factors into [latex](\\lambda+1)^2=0[\/latex], so the characteristic equation has a repeated real root [latex]\\lambda=-1[\/latex] (step 3). Then the general solution to the differential equation is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572468885\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]y(t)=c_1e^{-t}+c_2te^{-t}[\/latex] (step 4).<\/p>\n<\/div>\n<\/li>\n<li>The characteristic equation is [latex]\\lambda^2-5\\lambda=0[\/latex] (step 2). This factors into [latex]\\lambda(\\lambda-5)=0[\/latex], so the roots of the characteristic equation are [latex]\\lambda_1=0[\/latex] and [latex]\\lambda_2=5[\/latex] (step 3). Note that [latex]e^{0x}=e^0=1[\/latex], so our first solution is just a constant. Then the general solution to the differential equation is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571531346\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]y(x)=c_1+c_2e^{5x}[\/latex] (step 4).<\/p>\n<\/div>\n<\/li>\n<li>The characteristic equation is [latex]\\lambda^2-16=0[\/latex] (step 2). This factors into [latex](\\lambda+4)(\\lambda-4)=0[\/latex], so the roots of the characteristic equation are [latex]\\lambda_1=4[\/latex] and [latex]\\lambda_2=-4[\/latex] (step 3). Then the general solution to the differential equation is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572306377\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]y(x)=c_1e^{4x}+c_2e^{-4x}[\/latex] (step 4).<\/p>\n<\/div>\n<\/li>\n<li>The characteristic equation is [latex]\\lambda^2+16=0[\/latex] (step 2). This has complex conjugate roots [latex]\\pm4i[\/latex] (step 3). Note that [latex]e^{0x}=e^0=1[\/latex], so the exponential term in our solution is just a constant. Then the general solution to the differential equation is<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center;\">[latex]y(t)=c_1\\cos4t+c_2\\sin4t[\/latex] (step 4).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q283472981\">Show Solution<\/span><\/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<\/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=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\"><\/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<\/div>\n<\/div>\n<\/section>\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-5519\">\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 7.6. <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":349141,"menu_order":4,"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 7.6\",\"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-5519","chapter","type-chapter","status-publish","hentry"],"part":25,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5519","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\/349141"}],"version-history":[{"count":16,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5519\/revisions"}],"predecessor-version":[{"id":6413,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5519\/revisions\/6413"}],"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\/5519\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=5519"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=5519"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=5519"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=5519"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}