{"id":5517,"date":"2022-06-02T19:04:05","date_gmt":"2022-06-02T19:04:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5517"},"modified":"2022-11-01T22:36:33","modified_gmt":"2022-11-01T22:36:33","slug":"homogeneous-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/homogeneous-linear-equations\/","title":{"raw":"Homogeneous Linear Equations","rendered":"Homogeneous Linear Equations"},"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\">Recognize homogeneous and nonhomogeneous linear differential equations.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Determine the characteristic equation of a homogeneous linear equation.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1170571769962\" data-depth=\"1\">\r\n<p id=\"fs-id1170571585385\">Consider the second-order differential equation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{xy^{\\prime\\prime}+2x^2y^\\prime+5x^3y=0}[\/latex].<\/p>\r\n<p id=\"fs-id1170571602172\">Notice that [latex]y[\/latex] and its derivatives appear in a relatively simple form. They are multiplied by functions of [latex]x[\/latex], but are not raised to any powers themselves, nor are they multiplied together. As discussed in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/basics-of-differential-equations\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"1d39a348-071f-4537-85b6-c98912458c3c\" data-page-slug=\"4-introduction\">Introduction to Differential Equations<\/a>, first-order equations with similar characteristics are said to be linear. The same is true of second-order equations. Also note that all the terms in this differential equation involve either [latex]y[\/latex] or one of its derivatives. There are no terms involving only functions of [latex]x[\/latex]. Equations like this, in which every term contains [latex]y[\/latex] or one of its derivatives, are called homogeneous.<\/p>\r\n<p id=\"fs-id1170572582703\">Not all differential equations are homogeneous. Consider the differential equation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{xy^{\\prime\\prime}+2x^2y^\\prime+5x^3y=x^2}[\/latex].<\/p>\r\n<p id=\"fs-id1170572169093\">The [latex]x^{2}[\/latex] term on the right side of the equal sign does not contain [latex]y[\/latex] or any of its derivatives. Therefore, this differential equation is nonhomogeneous.<\/p>\r\n\r\n<\/section>\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572215852\">A second-order differential equation is linear if it can be written in the form<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{a_2(x)y^{\\prime\\prime}+a_1(x)y^{\\prime}+a_0(x)y=r(x)}[\/latex],<\/p>\r\n<p id=\"fs-id1170572404274\">where [latex]a_2(x)[\/latex], [latex]a_1(x)[\/latex], [latex]a_0(x)[\/latex], and [latex]r(x)[\/latex] are real-valued functions and [latex]a_2(x)[\/latex] is not identically zero. If [latex]r(x)=0[\/latex]\u2014in other words, if [latex]r(x)=0[\/latex] for every value of [latex]x[\/latex]\u2014the equation is said to be a\u00a0<strong><span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term295\" data-type=\"term\">homogeneous linear equation<\/span><\/strong>. If [latex]r(x)\\ne0[\/latex] for some value of [latex]x[\/latex], the equation is said to be a\u00a0<strong><span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term296\" data-type=\"term\">nonhomogeneous linear equation<\/span>.<\/strong><\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\nVisit 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.\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n<p id=\"fs-id1170572280427\">In linear differential equations, [latex]y[\/latex] and its derivatives can be raised only to the first power and they may not be multiplied by one another. Terms involving [latex]y^{2}[\/latex] or [latex]\\sqrt{y^\\prime}[\/latex] make the equation nonlinear. Functions of [latex]y[\/latex] and its derivatives, such as [latex]\\sin y[\/latex] or [latex]e^{y^\\prime}[\/latex] are similarly prohibited in linear differential equations.<\/p>\r\n<p id=\"fs-id1170571635343\">Note that equations may not always be given in standard form (the form shown in the definition). It can be helpful to rewrite them in that form to decide whether they are linear, or whether a linear equation is homogeneous.<\/p>\r\n\r\n<div id=\"fs-id1170571650370\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Classifying second-order equations<\/h3>\r\n<p id=\"fs-id1170572307004\">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-id1170572547853\" type=\"a\">\r\n \t<li>[latex]y''+3x^{4}y'+x^{2}y^{2}=x^{3}[\/latex]<\/li>\r\n \t<li>[latex](\\sin x)y''+(\\cos x)y'+3y=0[\/latex]<\/li>\r\n \t<li>[latex]4t^{2}x''+3txx'+4x=0[\/latex]<\/li>\r\n \t<li>[latex]5y''+y=4x^{5}[\/latex]<\/li>\r\n \t<li>[latex](\\cos x)y''-\\sin y'+(\\sin x)y-\\cos x=0[\/latex]<\/li>\r\n \t<li>[latex]8ty''-6t^{2}y'+4ty-3t^2=0[\/latex]<\/li>\r\n \t<li>[latex]\\sin (x^{2})y''-(\\cos x)y'+x^{2}y=y'-3[\/latex]<\/li>\r\n \t<li>[latex]y''+5xy'-3y=\\cos y[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"924732071\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"924732071\"]\r\n<ol id=\"fs-id1170572337089\" type=\"a\">\r\n \t<li>This equation is nonlinear because of the [latex]y^{2}[\/latex] term.<\/li>\r\n \t<li>This equation is linear. There is no term involving a power or function of [latex]y[\/latex], and the coefficients are all functions of [latex]x[\/latex]. The equation is already written in standard form, and [latex]r(x)[\/latex] is identically zero, so the equation is homogeneous.<\/li>\r\n \t<li>This equation is nonlinear. Note that, in this case, [latex]x[\/latex] is the dependent variable and [latex]t[\/latex] is the independent variable. The second term involves the product of [latex]x[\/latex] and [latex]x'[\/latex], so the equation is nonlinear.<\/li>\r\n \t<li>This equation is linear. Since [latex]r(x)=4x^{5}[\/latex], the equation is nonhomogeneous.<\/li>\r\n \t<li>This equation is nonlinear, because of the [latex]\\sin y'[\/latex] term.<\/li>\r\n \t<li>This equation is linear. Rewriting it in standard form gives\r\n<span data-type=\"newline\">[latex]8t^{2}-6t^{2}y'+4ty=3t^{2}[\/latex].\r\n<\/span>With the equation in standard form, we can see that [latex]r(t)=3t^2[\/latex] so the equation is nonhomogeneous.<\/li>\r\n \t<li>This equation looks like it\u2019s linear, but we should rewrite it in standard form to be sure. We get\r\n[latex]\\sin(x^{2})y''-(\\cos x+1)y'+x^{2}y=-3[\/latex].This equation is, indeed, linear. With [latex]r(x)=-3[\/latex], it is nonhomogeneous.<\/li>\r\n \t<li>This equation is nonlinear because of the [latex]\\cos y[\/latex] term.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\nVisit this\u00a0<a href=\"http:\/\/www.sosmath.com\/diffeq\/diffeq.html\" target=\"_blank\" rel=\"noopener nofollow\">website<\/a>\u00a0that discusses second-order differential equations.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\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<\/div>\r\n<p id=\"fs-id1170572553210\">Later in this section, we will see some techniques for solving specific types of differential equations. Before we get to that, however, let\u2019s get a feel for how solutions to linear differential equations behave. In many cases, solving differential equations depends on making educated guesses about what the solution might look like. Knowing how various types of solutions behave will be helpful.<\/p>\r\n\r\n<div id=\"fs-id1170572560326\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: verifying a solution<\/h3>\r\n<p id=\"fs-id1170572106755\">Consider the linear, homogeneous differential equation<\/p>\r\n[latex]x^{2}y''-xy'-3y=0[\/latex].\r\n<p id=\"fs-id1170572307137\">Looking at this equation, notice that the coefficient functions are polynomials, with higher powers of [latex]x[\/latex] associated with higher-order derivatives of [latex]y[\/latex]. Show that [latex]y=x^{3}[\/latex] is a solution to this differential equation.<\/p>\r\n[reveal-answer q=\"731671820\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"731671820\"]\r\n<p id=\"fs-id1170572225198\">Let [latex]y=x^{3}[\/latex]. Then [latex]y'=3x^{2}[\/latex] and [latex]y''=6(x)[\/latex]. Substituting into the differential equation, we see that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx^{2}y''-xy'-3y&amp;=x^2(6x)-x(3x^2)-3(x^3) \\\\\r\n&amp;=6x^3-3x^3-3x^3 \\\\\r\n&amp;=0\r\n\\end{aligned}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\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<\/div>\r\n<p id=\"fs-id1170572505444\">Although simply finding any solution to a differential equation is important, mathematicians and engineers often want to go beyond finding\u00a0<em data-effect=\"italics\">one<\/em>\u00a0solution to a differential equation to finding\u00a0<em data-effect=\"italics\">all<\/em>\u00a0solutions to a differential equation. In other words, we want to find a general solution<strong data-effect=\"bold\">.<\/strong>\u00a0Just as with first-order differential equations, a general solution (or family of solutions) gives the entire set of solutions to a differential equation. An important difference between first-order and second-order equations is that, with second-order equations, we typically need to find two different solutions to the equation to find the general solution. If we find two solutions, then any linear combination of these solutions is also a solution. We state this fact as the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1170572430386\" class=\"theorem ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Theorem: superposition principle<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170571613027\">If [latex]y_1(x)[\/latex] and [latex]y_2(x)[\/latex] are solutions to a linear homogeneous differential equation, then the function<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1y_1(x)+c_2y_2(x)}[\/latex],<\/p>\r\nwhere [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants, is also a solution.\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572215710\">The proof of this\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term297\" class=\"no-emphasis\" data-type=\"term\">superposition principle<\/span>\u00a0theorem is left as an exercise.<\/p>\r\n\r\n<div id=\"fs-id1170572167258\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: verifying the superposition principle<\/h3>\r\n<p id=\"fs-id1170571627560\">Consider the differential equation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y^{\\prime\\prime}-4y^\\prime-5y=0}[\/latex].<\/p>\r\n<p id=\"fs-id1170572366760\">Given that [latex]e^{-x}[\/latex] and [latex]e^{5x}[\/latex] are solutions to this differential equation, show that [latex]4e^{-x}+e^{5x}[\/latex] is a solution.<\/p>\r\n[reveal-answer q=\"830429349\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"830429349\"]\r\n<p id=\"fs-id1170572608709\">We have<\/p>\r\n<p style=\"text-align: center;\">[latex]y(x)=4e^{-x}+e^{5x},\\text{ so }y^\\prime(x)=-4e^{-x}+5e^{5x}\\text{ and }y^{\\prime\\prime}=4e^{-x}+25e^{5x}[\/latex].<\/p>\r\n<p id=\"fs-id1170572551127\">Then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny^{\\prime\\prime}-4y^\\prime-5y&amp;=\\left(4e^{-x}+25e^{5x}\\right)-4\\left(-4e^{-x}+5e^{5x}\\right)-5\\left(4e^{-x}+e^{5x}\\right) \\\\\r\n&amp;=4e^{-x}+25e^{5x}+16e^{-x}-20e^{5x}-20e^{-x}-5e^{5x} \\\\\r\n&amp;=0\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170571660196\">Thus, [latex]y(x)=4e^{-x}+e^{5x}[\/latex] is a solution.<\/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-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<\/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=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\"><\/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>\r\n<p id=\"fs-id1170572089292\">Unfortunately, to find the general solution to a second-order differential equation, it is not enough to find any two solutions and then combine them. Consider the differential equation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{x''+7x'+12x=0}[\/latex].<\/p>\r\n<p id=\"fs-id1170571825080\">Both [latex]e^{-3t}[\/latex] and [latex]2e^{-3t}[\/latex] are solutions (check this). However, [latex]x(t)=c_1e^{-3y}+c_2\\left(2e^{-3t}\\right)[\/latex] is\u00a0<em data-effect=\"italics\">not<\/em>\u00a0the general solution. This expression does not account for all solutions to the differential equation. In particular, it fails to account for the function [latex]e^{-4t}[\/latex], which is also a solution to the differential equation.<\/p>\r\n<p id=\"fs-id1170571609148\">It turns out that to find the general solution to a second-order differential equation, we must find two linearly independent solutions. We define that terminology here.<\/p>\r\n\r\n<div id=\"fs-id1170571609153\" class=\"ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nA set of functions [latex]f_1(x),f_2(x),\\ldots,f_n(x)[\/latex] is said to be\u00a0<strong><span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term298\" data-type=\"term\">linearly dependent<\/span><\/strong>\u00a0if there are constants [latex]c_1,c_2,\\ldots c_n[\/latex], not all zero, such that [latex]c_1f_1(x)+c_2f_2(x)+\\cdots c_nf_n(x)=0[\/latex] for all [latex]x[\/latex] over the interval of interest. A set of functions that is not linearly dependent is said to be<strong>\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term299\" data-type=\"term\">linearly independent<\/span>.<\/strong>\r\n\r\n<\/div>\r\n<div>\r\n<p id=\"fs-id1170571770955\">In this chapter, we usually test sets of only two functions for linear independence, which allows us to simplify this definition. From a practical perspective, we see that two functions are linearly dependent if either one of them is identically zero or if they are constant multiples of each other.<\/p>\r\n<p id=\"fs-id1170571770961\">First we show that if the functions meet the conditions given previously, then they are linearly dependent. If one of the functions is identically zero\u2014say, [latex]f_2(x)\\equiv0[\/latex]\u2014then choose [latex]c_1=0[\/latex] and [latex]c_2=1[\/latex], and the condition for linear dependence is satisfied. If, on the other hand, neither [latex]f_1(x)[\/latex] nor [latex]f_2(x)[\/latex] is identically zero, but [latex]f_1(x)=Cf_2(x)[\/latex] for some constant [latex]C[\/latex], then choose [latex]c_1=\\frac{1}{C}[\/latex] and [latex]c_2=-1[\/latex], and again, the condition is satisfied.<\/p>\r\n<p id=\"fs-id1170572572984\">Next, we show that if two functions are linearly dependent, then either one is identically zero or they are constant multiples of one another. Assume [latex]f_1(x)[\/latex] and [latex]f_2(x)[\/latex] are linearly independent. Then, there are constants, [latex]c_1[\/latex] and [latex]c_2[\/latex], not both zero, such that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{c_1f_1(x)+c_2f_2(x)=0}[\/latex]<\/p>\r\n<p id=\"fs-id1170571728246\">for all [latex]x[\/latex] over the interval of interest. Then,<\/p>\r\n\r\n<\/div>\r\n<p style=\"text-align: center;\">[latex]\\large{c_1f_1(x)=-c_2f_2(x)}[\/latex].<\/p>\r\n\r\n<div>\r\n<p id=\"fs-id1170571673284\">Now, since we stated that [latex]c_1[\/latex] and [latex]c_2[\/latex] can\u2019t both be zero, assume [latex]c_2\\ne0[\/latex]. Then, there are two cases: either [latex]c_1=0[\/latex] or [latex]c_1\\ne0[\/latex]. If [latex]c_1=0[\/latex], then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n0&amp;=-c_2f_2(x) \\\\\r\n0&amp;=f_2(x)\r\n\\end{aligned}[\/latex],<\/p>\r\n<p id=\"fs-id1170571813996\">so one of the functions is identically zero. Now suppose [latex]c_1\\ne0[\/latex]. Then,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{f_1(x)=\\left(-\\frac{c_2}{c_1}\\right)f_2(x)}[\/latex]<\/p>\r\n<p id=\"fs-id1170572448483\">and we see that the functions are constant multiples of one another.<\/p>\r\n\r\n<div id=\"fs-id1170572448486\" class=\"theorem ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: linear dependence of two functions<\/h3>\r\n\r\n<hr \/>\r\n\r\nTwo functions, [latex]f_1(x)[\/latex] and [latex]f_2(x)[\/latex], are said to be linearly dependent if either one of them is identically zero or if [latex]f_1(x)=Cf_2(x)[\/latex] for some constant [latex]C[\/latex] and for all [latex]x[\/latex] over the interval of interest. Functions that are not linearly dependent are said to be\u00a0<em data-effect=\"italics\">linearly independent<\/em>.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: testing for linear dependence<\/h3>\r\n<p id=\"fs-id1170572630843\">Determine whether the following pairs of functions are linearly dependent or linearly independent.<\/p>\r\n\r\n<ol id=\"fs-id1170572630846\" type=\"a\">\r\n \t<li>[latex]f_1(x)=x^{2}[\/latex], [latex]f_2(x)=5x^{2}[\/latex]<\/li>\r\n \t<li>[latex]f_1(x)=\\sin x[\/latex],\u00a0[latex]f_2(x)=\\cos x[\/latex]<\/li>\r\n \t<li>[latex]f_1(x)=e^{3x}[\/latex],\u00a0[latex]f_2(x)=e^{-3x}[\/latex]<\/li>\r\n \t<li>[latex]f_1(x)=3x[\/latex],\u00a0[latex]f_2(x)=3x+1[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"465788210\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"465788210\"]\r\n<ol id=\"fs-id1170572522330\" type=\"a\">\r\n \t<li>[latex]f_2(x)=5f_1(x)[\/latex], so the functions are linearly dependent.<\/li>\r\n \t<li>There is no constant [latex]C[\/latex] such that [latex]f_1(x)=Cf_2(x)[\/latex], so the functions are linearly independent.<\/li>\r\n \t<li>There is no constant [latex]C[\/latex] such that [latex]f_1(x)=Cf_2(x)[\/latex], so the functions are linearly independent. Don\u2019t get confused by the fact that the exponents are constant multiples of each other. With two exponential functions, unless the exponents are equal, the functions are linearly independent.<\/li>\r\n \t<li>There is no constant [latex]C[\/latex] such that [latex]f_1(x)=Cf_2(x)[\/latex], so the functions are linearly independent.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nDetermine 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].\r\n\r\n[reveal-answer q=\"283742979\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"283742979\"]\r\n\r\nLinearly independent.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572507380\">If we are able to find two linearly independent solutions to a second-order differential equation, then we can combine them to find the general solution. This result is formally stated in the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1170572507385\" class=\"theorem ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: general solution to a homogeneous equation<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572507391\">If [latex]y_1(x)[\/latex] and\u00a0[latex]y_2(x)[\/latex] are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by<\/p>\r\n[latex]y(x)=c_1y_1(x)+c_2y_2(x)[\/latex],\r\n<p id=\"fs-id1170571679863\">where[latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572520510\">When we say a family of functions is the\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term300\" class=\"no-emphasis\" data-type=\"term\">general solution to a differential equation<\/span>, we mean that (1) every expression of that form is a solution and (2) every solution to the differential equation can be written in that form, which makes this theorem extremely powerful. If we can find two linearly independent solutions to a differential equation, we have, effectively, found\u00a0<em data-effect=\"italics\">all<\/em>\u00a0solutions to the differential equation\u2014quite a remarkable statement. The proof of this theorem is beyond the scope of this text.<\/p>\r\n\r\n<div id=\"fs-id1170572520528\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: writing the general solution<\/h3>\r\nIf [latex]y_1(t)=e^{3t}[\/latex] and [latex]y_2(t)=e^{-3t}[\/latex] are solutions to [latex]y''-9y=0[\/latex], what is the general solution?\r\n\r\n[reveal-answer q=\"423749834\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"423749834\"]\r\n\r\nNote that [latex]y_1[\/latex] and\u00a0[latex]y_2[\/latex] are not constant multiples of one another, so they are linearly independent. Then, the general solution to the differential equation is [latex]y(t)=c_1e^{3t}+c_2e^{-3t}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nIf [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?\r\n\r\n[reveal-answer q=\"342894572\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"342894572\"]\r\n\r\n[latex]y(x)=c_1e^{3t}+c_2xe^{3t}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Recognize homogeneous and nonhomogeneous linear differential equations.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Determine the characteristic equation of a homogeneous linear equation.<\/span><\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1170571769962\" data-depth=\"1\">\n<p id=\"fs-id1170571585385\">Consider the second-order differential equation<\/p>\n<p style=\"text-align: center;\">[latex]\\large{xy^{\\prime\\prime}+2x^2y^\\prime+5x^3y=0}[\/latex].<\/p>\n<p id=\"fs-id1170571602172\">Notice that [latex]y[\/latex] and its derivatives appear in a relatively simple form. They are multiplied by functions of [latex]x[\/latex], but are not raised to any powers themselves, nor are they multiplied together. As discussed in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/basics-of-differential-equations\/\" target=\"_blank\" rel=\"noopener\" data-book-uuid=\"1d39a348-071f-4537-85b6-c98912458c3c\" data-page-slug=\"4-introduction\">Introduction to Differential Equations<\/a>, first-order equations with similar characteristics are said to be linear. The same is true of second-order equations. Also note that all the terms in this differential equation involve either [latex]y[\/latex] or one of its derivatives. There are no terms involving only functions of [latex]x[\/latex]. Equations like this, in which every term contains [latex]y[\/latex] or one of its derivatives, are called homogeneous.<\/p>\n<p id=\"fs-id1170572582703\">Not all differential equations are homogeneous. Consider the differential equation<\/p>\n<p style=\"text-align: center;\">[latex]\\large{xy^{\\prime\\prime}+2x^2y^\\prime+5x^3y=x^2}[\/latex].<\/p>\n<p id=\"fs-id1170572169093\">The [latex]x^{2}[\/latex] term on the right side of the equal sign does not contain [latex]y[\/latex] or any of its derivatives. Therefore, this differential equation is nonhomogeneous.<\/p>\n<\/section>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170572215852\">A second-order differential equation is linear if it can be written in the form<\/p>\n<p style=\"text-align: center;\">[latex]\\large{a_2(x)y^{\\prime\\prime}+a_1(x)y^{\\prime}+a_0(x)y=r(x)}[\/latex],<\/p>\n<p id=\"fs-id1170572404274\">where [latex]a_2(x)[\/latex], [latex]a_1(x)[\/latex], [latex]a_0(x)[\/latex], and [latex]r(x)[\/latex] are real-valued functions and [latex]a_2(x)[\/latex] is not identically zero. If [latex]r(x)=0[\/latex]\u2014in other words, if [latex]r(x)=0[\/latex] for every value of [latex]x[\/latex]\u2014the equation is said to be a\u00a0<strong><span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term295\" data-type=\"term\">homogeneous linear equation<\/span><\/strong>. If [latex]r(x)\\ne0[\/latex] for some value of [latex]x[\/latex], the equation is said to be a\u00a0<strong><span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term296\" data-type=\"term\">nonhomogeneous linear equation<\/span>.<\/strong><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p>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.<\/p>\n<\/div>\n<div data-type=\"note\">\n<p id=\"fs-id1170572280427\">In linear differential equations, [latex]y[\/latex] and its derivatives can be raised only to the first power and they may not be multiplied by one another. Terms involving [latex]y^{2}[\/latex] or [latex]\\sqrt{y^\\prime}[\/latex] make the equation nonlinear. Functions of [latex]y[\/latex] and its derivatives, such as [latex]\\sin y[\/latex] or [latex]e^{y^\\prime}[\/latex] are similarly prohibited in linear differential equations.<\/p>\n<p id=\"fs-id1170571635343\">Note that equations may not always be given in standard form (the form shown in the definition). It can be helpful to rewrite them in that form to decide whether they are linear, or whether a linear equation is homogeneous.<\/p>\n<div id=\"fs-id1170571650370\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: Classifying second-order equations<\/h3>\n<p id=\"fs-id1170572307004\">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-id1170572547853\" type=\"a\">\n<li>[latex]y''+3x^{4}y'+x^{2}y^{2}=x^{3}[\/latex]<\/li>\n<li>[latex](\\sin x)y''+(\\cos x)y'+3y=0[\/latex]<\/li>\n<li>[latex]4t^{2}x''+3txx'+4x=0[\/latex]<\/li>\n<li>[latex]5y''+y=4x^{5}[\/latex]<\/li>\n<li>[latex](\\cos x)y''-\\sin y'+(\\sin x)y-\\cos x=0[\/latex]<\/li>\n<li>[latex]8ty''-6t^{2}y'+4ty-3t^2=0[\/latex]<\/li>\n<li>[latex]\\sin (x^{2})y''-(\\cos x)y'+x^{2}y=y'-3[\/latex]<\/li>\n<li>[latex]y''+5xy'-3y=\\cos y[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q924732071\">Show Solution<\/span><\/p>\n<div id=\"q924732071\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572337089\" type=\"a\">\n<li>This equation is nonlinear because of the [latex]y^{2}[\/latex] term.<\/li>\n<li>This equation is linear. There is no term involving a power or function of [latex]y[\/latex], and the coefficients are all functions of [latex]x[\/latex]. The equation is already written in standard form, and [latex]r(x)[\/latex] is identically zero, so the equation is homogeneous.<\/li>\n<li>This equation is nonlinear. Note that, in this case, [latex]x[\/latex] is the dependent variable and [latex]t[\/latex] is the independent variable. The second term involves the product of [latex]x[\/latex] and [latex]x'[\/latex], so the equation is nonlinear.<\/li>\n<li>This equation is linear. Since [latex]r(x)=4x^{5}[\/latex], the equation is nonhomogeneous.<\/li>\n<li>This equation is nonlinear, because of the [latex]\\sin y'[\/latex] term.<\/li>\n<li>This equation is linear. Rewriting it in standard form gives<br \/>\n<span data-type=\"newline\">[latex]8t^{2}-6t^{2}y'+4ty=3t^{2}[\/latex].<br \/>\n<\/span>With the equation in standard form, we can see that [latex]r(t)=3t^2[\/latex] so the equation is nonhomogeneous.<\/li>\n<li>This equation looks like it\u2019s linear, but we should rewrite it in standard form to be sure. We get<br \/>\n[latex]\\sin(x^{2})y''-(\\cos x+1)y'+x^{2}y=-3[\/latex].This equation is, indeed, linear. With [latex]r(x)=-3[\/latex], it is nonhomogeneous.<\/li>\n<li>This equation is nonlinear because of the [latex]\\cos y[\/latex] term.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Interactive<\/h3>\n<p>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.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q374981011\">Show Solution<\/span><\/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<\/div>\n<p id=\"fs-id1170572553210\">Later in this section, we will see some techniques for solving specific types of differential equations. Before we get to that, however, let\u2019s get a feel for how solutions to linear differential equations behave. In many cases, solving differential equations depends on making educated guesses about what the solution might look like. Knowing how various types of solutions behave will be helpful.<\/p>\n<div id=\"fs-id1170572560326\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: verifying a solution<\/h3>\n<p id=\"fs-id1170572106755\">Consider the linear, homogeneous differential equation<\/p>\n<p>[latex]x^{2}y''-xy'-3y=0[\/latex].<\/p>\n<p id=\"fs-id1170572307137\">Looking at this equation, notice that the coefficient functions are polynomials, with higher powers of [latex]x[\/latex] associated with higher-order derivatives of [latex]y[\/latex]. Show that [latex]y=x^{3}[\/latex] is a solution to this differential equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q731671820\">Show Solution<\/span><\/p>\n<div id=\"q731671820\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572225198\">Let [latex]y=x^{3}[\/latex]. Then [latex]y'=3x^{2}[\/latex] and [latex]y''=6(x)[\/latex]. Substituting into the differential equation, we see that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x^{2}y''-xy'-3y&=x^2(6x)-x(3x^2)-3(x^3) \\\\  &=6x^3-3x^3-3x^3 \\\\  &=0  \\end{aligned}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q688213714\">Show Solution<\/span><\/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<\/div>\n<p id=\"fs-id1170572505444\">Although simply finding any solution to a differential equation is important, mathematicians and engineers often want to go beyond finding\u00a0<em data-effect=\"italics\">one<\/em>\u00a0solution to a differential equation to finding\u00a0<em data-effect=\"italics\">all<\/em>\u00a0solutions to a differential equation. In other words, we want to find a general solution<strong data-effect=\"bold\">.<\/strong>\u00a0Just as with first-order differential equations, a general solution (or family of solutions) gives the entire set of solutions to a differential equation. An important difference between first-order and second-order equations is that, with second-order equations, we typically need to find two different solutions to the equation to find the general solution. If we find two solutions, then any linear combination of these solutions is also a solution. We state this fact as the following theorem.<\/p>\n<div id=\"fs-id1170572430386\" class=\"theorem ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Theorem: superposition principle<\/h3>\n<hr \/>\n<p id=\"fs-id1170571613027\">If [latex]y_1(x)[\/latex] and [latex]y_2(x)[\/latex] are solutions to a linear homogeneous differential equation, then the function<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1y_1(x)+c_2y_2(x)}[\/latex],<\/p>\n<p>where [latex]c_1[\/latex] and [latex]c_2[\/latex] are constants, is also a solution.<\/p>\n<\/div>\n<p id=\"fs-id1170572215710\">The proof of this\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term297\" class=\"no-emphasis\" data-type=\"term\">superposition principle<\/span>\u00a0theorem is left as an exercise.<\/p>\n<div id=\"fs-id1170572167258\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: verifying the superposition principle<\/h3>\n<p id=\"fs-id1170571627560\">Consider the differential equation<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y^{\\prime\\prime}-4y^\\prime-5y=0}[\/latex].<\/p>\n<p id=\"fs-id1170572366760\">Given that [latex]e^{-x}[\/latex] and [latex]e^{5x}[\/latex] are solutions to this differential equation, show that [latex]4e^{-x}+e^{5x}[\/latex] is a solution.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q830429349\">Show Solution<\/span><\/p>\n<div id=\"q830429349\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572608709\">We have<\/p>\n<p style=\"text-align: center;\">[latex]y(x)=4e^{-x}+e^{5x},\\text{ so }y^\\prime(x)=-4e^{-x}+5e^{5x}\\text{ and }y^{\\prime\\prime}=4e^{-x}+25e^{5x}[\/latex].<\/p>\n<p id=\"fs-id1170572551127\">Then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y^{\\prime\\prime}-4y^\\prime-5y&=\\left(4e^{-x}+25e^{5x}\\right)-4\\left(-4e^{-x}+5e^{5x}\\right)-5\\left(4e^{-x}+e^{5x}\\right) \\\\  &=4e^{-x}+25e^{5x}+16e^{-x}-20e^{5x}-20e^{-x}-5e^{5x} \\\\  &=0  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170571660196\">Thus, [latex]y(x)=4e^{-x}+e^{5x}[\/latex] is a solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q238127390\">Show Solution<\/span><\/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<\/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=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\"><\/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<p id=\"fs-id1170572089292\">Unfortunately, to find the general solution to a second-order differential equation, it is not enough to find any two solutions and then combine them. Consider the differential equation<\/p>\n<p style=\"text-align: center;\">[latex]\\large{x''+7x'+12x=0}[\/latex].<\/p>\n<p id=\"fs-id1170571825080\">Both [latex]e^{-3t}[\/latex] and [latex]2e^{-3t}[\/latex] are solutions (check this). However, [latex]x(t)=c_1e^{-3y}+c_2\\left(2e^{-3t}\\right)[\/latex] is\u00a0<em data-effect=\"italics\">not<\/em>\u00a0the general solution. This expression does not account for all solutions to the differential equation. In particular, it fails to account for the function [latex]e^{-4t}[\/latex], which is also a solution to the differential equation.<\/p>\n<p id=\"fs-id1170571609148\">It turns out that to find the general solution to a second-order differential equation, we must find two linearly independent solutions. We define that terminology here.<\/p>\n<div id=\"fs-id1170571609153\" class=\"ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p>A set of functions [latex]f_1(x),f_2(x),\\ldots,f_n(x)[\/latex] is said to be\u00a0<strong><span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term298\" data-type=\"term\">linearly dependent<\/span><\/strong>\u00a0if there are constants [latex]c_1,c_2,\\ldots c_n[\/latex], not all zero, such that [latex]c_1f_1(x)+c_2f_2(x)+\\cdots c_nf_n(x)=0[\/latex] for all [latex]x[\/latex] over the interval of interest. A set of functions that is not linearly dependent is said to be<strong>\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term299\" data-type=\"term\">linearly independent<\/span>.<\/strong><\/p>\n<\/div>\n<div>\n<p id=\"fs-id1170571770955\">In this chapter, we usually test sets of only two functions for linear independence, which allows us to simplify this definition. From a practical perspective, we see that two functions are linearly dependent if either one of them is identically zero or if they are constant multiples of each other.<\/p>\n<p id=\"fs-id1170571770961\">First we show that if the functions meet the conditions given previously, then they are linearly dependent. If one of the functions is identically zero\u2014say, [latex]f_2(x)\\equiv0[\/latex]\u2014then choose [latex]c_1=0[\/latex] and [latex]c_2=1[\/latex], and the condition for linear dependence is satisfied. If, on the other hand, neither [latex]f_1(x)[\/latex] nor [latex]f_2(x)[\/latex] is identically zero, but [latex]f_1(x)=Cf_2(x)[\/latex] for some constant [latex]C[\/latex], then choose [latex]c_1=\\frac{1}{C}[\/latex] and [latex]c_2=-1[\/latex], and again, the condition is satisfied.<\/p>\n<p id=\"fs-id1170572572984\">Next, we show that if two functions are linearly dependent, then either one is identically zero or they are constant multiples of one another. Assume [latex]f_1(x)[\/latex] and [latex]f_2(x)[\/latex] are linearly independent. Then, there are constants, [latex]c_1[\/latex] and [latex]c_2[\/latex], not both zero, such that<\/p>\n<p style=\"text-align: center;\">[latex]\\large{c_1f_1(x)+c_2f_2(x)=0}[\/latex]<\/p>\n<p id=\"fs-id1170571728246\">for all [latex]x[\/latex] over the interval of interest. Then,<\/p>\n<\/div>\n<p style=\"text-align: center;\">[latex]\\large{c_1f_1(x)=-c_2f_2(x)}[\/latex].<\/p>\n<div>\n<p id=\"fs-id1170571673284\">Now, since we stated that [latex]c_1[\/latex] and [latex]c_2[\/latex] can\u2019t both be zero, assume [latex]c_2\\ne0[\/latex]. Then, there are two cases: either [latex]c_1=0[\/latex] or [latex]c_1\\ne0[\/latex]. If [latex]c_1=0[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  0&=-c_2f_2(x) \\\\  0&=f_2(x)  \\end{aligned}[\/latex],<\/p>\n<p id=\"fs-id1170571813996\">so one of the functions is identically zero. Now suppose [latex]c_1\\ne0[\/latex]. Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\large{f_1(x)=\\left(-\\frac{c_2}{c_1}\\right)f_2(x)}[\/latex]<\/p>\n<p id=\"fs-id1170572448483\">and we see that the functions are constant multiples of one another.<\/p>\n<div id=\"fs-id1170572448486\" class=\"theorem ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: linear dependence of two functions<\/h3>\n<hr \/>\n<p>Two functions, [latex]f_1(x)[\/latex] and [latex]f_2(x)[\/latex], are said to be linearly dependent if either one of them is identically zero or if [latex]f_1(x)=Cf_2(x)[\/latex] for some constant [latex]C[\/latex] and for all [latex]x[\/latex] over the interval of interest. Functions that are not linearly dependent are said to be\u00a0<em data-effect=\"italics\">linearly independent<\/em>.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: testing for linear dependence<\/h3>\n<p id=\"fs-id1170572630843\">Determine whether the following pairs of functions are linearly dependent or linearly independent.<\/p>\n<ol id=\"fs-id1170572630846\" type=\"a\">\n<li>[latex]f_1(x)=x^{2}[\/latex], [latex]f_2(x)=5x^{2}[\/latex]<\/li>\n<li>[latex]f_1(x)=\\sin x[\/latex],\u00a0[latex]f_2(x)=\\cos x[\/latex]<\/li>\n<li>[latex]f_1(x)=e^{3x}[\/latex],\u00a0[latex]f_2(x)=e^{-3x}[\/latex]<\/li>\n<li>[latex]f_1(x)=3x[\/latex],\u00a0[latex]f_2(x)=3x+1[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q465788210\">Show Solution<\/span><\/p>\n<div id=\"q465788210\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572522330\" type=\"a\">\n<li>[latex]f_2(x)=5f_1(x)[\/latex], so the functions are linearly dependent.<\/li>\n<li>There is no constant [latex]C[\/latex] such that [latex]f_1(x)=Cf_2(x)[\/latex], so the functions are linearly independent.<\/li>\n<li>There is no constant [latex]C[\/latex] such that [latex]f_1(x)=Cf_2(x)[\/latex], so the functions are linearly independent. Don\u2019t get confused by the fact that the exponents are constant multiples of each other. With two exponential functions, unless the exponents are equal, the functions are linearly independent.<\/li>\n<li>There is no constant [latex]C[\/latex] such that [latex]f_1(x)=Cf_2(x)[\/latex], so the functions are linearly independent.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q283742979\">Show Solution<\/span><\/p>\n<div id=\"q283742979\" class=\"hidden-answer\" style=\"display: none\">\n<p>Linearly independent.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170572507380\">If we are able to find two linearly independent solutions to a second-order differential equation, then we can combine them to find the general solution. This result is formally stated in the following theorem.<\/p>\n<div id=\"fs-id1170572507385\" class=\"theorem ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: general solution to a homogeneous equation<\/h3>\n<hr \/>\n<p id=\"fs-id1170572507391\">If [latex]y_1(x)[\/latex] and\u00a0[latex]y_2(x)[\/latex] are linearly independent solutions to a second-order, linear, homogeneous differential equation, then the general solution is given by<\/p>\n<p>[latex]y(x)=c_1y_1(x)+c_2y_2(x)[\/latex],<\/p>\n<p id=\"fs-id1170571679863\">where[latex]c_1[\/latex] and [latex]c_2[\/latex] are constants.<\/p>\n<\/div>\n<p id=\"fs-id1170572520510\">When we say a family of functions is the\u00a0<span id=\"658907a3-724d-4f81-9468-fbde7bb4b09e_term300\" class=\"no-emphasis\" data-type=\"term\">general solution to a differential equation<\/span>, we mean that (1) every expression of that form is a solution and (2) every solution to the differential equation can be written in that form, which makes this theorem extremely powerful. If we can find two linearly independent solutions to a differential equation, we have, effectively, found\u00a0<em data-effect=\"italics\">all<\/em>\u00a0solutions to the differential equation\u2014quite a remarkable statement. The proof of this theorem is beyond the scope of this text.<\/p>\n<div id=\"fs-id1170572520528\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: writing the general solution<\/h3>\n<p>If [latex]y_1(t)=e^{3t}[\/latex] and [latex]y_2(t)=e^{-3t}[\/latex] are solutions to [latex]y''-9y=0[\/latex], what is the general solution?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q423749834\">Show Solution<\/span><\/p>\n<div id=\"q423749834\" class=\"hidden-answer\" style=\"display: none\">\n<p>Note that [latex]y_1[\/latex] and\u00a0[latex]y_2[\/latex] are not constant multiples of one another, so they are linearly independent. Then, the general solution to the differential equation is [latex]y(t)=c_1e^{3t}+c_2e^{-3t}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q342894572\">Show Solution<\/span><\/p>\n<div id=\"q342894572\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y(x)=c_1e^{3t}+c_2xe^{3t}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5517\">\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 7.3. <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":3,"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 7.3\",\"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-5517","chapter","type-chapter","status-publish","hentry"],"part":25,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5517","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":14,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5517\/revisions"}],"predecessor-version":[{"id":6461,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/5517\/revisions\/6461"}],"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\/5517\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=5517"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=5517"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=5517"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=5517"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}