{"id":1330,"date":"2021-11-24T06:40:56","date_gmt":"2021-11-24T06:40:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1330"},"modified":"2022-11-01T22:46:48","modified_gmt":"2022-11-01T22:46:48","slug":"nonhomogeneous-linear-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/nonhomogeneous-linear-equations\/","title":{"raw":"Nonhomogeneous Linear Equations","rendered":"Nonhomogeneous 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\">Write the general solution to a nonhomogeneous differential equation.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Solve a nonhomogeneous differential equation by the method of undetermined coefficients.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Solve a nonhomogeneous differential equation by the method of variation of parameters.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1170572624887\" data-depth=\"1\">\r\n<h2 data-type=\"title\">General Solution to a Nonhomogeneous Linear Equation<\/h2>\r\n<p id=\"fs-id1170572616368\">Consider the nonhomogeneous linear differential equation<\/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-id1170572346900\">The associated homogeneous equation<\/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=0}[\/latex]<\/p>\r\n<p id=\"fs-id1170571543157\">is called the\u00a0<strong><span id=\"e2cb0fd1-6cb0-486c-93d7-2edd081caff6_term308\" data-type=\"term\">complementary equation<\/span>.<\/strong> We will see that solving the complementary equation is an important step in solving a nonhomogeneous differential equation.<\/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\r\nA solution [latex]y_p(x)[\/latex] of a differential equation that contains no arbitrary constants is called a\u00a0<span id=\"e2cb0fd1-6cb0-486c-93d7-2edd081caff6_term309\" data-type=\"term\">particular solution<\/span>\u00a0to the equation.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Theorem: general solution to a nonhomogeneous equation<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572168730\">Let [latex]y_p(x)[\/latex] be any particular solution to the nonhomogeneous linear differential equation<\/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-id1170571711288\">Also, let [latex]c_1y_1(x)+c_2y_2(x)[\/latex] denote the general solution to the complementary equation. Then, the general solution to the nonhomogeneous equation is given by<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1y_1(x)+c_2y_2(x)+y_p(x)}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1170572544509\">To prove [latex]y(x)[\/latex] is the general solution, we must first show that it solves the differential equation and, second, that any solution to the differential equation can be written in that form. Substituting [latex]y(x)[\/latex] into the differential equation, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\na_2(x)y^{\\prime\\prime}+a_1(x)y^\\prime+a_0(x)y&amp;=a_2(x)(c_1y_1+c_2y_2+y_p)^{\\prime\\prime}+a_1(x)(c_1y_1+c_2y_2+y_p)^\\prime+a_0(x)(c_1y_1+c_2y_2+y_p) \\\\\r\n&amp;=[a_2(x)(c_1y_1+c_2y_2)^{\\prime\\prime}+a_1(x)(c_1y_1+c_2y_2)^\\prime+a_0(x)(c_1y_1+c_2y_2)]+a_2(x)y_p^{\\prime\\prime}+a_1(x)y_p^\\prime+a_0(x)y_p \\\\\r\n&amp;=0+r(x) \\\\\r\n&amp;=r(x)\r\n\\end{aligned}[\/latex].<\/p>\r\nSo [latex]y(x)[\/latex] is a solution.\r\n<p id=\"fs-id1170572415209\">Now, let [latex]z(x)[\/latex] be any solution to [latex]a_2(x)y^{\\prime\\prime}+a_1(x)y^\\prime+a_0(x)y=r(x)[\/latex]. Then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\na_2(x)(z-y_p)^{\\prime\\prime}+a_1(x)(z-y_p)^\\prime+a_0(x)(z-y_p)&amp;=(a_2(x)z^{\\prime\\prime}+a_1(x)z^\\prime+a_0(x)z)-(a_2(x)y_p^{\\prime\\prime}+a_1(x)y_p^\\prime+a_0(x)y_p) \\\\\r\n&amp;=r(x)-r(x) \\\\\r\n&amp;=0\r\n\\end{aligned}[\/latex],<\/p>\r\nso [latex]z(x)=y_p(x)[\/latex] is a solution to the complementary equation. But, [latex]c_1y_1(x)+c_2y_2(x)[\/latex] is the general solution to the complementary equation, so there are constants [latex]c_1[\/latex] and [latex]c_2[\/latex] such that\r\n<p style=\"text-align: center;\">[latex]\\large{z(x)=y_p(x)=c_1y_1(x)+c_2y_2(x)}[\/latex].<\/p>\r\n<p id=\"fs-id1170572481683\">Hence, we see that [latex]z(x)=c_1y_1(x)+c_2y_2(x)y_p(x)[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n<div class=\"textbox exercises\">\r\n<h3>Example: verifying the general solution<\/h3>\r\nGiven that [latex]y_p(x)=x[\/latex] is a particular solution to the differential equation [latex]y^{\\prime\\prime}+y=x[\/latex], write the general solution and check by verifying that the solution satisfies the equation.\r\n\r\n[reveal-answer q=\"284270134\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"284270134\"]\r\n<p id=\"fs-id1170572547637\">The complementary equation is [latex]y^{\\prime\\prime}+y=0[\/latex], which has the general solution [latex]c_1\\cos x+c_2\\sin x[\/latex]. So, the general solution to the nonhomogeneous equation is<\/p>\r\n<p style=\"text-align: center;\">[latex]y(x)=c_1\\cos x+c_2\\sin x+x[\/latex].<\/p>\r\n<p id=\"fs-id1170572212481\">To verify that this is a solution, substitute it into the differential equation. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]y^\\prime(x)=-c_1\\sin x+c_2\\cos x+1\\text{ and }y^{\\prime\\prime}=-c_1\\cos x-c_2\\sin x[\/latex].<\/p>\r\n<p id=\"fs-id1170572637641\">Then<\/p>\r\n<p style=\"text-align: center;\">[latex]y^{\\prime\\prime}(x)+y(x)=-c_1\\cos x-c_2\\sin x+c_1\\cos x+c_2\\sin x+x[\/latex].<\/p>\r\n<p id=\"fs-id1170572416526\">So, [latex]y(x)[\/latex] is a solution to [latex]y^{\\prime\\prime}+y=x[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nGiven that [latex]y_p(x)=-2[\/latex] is a particular solution to [latex]y^{\\prime\\prime}-3y^\\prime-4y=8[\/latex], write the general solution and verify that the general solution satisfies the equation.\r\n\r\n[reveal-answer q=\"039452983\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"039452983\"]\r\n\r\n[latex]y(x)=c_1e^{-x}+c_2e^{4x}-2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<section id=\"fs-id1170572624887\" data-depth=\"1\"><section id=\"fs-id1170571569017\" data-depth=\"2\">\r\n<p id=\"fs-id1170572240485\">In the preceding section, we learned how to solve homogeneous equations with constant coefficients. Therefore, for nonhomogeneous equations of the form [latex]ay^{\\prime\\prime}+by^\\prime+cy=r(x)[\/latex], we already know how to solve the complementary equation, and the problem boils down to finding a particular solution for the nonhomogeneous equation. We now examine two techniques for this: the method of undetermined coefficients and the method of variation of parameters.<\/p>\r\n\r\n<\/section><\/section><section id=\"fs-id1170571565966\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Undetermined Coefficients<\/h2>\r\n<p id=\"fs-id1170571630374\">The\u00a0<strong><span id=\"e2cb0fd1-6cb0-486c-93d7-2edd081caff6_term310\" data-type=\"term\">method of undetermined coefficients<\/span><\/strong>\u00a0involves making educated guesses about the form of the particular solution based on the form of [latex]r(x)[\/latex]. When we take derivatives of polynomials, exponential functions, sines, and cosines, we get polynomials, exponential functions, sines, and cosines. So when [latex]r(x)[\/latex] has one of these forms, it is possible that the solution to the nonhomogeneous differential equation might take that same form. Let\u2019s look at some examples to see how this works.<\/p>\r\n\r\n<\/section>\r\n<div data-type=\"note\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: undetermined coefficients when\u00a0[latex]r(x)[\/latex] is a polynomial<\/h3>\r\nFind the general solution to [latex]y^{\\prime\\prime}+4y^\\prime+3y=3x[\/latex].\r\n\r\n[reveal-answer q=\"204756223\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204756223\"]\r\n<p id=\"fs-id1170572204075\">The complementary equation is [latex]y^{\\prime\\prime}+4y^\\prime+3y=0[\/latex], with general solution [latex]c_1e^{-x}+c_2e^{-3x}[\/latex]. Since [latex]r(x)=3x[\/latex], the particular solution might have the form [latex]y_p(x)=Ax+B[\/latex]. If this is the case, then we have [latex]y_p^\\prime(x)=A[\/latex] and [latex]y_py^{\\prime\\prime}(x)=0[\/latex]. For [latex]y_p[\/latex] to be a solution to the differential equation, we must find values for [latex]A[\/latex] and [latex]B[\/latex] such that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny^{\\prime\\prime}+4^\\prime+3y&amp;=3x \\\\\r\n0+4(A)+3(Ax+B)&amp;=3x \\\\\r\n3Ax+(4A+3B)&amp;=3x\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170572481625\">Setting coefficients of like terms equal, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{aligned}\r\n3A&amp;=3 \\\\\r\n4A+3b&amp;=0\r\n\\end{aligned}\r\n[\/latex].<\/p>\r\n<p id=\"fs-id1170572643180\">Then, [latex]A=1[\/latex] and [latex]B=-\\frac{4}{3}[\/latex], so [latex]y_p(x)=x-\\frac{4}{3}[\/latex] and the general solution is<\/p>\r\n<p style=\"text-align: center;\">[latex]y(x)=c_1e^{-x}+c_2e^{-3x}+x-\\frac43[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn\u00a0Example \"Undetermined Coefficients When [latex]\ud835\udc5f(\ud835\udc65)[\/latex] Is a Polynomial\", notice that even though [latex]r(x)[\/latex] did not include a constant term, it was necessary for us to include the constant term in our guess. If we had assumed a solution of the form [latex]y_p=Ax[\/latex] (with no constant term), we would not have been able to find a solution. (Verify this!) If the function [latex]r(x)[\/latex] is a polynomial, our guess for the particular solution should be a polynomial of the same degree, and it must include all lower-order terms, regardless of whether they are present in [latex]r(x)[\/latex].\r\n<div class=\"textbox exercises\">\r\n<h3>Example: undetermined coefficients when\u00a0[latex]r(x)[\/latex] is an exponential<\/h3>\r\nFind the general solution to [latex]y^{\\prime\\prime}-y^\\prime-2y=2e^{3x}[\/latex].\r\n\r\n[reveal-answer q=\"083272901\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"083272901\"]\r\n<p id=\"fs-id1170572306234\">The complementary equation is [latex]y^{\\prime\\prime}-y^\\prime-2y=0[\/latex], with the general solution [latex]c_1e^{-x}+c_2e^{2x}[\/latex]. Since [latex]r(x)=2e^{3x}[\/latex], the particular solution might have the form [latex]y_p=Ae^{3x}[\/latex]. Then, we have [latex]y_p^\\prime(x)=3Ae^{3x}[\/latex] and [latex]y^{\\prime\\prime}=9Ae^{3x}[\/latex]. For [latex]y_p[\/latex] to be a solution to the differential equation, we must find a value for [latex]A[\/latex] such that<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny^{\\prime\\prime}-y^\\prime-2y&amp;=2e^{3x} \\\\\r\n9Ae^{3x}-3Ae^{3x}-2Ae^{3x}&amp;=2Ae^{3x} \\\\\r\n4Ae^{3x}&amp;=2Ae^{3x}\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170572221173\">So, [latex]4A=2[\/latex] and [latex]A=\\frac{1}{2}[\/latex]. Then, [latex]y_p(x)=\\left(\\frac12\\right)e^{3x}[\/latex], and the general solution is<\/p>\r\n<p style=\"text-align: center;\">[latex]y(x)=c_1e^{-x}+c_2e^{2x}+\\frac12e^{3x}[\/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\nFind the general solution to [latex]y^{\\prime\\prime}-4y^\\prime+4y=7\\sin t-\\cos t[\/latex].\r\n\r\n[reveal-answer q=\"028934924\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"028934924\"]\r\n\r\n[latex]y(t)=c_1e^{2t}+c_2te^{2t}+\\sin t+\\cos t[\/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=8250342&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=rQsUWvHvRSY&amp;video_target=tpm-plugin-hfu2pzjl-rQsUWvHvRSY\" 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\/CP7.11_transcript.html\">transcript for \u201cCP 7.11\u201d here (opens in new window).<\/a><\/center>In the previous checkpoint, [latex]r(x)[\/latex] included both sine and cosine terms. However, even if [latex]r(x)[\/latex] included a sine term only or a cosine term only, both terms must be present in the guess. The method of undetermined coefficients also works with products of polynomials, exponentials, sines, and cosines. Some of the key forms of [latex]r(x)[\/latex] and the associated guesses for [latex]y_p(x)[\/latex] are summarized in\u00a0Table 7.2\u00a0<span class=\"os-title\" data-type=\"title\">Key Forms for the Method of Undetermined Coefficients below.<\/span>\r\n\r\n<\/div>\r\n<div data-type=\"note\">\r\n<table style=\"height: 192px;\" data-id=\"fs-id1170572294713\">\r\n<thead>\r\n<tr style=\"height: 14px;\" valign=\"top\">\r\n<th style=\"height: 14px; width: 141.922px;\" scope=\"col\" data-align=\"left\">[latex]r(x)[\/latex]<\/th>\r\n<th style=\"height: 14px; width: 739.438px;\" scope=\"col\" data-align=\"left\">Initial guess for [latex]y_p(x)[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 28px;\" valign=\"top\">\r\n<td style=\"height: 28px; width: 141.922px;\" data-align=\"left\">[latex]k[\/latex] (a constant)<\/td>\r\n<td style=\"height: 28px; width: 739.438px;\" data-align=\"left\">[latex]A[\/latex] (a constant)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\" valign=\"top\">\r\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex]ax+b[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex]Ax+B[\/latex] (<em data-effect=\"italics\">Note<\/em>: The guess must include both terms even if [latex]b=0[\/latex].)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\" valign=\"top\">\r\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex]ax^{2}+bx+c[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex]Ax^{2}+Bx+C[\/latex] (<em data-effect=\"italics\">Note<\/em>: The guess must include all three terms even if [latex]b[\/latex] or [latex]c[\/latex] are zero.)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 28px;\" valign=\"top\">\r\n<td style=\"height: 28px; width: 141.922px;\" data-align=\"left\">Higher-order polynomials<\/td>\r\n<td style=\"height: 28px; width: 739.438px;\" data-align=\"left\">Polynomial of the same order as [latex]r(x)[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\" valign=\"top\">\r\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex]ae^{\\lambda x}[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex]Ae^{\\lambda x}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\" valign=\"top\">\r\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex]a\\cos\\beta x+b\\sin\\beta x[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex]A\\cos\\beta x+B\\sin\\beta x[\/latex] (<em data-effect=\"italics\">Note<\/em>: The guess must include both terms even if either [latex]a=0[\/latex] or [latex]b=0[\/latex].)<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\" valign=\"top\">\r\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex]ae^{\\alpha x}\\cos\\beta x+be^{\\alpha x}\\sin\\beta x[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex]Ae^{\\alpha x}\\cos\\beta x+Be^{\\alpha x}\\sin\\beta x[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\" valign=\"top\">\r\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex](ax^2+bx+c)e^{\\lambda x}[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex](Ax^2+Bx+C)e^{\\lambda x}[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\" valign=\"top\">\r\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex](a_2x^2+a_1x+a_0)\\cos\\beta x+(b_2x^2+b_1x+b_0)\\sin\\beta x[\/latex]<\/td>\r\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex](A_2x^2+A_1x+A_0)\\cos\\beta x+(B_2x^2+B_1x+B_0)\\sin\\beta x[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px;\" valign=\"top\">\r\n<td style=\"height: 24px; width: 141.922px;\" data-align=\"left\">[latex](a_2x^2+a_1x+a_0)e^{\\alpha x}\\cos\\beta x+(b_2x^2+b_1x+b_0)e^{\\alpha x}\\sin\\beta x[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 739.438px;\" data-align=\"left\">[latex](A_2x^2+A_1x+A_0)e^{\\alpha x}\\cos\\beta x+(B_2x^2+B_1x+B_0)e^{\\alpha x}\\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\">7.2<\/span>\u00a0<span class=\"os-title\" data-type=\"title\">Key Forms for the Method of Undetermined Coefficients<\/span><\/div>\r\n<\/div>\r\n<div data-type=\"note\"><\/div>\r\n<div data-type=\"note\">\r\n<p id=\"fs-id1170572304163\">Keep in mind that there is a key pitfall to this method. Consider the differential equation [latex]y^{\\prime\\prime}+5y^\\prime+6y=3e^{-2x}[\/latex]. Based on the form of [latex]r(x)[\/latex], we guess a particular solution of the form [latex]y_p(x)=Ae^{-2x}[\/latex]. But when we substitute this expression into the differential equation to find a value for [latex]A[\/latex], we run into a problem. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y_p^\\prime(x)=-2Ae^{-2x}}[\/latex]<\/p>\r\n<p id=\"fs-id1170572611802\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y_p^{\\prime\\prime}=4Ae^{-2x}}[\/latex],<\/p>\r\n<p id=\"fs-id1170572593508\">so we want<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny^{\\prime\\prime}+5y^\\prime+6y&amp;=3e^{-2x} \\\\\r\n4Ae^{-2x}+\\left(-2Ae^{-2x}\\right)+6Ae^{-2x}&amp;=3e^{-2x} \\\\\r\n4Ae^{-2x}-10Ae^{-2x}+6Ae^{-2x}&amp;=3e^{-2x} \\\\\r\n0&amp;=3e^{-2x}\r\n\\end{aligned}[\/latex],<\/p>\r\nwhich is not possible.\r\n<p id=\"fs-id1170572214728\">Looking closely, we see that, in this case, the general solution to the complementary equation is [latex]c_1e^{-2x}+c_2e^{-2x}[\/latex]. The exponential function in [latex]r(x)[\/latex] is actually a solution to the complementary equation, so, as we just saw, all the terms on the left side of the equation cancel out. We can still use the method of undetermined coefficients in this case, but we have to alter our guess by multiplying it by [latex]x[\/latex]. Using the new guess, [latex]y_p(x)=Ae^{-2x}[\/latex], we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y_p^\\prime(x)=A\\left(e^{-2x}-2xe^{-2x}\\right)}[\/latex]<\/p>\r\n<p id=\"fs-id1170572408339\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{y_p^{\\prime\\prime}(x)=-4Ae^{-2x}+4Axe^{-2x}}[\/latex].<\/p>\r\n<p id=\"fs-id1170572151641\">Substitution gives<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny^{\\prime\\prime}+5y^\\prime+6y&amp;=3e^{-2x} \\\\\r\n\\left(-4Ae^{-2x}+4Axe^{-2x}\\right)+5\\left(Ae^{-2x}-2Axe^{-2x}\\right)+6Axe^{-2x}&amp;=3e^{-2x} \\\\\r\n-4Ae^{-2x}+4Axe^{-2x}-10Axe^{-2x}+6Axe^{-2x}&amp;=3e^{-2x} \\\\\r\nAe^{-2x}&amp;=3e^{-2x}\r\n\\end{aligned}[\/latex],<\/p>\r\nSo, [latex]A=3[\/latex] and [latex]y_p(x)=3xe^{-2x}[\/latex]. This gives us the following general solution\r\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1e^{-2x}+c_2e^{-2x}+3xe^{-2x}}[\/latex].<\/p>\r\n<p id=\"fs-id1170572205861\">Note that if [latex]xe^{-2x}[\/latex] were also a solution to the complementary equation, we would have to multiply by [latex]x[\/latex] again, and we would try [latex]y_p(x)=Ax^2e^{-2x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"note\">\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: method of undetermined coefficients<\/h3>\r\n<ol id=\"fs-id1170572177553\" type=\"1\">\r\n \t<li>Solve the complementary equation and write down the general solution.<\/li>\r\n \t<li>Based on the form of [latex]r(x)[\/latex], make an initial guess for [latex]y_p(x)[\/latex].<\/li>\r\n \t<li>Check whether any term in the guess for [latex]y_p(x)[\/latex] is a solution to the complementary equation. If so, multiply the guess by [latex]x[\/latex]. Repeat this step until there are no terms in [latex]y_p(x)[\/latex] that solve the complementary equation.<\/li>\r\n \t<li>Substitute [latex]y_p(x)[\/latex] into the differential equation and equate like terms to find values for the unknown coefficients in [latex]y_p(x)[\/latex]<\/li>\r\n \t<li>Add the general solution to the complementary equation and the particular solution you just found to obtain the general solution to the nonhomogeneous equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example: solving nonhomogeneous equations<\/h3>\r\n<p id=\"fs-id1170571583170\">Find the general solutions to the following differential equations.<\/p>\r\n\r\n<ol id=\"fs-id1170571660171\" type=\"a\">\r\n \t<li>[latex]y^{\\prime\\prime}-9y=-6\\cos 3x[\/latex]<\/li>\r\n \t<li>[latex]x^{\\prime\\prime}+2x'+x=4e^{-t}[\/latex]<\/li>\r\n \t<li>[latex]y^{\\prime\\prime}-2y^\\prime+5y=10^{2}-3x-3[\/latex]<\/li>\r\n \t<li>[latex]y^{\\prime\\prime}-3y^\\prime=-12t[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"694379198\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"694379198\"]\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol type=\"a\">\r\n \t<li>The complementary equation is [latex]y^{\\prime\\prime}-9y=0[\/latex], which has the general solution [latex]c_1e^{3x}+c_2e^{-3x}[\/latex] (step 1). Based on the form of [latex]r(x)=-6\\cos3x[\/latex], our initial guess for the particular solution is [latex]y_p(x)=A\\cos3x+B\\sin3x[\/latex] (step 2). None of the terms in [latex]y_p(x)[\/latex] solve the complementary equation, so this is a valid guess (step 3).<span data-type=\"newline\">\r\n<\/span>Now we want to find values for [latex]A[\/latex] and [latex]B[\/latex], so substitute [latex]y_p[\/latex] into the differential equation. We have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\large{y_p^\\prime(x)=-3A\\sin3x+3B\\cos3x\\text{ and }y_p^{\\prime\\prime}(x)=-9A\\cos3x-9B\\sin3x}[\/latex],<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>so we want to find values of [latex]A[\/latex] and [latex]B[\/latex] such that<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny^{\\prime\\prime}-9y&amp;=-6\\cos 3x \\\\\r\n-9A\\cos3x-9B\\sin3x-9(A\\cos3x+B\\sin3x)&amp;=-6\\cos 3x \\\\\r\n-18A\\cos3x-18B\\sin3x&amp;=-6\\cos 3x\r\n\\end{aligned}[\/latex]<\/p>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\nTherefore,<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n-18A&amp;=-6 \\\\\r\n-18B&amp;=0\r\n\\end{aligned}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>This gives [latex]A=\\frac{1}{3}[\/latex] and [latex]B=0[\/latex], so [latex]y_p(x)=\\left(\\frac13\\right)\\cos3x[\/latex] (step 4).<span data-type=\"newline\">\r\n<\/span>Putting everything together, we have the general solution<span data-type=\"newline\">\r\n<\/span>\r\n\r\n<center>[latex]\\large{y(x)=c_1e^{3x}+c_2e^{-3x}+\\frac13\\cos3x}[\/latex].<\/center>\r\n<ol id=\"fs-id1170571890171\" type=\"a\">\r\n \t<li>The complementary equation is [latex]x^{\\prime\\prime}+2x^\\prime+x=0[\/latex], which has the general solution [latex]c_1e^{-t}+c_2te^{-t}[\/latex] (step 1). Based on the form [latex]r(t)=4e^{-t}[\/latex], our initial guess for the particular solution is [latex]x_p(t)=Ae^{-t}[\/latex] (step 2). However, we see that this guess solves the complementary equation, so we must multiply by [latex]t[\/latex], which gives a new guess: [latex]x_p(t)=Ate^{-t}[\/latex] (step 3). Checking this new guess, we see that it, too, solves the complementary equation, so we must multiply by [latex]t[\/latex] again, which gives [latex]x_p(t)=At^2e^{-t}[\/latex] (step 3 again). Now, checking this guess, we see that [latex]x_p(t)[\/latex] does not solve the complementary equation, so this is a valid guess (step 3 yet again).<span data-type=\"newline\">\r\n<\/span>We now want to find a value for [latex]A[\/latex], so we substitute [latex]x_p[\/latex] into the differential equation. We have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx_p(t)&amp;=At^2e^{-t} \\\\\r\nx_p^\\prime(t)&amp;=2Ate^{-t} -At^2e^{-t}\r\n\\end{aligned}[\/latex]<\/p>\r\nand [latex]x_p^{\\prime\\prime}=2Ae^{-t}-2Ate^{-t}-\\left(2Ate^{-t}-At^2e^{-t}\\right)=2Ae^{-t}-4Ate^{-t}+At^2e^{-t}[\/latex].\r\nSubstituting into the differential equation, we want to find a value of [latex]A[\/latex] so that<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx^{\\prime\\prime}+2x^\\prime+x&amp;=4e^{-t} \\\\\r\n2Ae^{-t}-4Ate^{-t}+At^2e^{-t}+2(2Ate^{-t}-At^2e^{-t})+At^2e^{-t}&amp;=4e^{-t} \\\\\r\n2Ae^{-t}=4e^{-t}\\end{aligned}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\nThis gives [latex]A=2[\/latex], so [latex]x_p(t)=2t^2e^{-t}[\/latex] (step 4). Putting everything together, we have the general solution\r\n<\/span>\r\n\r\n<center>[latex]\\large{x(t)=c_1e^{-t}+c_2te^{-t}+2t^2e^{-t}}[\/latex].<\/center><\/li>\r\n \t<li>The complementary equation is [latex]y^{\\prime\\prime}-2y^\\prime+5y=0[\/latex], which has the general solution [latex]c_1e^x\\cos2x+c_2e^x\\sin2x[\/latex] (step 1). Based on the form [latex]r(x)=10x^2-3x-3[\/latex], our initial guess for the particular solution is [latex]y_p(x)=Ax^2+Bx+C[\/latex] (step 2). None of the terms in [latex]y_p(x)[\/latex] solve the complementary equation, so this is a valid guess (step 3). We now want to find values for [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex], so we substitute [latex]y_p[\/latex] into the differential equation. We have [latex]y_p^\\prime(x)=2Ax+B[\/latex] and [latex]y_p^{\\prime\\prime}(x)=2A[\/latex], so we want to find values of [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] such that<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny^{\\prime\\prime}-2y^\\prime+5y&amp;=10x^2-3x-3 \\\\\r\n2A-2(2Ax+b)+5(Ax^2+Bx+C)&amp;=10x^2-3x-3 \\\\\r\n5Ax^2+(5B-4A)x+(5C-2B+2A)&amp;=10x^2-3x-3\r\n\\end{aligned}[\/latex].<\/p>\r\nTherefore,<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n5A&amp;=10 \\\\\r\n5B-4A&amp;=-3 \\\\\r\n5C-2B+2A&amp;=-3\r\n\\end{aligned}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>This gives [latex]A=2[\/latex], [latex]B=1[\/latex], and [latex]C=-1[\/latex], so [latex]y_p(x)=2x^2+x-1[\/latex] (step 4). Putting everything together, we have the general solution<span data-type=\"newline\">\r\n<\/span>\r\n\r\n<center>[latex]y(x)=c_1e^x\\cos2x+c_2e^x\\sin2x+2x^2+x-1[\/latex].<\/center><\/li>\r\n \t<li>The complementary equation is [latex]y^{\\prime\\prime}-3y^\\prime=0[\/latex], which has the general solution [latex]c_1e^{3t}+c_2[\/latex] (step 1). Based on the form [latex]r(t)=-12t[\/latex], our initial guess for the particular solution is [latex]y_p(t)=At+B[\/latex] (step 2). However, we see that the constant term in this guess solves the complementary equation, so we must multiply by [latex]t[\/latex], which gives a new guess: [latex]y_P(t)=At^2+Bt[\/latex] (step 3). Checking this new guess, we see that none of the terms in [latex]y_p(t)[\/latex] solve the complementary equation, so this is a valid guess (step 3 again). We now want to find values for [latex]A[\/latex] and [latex]B[\/latex], so we substitute [latex]y_p[\/latex] into the differential equation. We have [latex]y_p^\\prime(t)=2At+B[\/latex] and [latex]y_p^{\\prime\\prime}(t)=2A[\/latex], so we want to find values of [latex]A[\/latex] and [latex]B[\/latex] such that<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny^{\\prime\\prime}-3y^\\prime&amp;=-12t \\\\\r\n2A-3(2At+B)=-12t \\\\\r\n-6At+(2A-3B)&amp;=-12t\r\n\\end{aligned}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Therefore,<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n-6A&amp;=-12 \\\\\r\n2A-3B&amp;=0\r\n\\end{aligned}[\/latex].<\/p>\r\n<p style=\"text-align: center;\"><span data-type=\"newline\">\r\n<\/span>This gives [latex]A=2[\/latex] and [latex]b=4\/3[\/latex], so [latex]y_p(t)=2t^2+(4\/3)t[\/latex] (step 4). Putting everything together, we have the general solution\r\n[latex]y(t)=c_1e^{3t}+c_2+2t^2+\\frac43t[\/latex].<\/p>\r\n<\/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\n<p id=\"fs-id1170571715424\">Find the general solution to the following differential equations.<\/p>\r\n\r\n<ol id=\"fs-id1170571715427\" type=\"a\">\r\n \t<li>[latex]y^{\\prime\\prime}-5y^\\prime+4y=3e^x[\/latex]<\/li>\r\n \t<li>[latex]y^{\\prime\\prime}+y^\\prime-6y=52\\cos2t[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"597293813\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"597293813\"]\r\n\r\na.\u00a0[latex]y(x)=c_1e^{4x}+c_2e^x-xe^x[\/latex]\r\n\r\nb.\u00a0[latex]y(t)=c_1e^t+c_2te^t+te^t\\ln|t|[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Variation of Parameters<\/h2>\r\n<p id=\"fs-id1170572593832\">Sometimes, [latex]r(x)[\/latex] is not a combination of polynomials, exponentials, or sines and cosines. When this is the case, the method of undetermined coefficients does not work, and we have to use another approach to find a particular solution to the differential equation. We use an approach called the\u00a0<span id=\"e2cb0fd1-6cb0-486c-93d7-2edd081caff6_term311\" data-type=\"term\">method of variation of parameters<\/span>.<\/p>\r\n<p id=\"fs-id1170572593856\">To simplify our calculations a little, we are going to divide the differential equation through by [latex]a[\/latex], so we have a leading coefficient of 1. Then the differential equation has the form<\/p>\r\n\r\n<div id=\"fs-id1170572604296\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><center>[latex]y^{\\prime\\prime}+py^\\prime+qy=r(x)[\/latex],<\/center>\r\n<div><\/div>\r\nwhere [latex]p[\/latex] and [latex]q[\/latex] are constants.\r\n\r\n<\/div>\r\n<p id=\"fs-id1170572604354\">If the general solution to the complementary equation is given by [latex]c_1y_1(x)+c_2y_2(x)[\/latex], we are going to look for a particular solution of the form [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex]. In this case, we use the two linearly independent solutions to the complementary equation to form our particular solution. However, we are assuming the coefficients are functions of [latex]x[\/latex], rather than constants. We want to find functions [latex]u(x)[\/latex] and [latex]v(x)[\/latex] such that [latex]y_p(x)[\/latex] satisfies the differential equation. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny_p&amp;=uy_1+vy_2 \\\\\r\ny_p^\\prime&amp;=u^\\prime+uy_1^\\prime+v^\\prime y_2+vy_2^\\prime \\\\\r\ny_p^{\\prime\\prime}&amp;=(u^\\prime y_1+v^\\prime y_2)^\\prime+u^\\prime y_1^\\prime+uy_1^{\\prime\\prime}+v^\\prime y_2^\\prime+vy_2^{\\prime\\prime}\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170571579727\">Substituting into the differential equation, we obtain<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny_p^{\\prime\\prime}+py_p^\\prime+qy_p&amp;=[(u^\\prime y_1+v^\\prime y_2)+u^\\prime y_1^\\prime+uy_1^{\\prime\\prime}+v^\\prime y_2^\\prime+vy_2^{\\prime\\prime}]+p[u^\\prime y_1+uy_1^\\prime+v^\\prime y_2+vy_2^\\prime]+q[uy_1+vy_2] \\\\\r\n&amp;=u[y_1^{\\prime\\prime}+py_1^\\prime+qy_1]+v[y_2^{\\prime\\prime}+py_2^\\prime+qy_2]+(u^\\prime y_1+v^\\prime y_2)^\\prime+p(u^\\prime y_1+v^\\prime y_2)+(u^\\prime y_1^\\prime+v^\\prime y_2^\\prime)\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170572185030\">Note that [latex]y_1[\/latex] and [latex]y_2[\/latex] are solutions to the complementary equation, so the first two terms are zero. Thus, we have<\/p>\r\n<p style=\"text-align: center;\">[latex](u^\\prime y_1+v^\\prime y_2)^\\prime+p(u^\\prime y_1+v^\\prime y_2)+(u^\\prime y_1^\\prime+v^\\prime y_2^\\prime)=r(x)[\/latex].<\/p>\r\n<p id=\"fs-id1170571670359\">If we simplify this equation by imposing the additional condition [latex]u^\\prime y_1+v^\\prime y_2=0[\/latex], the first two terms are zero, and this reduces to [latex]u^\\prime y_1^\\prime+v^\\prime y_2=r(x)[\/latex]. So, with this additional condition, we have a system of two equations in two unknowns:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nu^\\prime y_1+v^\\prime y_2&amp;=0 \\\\\r\nu^\\prime y_1^\\prime+v^\\prime y_2^\\prime&amp;=r(x)\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170571613606\">Solving this system gives us [latex]u'[\/latex] and [latex]v'[\/latex], which we can integrate to find [latex]u[\/latex] and\u00a0[latex]v[\/latex].<\/p>\r\n<p id=\"fs-id1170571613638\">Then, [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex] is a particular solution to the differential equation. Solving this system of equations is sometimes challenging, so let\u2019s take this opportunity to review Cramer\u2019s rule, which allows us to solve the system of equations using determinants.<\/p>\r\n\r\n<div id=\"fs-id1170572548183\" class=\"ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">rule: cramer's rule<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572548189\">The system of equations<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\na_1z_1+b_1z_2=r_1 \\\\\r\na_2z_1+b_2z_2&amp;=r_2\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1170572387640\">has a unique solution if and only if the determinant of the coefficients is not zero. In this case, the solution is given by<\/p>\r\n<p style=\"text-align: center;\">[latex]z_1=\\frac{\\begin{vmatrix}r_1&amp;b_1 \\\\ r_2&amp;b_2\\end{vmatrix}}{\\begin{vmatrix}a_1&amp;b_1 \\\\ a_2&amp;b_2\\end{vmatrix}}[\/latex] and\u00a0[latex]z_2=\\frac{\\begin{vmatrix}a_1&amp;r_1 \\\\ a_2&amp;r_2\\end{vmatrix}}{\\begin{vmatrix}a_1&amp;b_1 \\\\ a_2&amp;b_2\\end{vmatrix}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: using cramer's rule<\/h3>\r\n<p id=\"fs-id1170572504528\">Use Cramer\u2019s rule to solve the following system of equations.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx^2z_1+2xz_2&amp;=0 \\\\\r\nz_1-3x^2z_2&amp;=2x\r\n\\end{aligned}[\/latex].<\/p>\r\n[reveal-answer q=\"923921637\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"923921637\"]\r\n<p id=\"fs-id1170572116382\">We have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\na_1(x)&amp;=x^2 \\\\\r\na_2(x)&amp;=1 \\\\\r\nb_1(x)&amp;=2x \\\\\r\nb_2(x)&amp;=-3x^2 \\\\\r\nr_1(x)&amp;=0 \\\\\r\nr_2(x)&amp;=2x\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170571748971\">Then,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{vmatrix}a_1&amp;b_1 \\\\ a_2&amp;b_2\\end{vmatrix}=\\begin{vmatrix}x^2&amp;2x \\\\ 1&amp;-3x^2\\end{vmatrix}=-3x^4-2x[\/latex]<\/p>\r\n<p id=\"fs-id1170572088607\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{vmatrix}r_1&amp;b_1 \\\\ r_2&amp;b_2\\end{vmatrix}=\\begin{vmatrix}0&amp;2x \\\\ 2x&amp;-3x^2\\end{vmatrix}=0-4x^2=-4x^2[\/latex].<\/p>\r\n<p id=\"fs-id1170571688223\">Thus,<\/p>\r\n<p style=\"text-align: center;\">[latex]z_1=\\frac{\\begin{vmatrix}r_1&amp;b_1 \\\\ r_2&amp;b_2\\end{vmatrix}}{\\begin{vmatrix}a_1&amp;b_1 \\\\ a_2&amp;b_2\\end{vmatrix}}=\\frac{-4x^2}{-3x^4-2x}=\\frac{4x}{3x^3+2}[\/latex].<\/p>\r\n<p id=\"fs-id1170571611310\">In addition,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{vmatrix}a_1&amp;r_1 \\\\ a_2&amp;r_2\\end{vmatrix}=\\begin{vmatrix}x^2&amp;0 \\\\1&amp;2x\\end{vmatrix}=2x^3-0=2x^3[\/latex].<\/p>\r\n<p id=\"fs-id1170572330219\">Thus,<\/p>\r\n<p style=\"text-align: center;\">[latex]z_2=\\frac{\\begin{vmatrix}a_1&amp;r_1 \\\\ a_2&amp;r_2\\end{vmatrix}}{\\begin{vmatrix}a_1&amp;b_1 \\\\ a_2&amp;b_2\\end{vmatrix}}=\\frac{2x^3}{-3x^4-2x}=\\frac{-2x^2}{3x^3+2}[\/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-id1170571553660\">Use Cramer\u2019s rule to solve the following system of equations.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n2xz_1-3z_2&amp;=0 \\\\\r\nx^2z_1+4xz_2&amp;=x+1\r\n\\end{aligned}[\/latex].<\/p>\r\n[reveal-answer q=\"734571894\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"734571894\"]\r\n<p style=\"text-align: center;\">[latex]z_1=\\frac{3x+3}{11x^2}, \\ z_2\\frac{2x+2}{11x}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\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: method of variation of parameters<\/h3>\r\n<ol id=\"fs-id1170571789088\" type=\"1\">\r\n \t<li>Solve the complementary equation and write down the general solution<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571789098\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]c_1y_1(x)+c_2y_2(x)[\/latex].<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li>Use Cramer\u2019s rule or another suitable technique to find functions [latex]u^\\prime(x)[\/latex] and [latex]v^\\prime(x)[\/latex] satisfying<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nu^\\prime y_1+v^\\prime y_2&amp;=0 \\\\\r\nu^\\prime y_1^\\prime+v^\\prime y_2^\\prime&amp;=r(x)\r\n\\end{aligned}[\/latex]<\/p>\r\n<\/li>\r\n \t<li>Integrate [latex]u^\\prime[\/latex] and [latex]v^\\prime[\/latex] to find [latex]u(x)[\/latex] and [latex]v(x)[\/latex]. Then, [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex] is a particular solution to the equation.<\/li>\r\n \t<li>Add the general solution to the complementary equation and the particular solution found in step 3 to obtain the general solution to the nonhomogeneous equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>example: using the method of variation of parameters<\/h3>\r\n<p id=\"fs-id1170571823253\">Find the general solution to the following differential equations.<\/p>\r\n\r\n<ol id=\"fs-id1170571823257\" type=\"a\">\r\n \t<li>[latex]y^{\\prime\\prime}-2y^\\prime+y=\\frac{e^t}{t^2}[\/latex]<\/li>\r\n \t<li>[latex]y^{\\prime\\prime}+y=3\\sin^2x[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"982348083\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"982348083\"]\r\n<ol id=\"fs-id1170572589220\" type=\"a\">\r\n \t<li>The complementary equation is [latex]y^{\\prime\\prime}-2y^\\prime+y=0[\/latex] with associated general solution [latex]c_1e^t+c_2te^t[\/latex]. Therefore, [latex]y_1(t)=e^t[\/latex] and [latex]y_2(t)=te^t[\/latex]. Calculating the derivatives, we get [latex]y_1^\\prime(t)=e^t[\/latex] and [latex]y_2^\\prime(t)=e^t+te^t[\/latex] (step 1). Then, we want to find functions [latex]u'(t)[\/latex] and [latex]v'(t)[\/latex] so that<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nu^\\prime e^t+v^\\prime te^t&amp;=0 \\\\\r\nu^\\prime e^t+v^\\prime(e^t+te^t)=\\frac{e^t}{t^2}\r\n\\end{aligned}[\/latex]<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Applying Cramer\u2019s rule, we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]u^\\prime=\\frac{\\begin{vmatrix}0&amp;te^t \\\\ \\frac{e^t}{t^2}&amp;e^t+te^t\\end{vmatrix}}{\\begin{vmatrix}e^t&amp;te^t \\\\e^t&amp;e^t+te^t\\end{vmatrix}}=\\frac{0-te^t\\left(\\frac{e^t}{t^2}\\right)}{e^t(e^t+te^t)-e^tte^t}=\\frac{-\\frac{e^{2t}}{t}}{e^{2t}}=-\\frac1t[\/latex]<\/p>\r\nand<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572296671\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]v^\\prime=\\frac{\\begin{vmatrix}e^t&amp;0\\\\e^t&amp;\\frac{e^t}{t^2}\\end{vmatrix}}{\\begin{vmatrix}e^t&amp;te^t\\\\e^t&amp;e^t+te^t\\end{vmatrix}}=\\frac{e^t\\left(\\frac{e^t}{t^2}\\right)}{e^{2t}}=\\frac{1}{t^2}[\/latex] (step 2).<\/p>\r\n\r\n<\/div>\r\nIntegrating, we get<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nu&amp;=-\\displaystyle\\int\\frac1t \\ dt=-\\ln|t| \\\\\r\nv&amp;=\\displaystyle\\int\\frac{1}{t^2}dt=-\\frac{1}{t}\\text{(step 3)}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Then we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny_p&amp;=-e^t\\ln|t|-\\frac{1}{t}te^t \\\\\r\n&amp;=-e^t\\ln|t|-e^t \\text{(step 4)}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>The [latex]e^{t}[\/latex] term is a solution to the complementary equation, so we don\u2019t need to carry that term into our general solution explicitly. The general solution is<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]y(t)=c_1e^t+c_2te^t-e^t\\ln|t|[\/latex] (step 5).<\/p>\r\n<\/li>\r\n \t<li>The complementary equation is [latex]y^{\\prime\\prime}+y+0[\/latex] with associated general solution [latex]c_1\\cos x+c_2\\sin x[\/latex]. So, [latex]y_1(x)=\\cos x[\/latex] and [latex]y_2(x)=\\sin x[\/latex] (step 1). Then, we want to find functions [latex]u'(x)[\/latex] and [latex]v'(x)[\/latex] such that<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}<\/p>\r\nu^\\prime\\cos x+v^\\prime\\sin x&amp;=0 \\\\\r\n-y^\\prime\\sin x+v^\\prime\\cos x&amp;=3\\sin^2x\r\n\\end{aligned}[\/latex].\r\n\r\n<span data-type=\"newline\">\r\n<\/span>Applying Cramer\u2019s rule, we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]u^\\prime=\\frac{\\begin{vmatrix}0&amp;\\sin x \\\\3\\sin^2x&amp;\\cos x\\end{vmatrix}}{\\begin{vmatrix}\\cos x&amp;\\sin x \\\\ -\\sin x&amp;\\cos x\\end{vmatrix}}=\\frac{0-3\\sin^3x}{\\cos^2x+\\sin^2x}=-3\\sin^3x[\/latex]<\/p>\r\nand<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]v^\\prime=\\frac{\\begin{vmatrix}\\cos x&amp;0\\\\ -\\sin x&amp;3\\sin^2x\\end{vmatrix}}{\\begin{vmatrix}\\cos x&amp;\\sin x \\\\-\\sin x&amp;\\cos x\\end{vmatrix}}=\\frac{3\\sin^2x\\cos x}{1}=3\\sin^2x\\cos x[\/latex] (step 2).<\/p>\r\nIntegrating first to find [latex]u[\/latex], we get<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]u=\\displaystyle\\int-3\\sin^3xdx=-3\\left[-\\frac13\\sin^2x\\cos x+\\frac23\\displaystyle\\int\\sin xdx\\right]=\\sin^2x\\cos x+2\\cos x[\/latex] (step 3).<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Now, we integrate to find [latex]v[\/latex]. Using substitution (with [latex]w=\\sin x[\/latex]), we get<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]v=\\displaystyle\\int3\\sin^2x\\cos xdx=\\displaystyle\\int3w^2dw=w^3=\\sin^3x[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Then,<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\r\n\\begin{aligned}\r\ny_p&amp;=(\\sin^2x\\cos x+2\\cos x)\\cos x+(\\sin^3x)\\sin x \\\\\r\n&amp;=\\sin^2x\\cos^2x+2\\cos^2x+\\sin^4x \\\\\r\n&amp;=2\\cos^2x+\\sin^2x(\\cos^2x+\\sin^2x) \\ \\ \\ \\ (step 4). \\\\\r\n&amp;=2\\cos^2x+\\sin^2 \\\\\r\n&amp;=\\cos^2x+1\r\n\\end{aligned}\r\n[\/latex]<\/p>\r\nThe general solution is\r\n<p style=\"text-align: center;\">[latex]y(x)=c_1\\cos x+c_2\\sin x+1+\\cos^2x[\/latex] (step 5).<\/p>\r\n[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<p id=\"fs-id1170572554234\">Find the general solution to the following differential equations.<\/p>\r\n\r\n<ol id=\"fs-id1170572554237\" type=\"a\">\r\n \t<li>[latex]y^{\\prime\\prime}+y=\\sec x[\/latex]<\/li>\r\n \t<li>[latex]x^{\\prime\\prime}-2x^\\prime+x=\\frac{e^t}t[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"687247112\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"687247112\"]\r\n\r\na.\u00a0[latex]y(x)=c_1\\cos x+c_2\\sin x+\\cos x\\ln|\\cos x|+x\\sin x[\/latex]\r\n\r\nb.\u00a0[latex]x(t)=c_1e^t+c_2te^t+te^t\\ln|t|[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following videos to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250341&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=8N2A8lhcxPI&amp;video_target=tpm-plugin-y9gjmauh-8N2A8lhcxPI\" 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\/CP7.14a_transcript.html\">transcript for \u201cCP 7.14a\u201d here (opens in new window).<\/a><\/center><center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250340&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=HCMJodmpYKg&amp;video_target=tpm-plugin-4khktbip-HCMJodmpYKg\" 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\/CP7.14b_transcript.html\">transcript for \u201cCP 7.14b\u201d here (opens in new window).<\/a><\/center><\/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\">Write the general solution to a nonhomogeneous differential equation.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Solve a nonhomogeneous differential equation by the method of undetermined coefficients.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Solve a nonhomogeneous differential equation by the method of variation of parameters.<\/span><\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1170572624887\" data-depth=\"1\">\n<h2 data-type=\"title\">General Solution to a Nonhomogeneous Linear Equation<\/h2>\n<p id=\"fs-id1170572616368\">Consider the nonhomogeneous linear differential equation<\/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-id1170572346900\">The associated homogeneous equation<\/p>\n<p style=\"text-align: center;\">[latex]\\large{a_2(x)y^{\\prime\\prime}+a_1(x)y^\\prime+a_0(x)y=0}[\/latex]<\/p>\n<p id=\"fs-id1170571543157\">is called the\u00a0<strong><span id=\"e2cb0fd1-6cb0-486c-93d7-2edd081caff6_term308\" data-type=\"term\">complementary equation<\/span>.<\/strong> We will see that solving the complementary equation is an important step in solving a nonhomogeneous differential equation.<\/p>\n<\/section>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p>A solution [latex]y_p(x)[\/latex] of a differential equation that contains no arbitrary constants is called a\u00a0<span id=\"e2cb0fd1-6cb0-486c-93d7-2edd081caff6_term309\" data-type=\"term\">particular solution<\/span>\u00a0to the equation.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Theorem: general solution to a nonhomogeneous equation<\/h3>\n<hr \/>\n<p id=\"fs-id1170572168730\">Let [latex]y_p(x)[\/latex] be any particular solution to the nonhomogeneous linear differential equation<\/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-id1170571711288\">Also, let [latex]c_1y_1(x)+c_2y_2(x)[\/latex] denote the general solution to the complementary equation. Then, the general solution to the nonhomogeneous equation is given by<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1y_1(x)+c_2y_2(x)+y_p(x)}[\/latex].<\/p>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1170572544509\">To prove [latex]y(x)[\/latex] is the general solution, we must first show that it solves the differential equation and, second, that any solution to the differential equation can be written in that form. Substituting [latex]y(x)[\/latex] into the differential equation, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  a_2(x)y^{\\prime\\prime}+a_1(x)y^\\prime+a_0(x)y&=a_2(x)(c_1y_1+c_2y_2+y_p)^{\\prime\\prime}+a_1(x)(c_1y_1+c_2y_2+y_p)^\\prime+a_0(x)(c_1y_1+c_2y_2+y_p) \\\\  &=[a_2(x)(c_1y_1+c_2y_2)^{\\prime\\prime}+a_1(x)(c_1y_1+c_2y_2)^\\prime+a_0(x)(c_1y_1+c_2y_2)]+a_2(x)y_p^{\\prime\\prime}+a_1(x)y_p^\\prime+a_0(x)y_p \\\\  &=0+r(x) \\\\  &=r(x)  \\end{aligned}[\/latex].<\/p>\n<p>So [latex]y(x)[\/latex] is a solution.<\/p>\n<p id=\"fs-id1170572415209\">Now, let [latex]z(x)[\/latex] be any solution to [latex]a_2(x)y^{\\prime\\prime}+a_1(x)y^\\prime+a_0(x)y=r(x)[\/latex]. Then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  a_2(x)(z-y_p)^{\\prime\\prime}+a_1(x)(z-y_p)^\\prime+a_0(x)(z-y_p)&=(a_2(x)z^{\\prime\\prime}+a_1(x)z^\\prime+a_0(x)z)-(a_2(x)y_p^{\\prime\\prime}+a_1(x)y_p^\\prime+a_0(x)y_p) \\\\  &=r(x)-r(x) \\\\  &=0  \\end{aligned}[\/latex],<\/p>\n<p>so [latex]z(x)=y_p(x)[\/latex] is a solution to the complementary equation. But, [latex]c_1y_1(x)+c_2y_2(x)[\/latex] is the general solution to the complementary equation, so there are constants [latex]c_1[\/latex] and [latex]c_2[\/latex] such that<\/p>\n<p style=\"text-align: center;\">[latex]\\large{z(x)=y_p(x)=c_1y_1(x)+c_2y_2(x)}[\/latex].<\/p>\n<p id=\"fs-id1170572481683\">Hence, we see that [latex]z(x)=c_1y_1(x)+c_2y_2(x)y_p(x)[\/latex].<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: verifying the general solution<\/h3>\n<p>Given that [latex]y_p(x)=x[\/latex] is a particular solution to the differential equation [latex]y^{\\prime\\prime}+y=x[\/latex], write the general solution and check by verifying that the solution satisfies the equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q284270134\">Show Solution<\/span><\/p>\n<div id=\"q284270134\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572547637\">The complementary equation is [latex]y^{\\prime\\prime}+y=0[\/latex], which has the general solution [latex]c_1\\cos x+c_2\\sin x[\/latex]. So, the general solution to the nonhomogeneous equation is<\/p>\n<p style=\"text-align: center;\">[latex]y(x)=c_1\\cos x+c_2\\sin x+x[\/latex].<\/p>\n<p id=\"fs-id1170572212481\">To verify that this is a solution, substitute it into the differential equation. We have<\/p>\n<p style=\"text-align: center;\">[latex]y^\\prime(x)=-c_1\\sin x+c_2\\cos x+1\\text{ and }y^{\\prime\\prime}=-c_1\\cos x-c_2\\sin x[\/latex].<\/p>\n<p id=\"fs-id1170572637641\">Then<\/p>\n<p style=\"text-align: center;\">[latex]y^{\\prime\\prime}(x)+y(x)=-c_1\\cos x-c_2\\sin x+c_1\\cos x+c_2\\sin x+x[\/latex].<\/p>\n<p id=\"fs-id1170572416526\">So, [latex]y(x)[\/latex] is a solution to [latex]y^{\\prime\\prime}+y=x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Given that [latex]y_p(x)=-2[\/latex] is a particular solution to [latex]y^{\\prime\\prime}-3y^\\prime-4y=8[\/latex], write the general solution and verify that the general solution satisfies the equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q039452983\">Show Solution<\/span><\/p>\n<div id=\"q039452983\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y(x)=c_1e^{-x}+c_2e^{4x}-2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1170572624887\" data-depth=\"1\">\n<section id=\"fs-id1170571569017\" data-depth=\"2\">\n<p id=\"fs-id1170572240485\">In the preceding section, we learned how to solve homogeneous equations with constant coefficients. Therefore, for nonhomogeneous equations of the form [latex]ay^{\\prime\\prime}+by^\\prime+cy=r(x)[\/latex], we already know how to solve the complementary equation, and the problem boils down to finding a particular solution for the nonhomogeneous equation. We now examine two techniques for this: the method of undetermined coefficients and the method of variation of parameters.<\/p>\n<\/section>\n<\/section>\n<section id=\"fs-id1170571565966\" data-depth=\"1\">\n<h2 data-type=\"title\">Undetermined Coefficients<\/h2>\n<p id=\"fs-id1170571630374\">The\u00a0<strong><span id=\"e2cb0fd1-6cb0-486c-93d7-2edd081caff6_term310\" data-type=\"term\">method of undetermined coefficients<\/span><\/strong>\u00a0involves making educated guesses about the form of the particular solution based on the form of [latex]r(x)[\/latex]. When we take derivatives of polynomials, exponential functions, sines, and cosines, we get polynomials, exponential functions, sines, and cosines. So when [latex]r(x)[\/latex] has one of these forms, it is possible that the solution to the nonhomogeneous differential equation might take that same form. Let\u2019s look at some examples to see how this works.<\/p>\n<\/section>\n<div data-type=\"note\">\n<div class=\"textbox exercises\">\n<h3>Example: undetermined coefficients when\u00a0[latex]r(x)[\/latex] is a polynomial<\/h3>\n<p>Find the general solution to [latex]y^{\\prime\\prime}+4y^\\prime+3y=3x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q204756223\">Show Solution<\/span><\/p>\n<div id=\"q204756223\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572204075\">The complementary equation is [latex]y^{\\prime\\prime}+4y^\\prime+3y=0[\/latex], with general solution [latex]c_1e^{-x}+c_2e^{-3x}[\/latex]. Since [latex]r(x)=3x[\/latex], the particular solution might have the form [latex]y_p(x)=Ax+B[\/latex]. If this is the case, then we have [latex]y_p^\\prime(x)=A[\/latex] and [latex]y_py^{\\prime\\prime}(x)=0[\/latex]. For [latex]y_p[\/latex] to be a solution to the differential equation, we must find values for [latex]A[\/latex] and [latex]B[\/latex] such that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y^{\\prime\\prime}+4^\\prime+3y&=3x \\\\  0+4(A)+3(Ax+B)&=3x \\\\  3Ax+(4A+3B)&=3x  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170572481625\">Setting coefficients of like terms equal, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  3A&=3 \\\\  4A+3b&=0  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170572643180\">Then, [latex]A=1[\/latex] and [latex]B=-\\frac{4}{3}[\/latex], so [latex]y_p(x)=x-\\frac{4}{3}[\/latex] and the general solution is<\/p>\n<p style=\"text-align: center;\">[latex]y(x)=c_1e^{-x}+c_2e^{-3x}+x-\\frac43[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In\u00a0Example &#8220;Undetermined Coefficients When [latex]\ud835\udc5f(\ud835\udc65)[\/latex] Is a Polynomial&#8221;, notice that even though [latex]r(x)[\/latex] did not include a constant term, it was necessary for us to include the constant term in our guess. If we had assumed a solution of the form [latex]y_p=Ax[\/latex] (with no constant term), we would not have been able to find a solution. (Verify this!) If the function [latex]r(x)[\/latex] is a polynomial, our guess for the particular solution should be a polynomial of the same degree, and it must include all lower-order terms, regardless of whether they are present in [latex]r(x)[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: undetermined coefficients when\u00a0[latex]r(x)[\/latex] is an exponential<\/h3>\n<p>Find the general solution to [latex]y^{\\prime\\prime}-y^\\prime-2y=2e^{3x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q083272901\">Show Solution<\/span><\/p>\n<div id=\"q083272901\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572306234\">The complementary equation is [latex]y^{\\prime\\prime}-y^\\prime-2y=0[\/latex], with the general solution [latex]c_1e^{-x}+c_2e^{2x}[\/latex]. Since [latex]r(x)=2e^{3x}[\/latex], the particular solution might have the form [latex]y_p=Ae^{3x}[\/latex]. Then, we have [latex]y_p^\\prime(x)=3Ae^{3x}[\/latex] and [latex]y^{\\prime\\prime}=9Ae^{3x}[\/latex]. For [latex]y_p[\/latex] to be a solution to the differential equation, we must find a value for [latex]A[\/latex] such that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y^{\\prime\\prime}-y^\\prime-2y&=2e^{3x} \\\\  9Ae^{3x}-3Ae^{3x}-2Ae^{3x}&=2Ae^{3x} \\\\  4Ae^{3x}&=2Ae^{3x}  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170572221173\">So, [latex]4A=2[\/latex] and [latex]A=\\frac{1}{2}[\/latex]. Then, [latex]y_p(x)=\\left(\\frac12\\right)e^{3x}[\/latex], and the general solution is<\/p>\n<p style=\"text-align: center;\">[latex]y(x)=c_1e^{-x}+c_2e^{2x}+\\frac12e^{3x}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the general solution to [latex]y^{\\prime\\prime}-4y^\\prime+4y=7\\sin t-\\cos t[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q028934924\">Show Solution<\/span><\/p>\n<div id=\"q028934924\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y(t)=c_1e^{2t}+c_2te^{2t}+\\sin t+\\cos t[\/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=8250342&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=rQsUWvHvRSY&amp;video_target=tpm-plugin-hfu2pzjl-rQsUWvHvRSY\" 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\/CP7.11_transcript.html\">transcript for \u201cCP 7.11\u201d here (opens in new window).<\/a><\/div>\n<p>In the previous checkpoint, [latex]r(x)[\/latex] included both sine and cosine terms. However, even if [latex]r(x)[\/latex] included a sine term only or a cosine term only, both terms must be present in the guess. The method of undetermined coefficients also works with products of polynomials, exponentials, sines, and cosines. Some of the key forms of [latex]r(x)[\/latex] and the associated guesses for [latex]y_p(x)[\/latex] are summarized in\u00a0Table 7.2\u00a0<span class=\"os-title\" data-type=\"title\">Key Forms for the Method of Undetermined Coefficients below.<\/span><\/p>\n<\/div>\n<div data-type=\"note\">\n<table style=\"height: 192px;\" data-id=\"fs-id1170572294713\">\n<thead>\n<tr style=\"height: 14px;\" valign=\"top\">\n<th style=\"height: 14px; width: 141.922px;\" scope=\"col\" data-align=\"left\">[latex]r(x)[\/latex]<\/th>\n<th style=\"height: 14px; width: 739.438px;\" scope=\"col\" data-align=\"left\">Initial guess for [latex]y_p(x)[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 28px;\" valign=\"top\">\n<td style=\"height: 28px; width: 141.922px;\" data-align=\"left\">[latex]k[\/latex] (a constant)<\/td>\n<td style=\"height: 28px; width: 739.438px;\" data-align=\"left\">[latex]A[\/latex] (a constant)<\/td>\n<\/tr>\n<tr style=\"height: 14px;\" valign=\"top\">\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex]ax+b[\/latex]<\/td>\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex]Ax+B[\/latex] (<em data-effect=\"italics\">Note<\/em>: The guess must include both terms even if [latex]b=0[\/latex].)<\/td>\n<\/tr>\n<tr style=\"height: 14px;\" valign=\"top\">\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex]ax^{2}+bx+c[\/latex]<\/td>\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex]Ax^{2}+Bx+C[\/latex] (<em data-effect=\"italics\">Note<\/em>: The guess must include all three terms even if [latex]b[\/latex] or [latex]c[\/latex] are zero.)<\/td>\n<\/tr>\n<tr style=\"height: 28px;\" valign=\"top\">\n<td style=\"height: 28px; width: 141.922px;\" data-align=\"left\">Higher-order polynomials<\/td>\n<td style=\"height: 28px; width: 739.438px;\" data-align=\"left\">Polynomial of the same order as [latex]r(x)[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\" valign=\"top\">\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex]ae^{\\lambda x}[\/latex]<\/td>\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex]Ae^{\\lambda x}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\" valign=\"top\">\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex]a\\cos\\beta x+b\\sin\\beta x[\/latex]<\/td>\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex]A\\cos\\beta x+B\\sin\\beta x[\/latex] (<em data-effect=\"italics\">Note<\/em>: The guess must include both terms even if either [latex]a=0[\/latex] or [latex]b=0[\/latex].)<\/td>\n<\/tr>\n<tr style=\"height: 14px;\" valign=\"top\">\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex]ae^{\\alpha x}\\cos\\beta x+be^{\\alpha x}\\sin\\beta x[\/latex]<\/td>\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex]Ae^{\\alpha x}\\cos\\beta x+Be^{\\alpha x}\\sin\\beta x[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\" valign=\"top\">\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex](ax^2+bx+c)e^{\\lambda x}[\/latex]<\/td>\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex](Ax^2+Bx+C)e^{\\lambda x}[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 14px;\" valign=\"top\">\n<td style=\"height: 14px; width: 141.922px;\" data-align=\"left\">[latex](a_2x^2+a_1x+a_0)\\cos\\beta x+(b_2x^2+b_1x+b_0)\\sin\\beta x[\/latex]<\/td>\n<td style=\"height: 14px; width: 739.438px;\" data-align=\"left\">[latex](A_2x^2+A_1x+A_0)\\cos\\beta x+(B_2x^2+B_1x+B_0)\\sin\\beta x[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 24px;\" valign=\"top\">\n<td style=\"height: 24px; width: 141.922px;\" data-align=\"left\">[latex](a_2x^2+a_1x+a_0)e^{\\alpha x}\\cos\\beta x+(b_2x^2+b_1x+b_0)e^{\\alpha x}\\sin\\beta x[\/latex]<\/td>\n<td style=\"height: 24px; width: 739.438px;\" data-align=\"left\">[latex](A_2x^2+A_1x+A_0)e^{\\alpha x}\\cos\\beta x+(B_2x^2+B_1x+B_0)e^{\\alpha x}\\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\">7.2<\/span>\u00a0<span class=\"os-title\" data-type=\"title\">Key Forms for the Method of Undetermined Coefficients<\/span><\/div>\n<\/div>\n<div data-type=\"note\"><\/div>\n<div data-type=\"note\">\n<p id=\"fs-id1170572304163\">Keep in mind that there is a key pitfall to this method. Consider the differential equation [latex]y^{\\prime\\prime}+5y^\\prime+6y=3e^{-2x}[\/latex]. Based on the form of [latex]r(x)[\/latex], we guess a particular solution of the form [latex]y_p(x)=Ae^{-2x}[\/latex]. But when we substitute this expression into the differential equation to find a value for [latex]A[\/latex], we run into a problem. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y_p^\\prime(x)=-2Ae^{-2x}}[\/latex]<\/p>\n<p id=\"fs-id1170572611802\">and<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y_p^{\\prime\\prime}=4Ae^{-2x}}[\/latex],<\/p>\n<p id=\"fs-id1170572593508\">so we want<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y^{\\prime\\prime}+5y^\\prime+6y&=3e^{-2x} \\\\  4Ae^{-2x}+\\left(-2Ae^{-2x}\\right)+6Ae^{-2x}&=3e^{-2x} \\\\  4Ae^{-2x}-10Ae^{-2x}+6Ae^{-2x}&=3e^{-2x} \\\\  0&=3e^{-2x}  \\end{aligned}[\/latex],<\/p>\n<p>which is not possible.<\/p>\n<p id=\"fs-id1170572214728\">Looking closely, we see that, in this case, the general solution to the complementary equation is [latex]c_1e^{-2x}+c_2e^{-2x}[\/latex]. The exponential function in [latex]r(x)[\/latex] is actually a solution to the complementary equation, so, as we just saw, all the terms on the left side of the equation cancel out. We can still use the method of undetermined coefficients in this case, but we have to alter our guess by multiplying it by [latex]x[\/latex]. Using the new guess, [latex]y_p(x)=Ae^{-2x}[\/latex], we have<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y_p^\\prime(x)=A\\left(e^{-2x}-2xe^{-2x}\\right)}[\/latex]<\/p>\n<p id=\"fs-id1170572408339\">and<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y_p^{\\prime\\prime}(x)=-4Ae^{-2x}+4Axe^{-2x}}[\/latex].<\/p>\n<p id=\"fs-id1170572151641\">Substitution gives<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y^{\\prime\\prime}+5y^\\prime+6y&=3e^{-2x} \\\\  \\left(-4Ae^{-2x}+4Axe^{-2x}\\right)+5\\left(Ae^{-2x}-2Axe^{-2x}\\right)+6Axe^{-2x}&=3e^{-2x} \\\\  -4Ae^{-2x}+4Axe^{-2x}-10Axe^{-2x}+6Axe^{-2x}&=3e^{-2x} \\\\  Ae^{-2x}&=3e^{-2x}  \\end{aligned}[\/latex],<\/p>\n<p>So, [latex]A=3[\/latex] and [latex]y_p(x)=3xe^{-2x}[\/latex]. This gives us the following general solution<\/p>\n<p style=\"text-align: center;\">[latex]\\large{y(x)=c_1e^{-2x}+c_2e^{-2x}+3xe^{-2x}}[\/latex].<\/p>\n<p id=\"fs-id1170572205861\">Note that if [latex]xe^{-2x}[\/latex] were also a solution to the complementary equation, we would have to multiply by [latex]x[\/latex] again, and we would try [latex]y_p(x)=Ax^2e^{-2x}[\/latex].<\/p>\n<\/div>\n<div data-type=\"note\">\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: method of undetermined coefficients<\/h3>\n<ol id=\"fs-id1170572177553\" type=\"1\">\n<li>Solve the complementary equation and write down the general solution.<\/li>\n<li>Based on the form of [latex]r(x)[\/latex], make an initial guess for [latex]y_p(x)[\/latex].<\/li>\n<li>Check whether any term in the guess for [latex]y_p(x)[\/latex] is a solution to the complementary equation. If so, multiply the guess by [latex]x[\/latex]. Repeat this step until there are no terms in [latex]y_p(x)[\/latex] that solve the complementary equation.<\/li>\n<li>Substitute [latex]y_p(x)[\/latex] into the differential equation and equate like terms to find values for the unknown coefficients in [latex]y_p(x)[\/latex]<\/li>\n<li>Add the general solution to the complementary equation and the particular solution you just found to obtain the general solution to the nonhomogeneous equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example: solving nonhomogeneous equations<\/h3>\n<p id=\"fs-id1170571583170\">Find the general solutions to the following differential equations.<\/p>\n<ol id=\"fs-id1170571660171\" type=\"a\">\n<li>[latex]y^{\\prime\\prime}-9y=-6\\cos 3x[\/latex]<\/li>\n<li>[latex]x^{\\prime\\prime}+2x'+x=4e^{-t}[\/latex]<\/li>\n<li>[latex]y^{\\prime\\prime}-2y^\\prime+5y=10^{2}-3x-3[\/latex]<\/li>\n<li>[latex]y^{\\prime\\prime}-3y^\\prime=-12t[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q694379198\">Show Solution<\/span><\/p>\n<div id=\"q694379198\" class=\"hidden-answer\" style=\"display: none\">\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ol type=\"a\">\n<li>The complementary equation is [latex]y^{\\prime\\prime}-9y=0[\/latex], which has the general solution [latex]c_1e^{3x}+c_2e^{-3x}[\/latex] (step 1). Based on the form of [latex]r(x)=-6\\cos3x[\/latex], our initial guess for the particular solution is [latex]y_p(x)=A\\cos3x+B\\sin3x[\/latex] (step 2). None of the terms in [latex]y_p(x)[\/latex] solve the complementary equation, so this is a valid guess (step 3).<span data-type=\"newline\"><br \/>\n<\/span>Now we want to find values for [latex]A[\/latex] and [latex]B[\/latex], so substitute [latex]y_p[\/latex] into the differential equation. We have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{y_p^\\prime(x)=-3A\\sin3x+3B\\cos3x\\text{ and }y_p^{\\prime\\prime}(x)=-9A\\cos3x-9B\\sin3x}[\/latex],<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>so we want to find values of [latex]A[\/latex] and [latex]B[\/latex] such that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y^{\\prime\\prime}-9y&=-6\\cos 3x \\\\  -9A\\cos3x-9B\\sin3x-9(A\\cos3x+B\\sin3x)&=-6\\cos 3x \\\\  -18A\\cos3x-18B\\sin3x&=-6\\cos 3x  \\end{aligned}[\/latex]<\/p>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p>Therefore,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  -18A&=-6 \\\\  -18B&=0  \\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>This gives [latex]A=\\frac{1}{3}[\/latex] and [latex]B=0[\/latex], so [latex]y_p(x)=\\left(\\frac13\\right)\\cos3x[\/latex] (step 4).<span data-type=\"newline\"><br \/>\n<\/span>Putting everything together, we have the general solution<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div style=\"text-align: center;\">[latex]\\large{y(x)=c_1e^{3x}+c_2e^{-3x}+\\frac13\\cos3x}[\/latex].<\/div>\n<ol id=\"fs-id1170571890171\" type=\"a\">\n<li>The complementary equation is [latex]x^{\\prime\\prime}+2x^\\prime+x=0[\/latex], which has the general solution [latex]c_1e^{-t}+c_2te^{-t}[\/latex] (step 1). Based on the form [latex]r(t)=4e^{-t}[\/latex], our initial guess for the particular solution is [latex]x_p(t)=Ae^{-t}[\/latex] (step 2). However, we see that this guess solves the complementary equation, so we must multiply by [latex]t[\/latex], which gives a new guess: [latex]x_p(t)=Ate^{-t}[\/latex] (step 3). Checking this new guess, we see that it, too, solves the complementary equation, so we must multiply by [latex]t[\/latex] again, which gives [latex]x_p(t)=At^2e^{-t}[\/latex] (step 3 again). Now, checking this guess, we see that [latex]x_p(t)[\/latex] does not solve the complementary equation, so this is a valid guess (step 3 yet again).<span data-type=\"newline\"><br \/>\n<\/span>We now want to find a value for [latex]A[\/latex], so we substitute [latex]x_p[\/latex] into the differential equation. We have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x_p(t)&=At^2e^{-t} \\\\  x_p^\\prime(t)&=2Ate^{-t} -At^2e^{-t}  \\end{aligned}[\/latex]<\/p>\n<p>and [latex]x_p^{\\prime\\prime}=2Ae^{-t}-2Ate^{-t}-\\left(2Ate^{-t}-At^2e^{-t}\\right)=2Ae^{-t}-4Ate^{-t}+At^2e^{-t}[\/latex].<br \/>\nSubstituting into the differential equation, we want to find a value of [latex]A[\/latex] so that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x^{\\prime\\prime}+2x^\\prime+x&=4e^{-t} \\\\  2Ae^{-t}-4Ate^{-t}+At^2e^{-t}+2(2Ate^{-t}-At^2e^{-t})+At^2e^{-t}&=4e^{-t} \\\\  2Ae^{-t}=4e^{-t}\\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\nThis gives [latex]A=2[\/latex], so [latex]x_p(t)=2t^2e^{-t}[\/latex] (step 4). Putting everything together, we have the general solution<br \/>\n<\/span><\/p>\n<div style=\"text-align: center;\">[latex]\\large{x(t)=c_1e^{-t}+c_2te^{-t}+2t^2e^{-t}}[\/latex].<\/div>\n<\/li>\n<li>The complementary equation is [latex]y^{\\prime\\prime}-2y^\\prime+5y=0[\/latex], which has the general solution [latex]c_1e^x\\cos2x+c_2e^x\\sin2x[\/latex] (step 1). Based on the form [latex]r(x)=10x^2-3x-3[\/latex], our initial guess for the particular solution is [latex]y_p(x)=Ax^2+Bx+C[\/latex] (step 2). None of the terms in [latex]y_p(x)[\/latex] solve the complementary equation, so this is a valid guess (step 3). We now want to find values for [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex], so we substitute [latex]y_p[\/latex] into the differential equation. We have [latex]y_p^\\prime(x)=2Ax+B[\/latex] and [latex]y_p^{\\prime\\prime}(x)=2A[\/latex], so we want to find values of [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] such that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y^{\\prime\\prime}-2y^\\prime+5y&=10x^2-3x-3 \\\\  2A-2(2Ax+b)+5(Ax^2+Bx+C)&=10x^2-3x-3 \\\\  5Ax^2+(5B-4A)x+(5C-2B+2A)&=10x^2-3x-3  \\end{aligned}[\/latex].<\/p>\n<p>Therefore,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  5A&=10 \\\\  5B-4A&=-3 \\\\  5C-2B+2A&=-3  \\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>This gives [latex]A=2[\/latex], [latex]B=1[\/latex], and [latex]C=-1[\/latex], so [latex]y_p(x)=2x^2+x-1[\/latex] (step 4). Putting everything together, we have the general solution<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div style=\"text-align: center;\">[latex]y(x)=c_1e^x\\cos2x+c_2e^x\\sin2x+2x^2+x-1[\/latex].<\/div>\n<\/li>\n<li>The complementary equation is [latex]y^{\\prime\\prime}-3y^\\prime=0[\/latex], which has the general solution [latex]c_1e^{3t}+c_2[\/latex] (step 1). Based on the form [latex]r(t)=-12t[\/latex], our initial guess for the particular solution is [latex]y_p(t)=At+B[\/latex] (step 2). However, we see that the constant term in this guess solves the complementary equation, so we must multiply by [latex]t[\/latex], which gives a new guess: [latex]y_P(t)=At^2+Bt[\/latex] (step 3). Checking this new guess, we see that none of the terms in [latex]y_p(t)[\/latex] solve the complementary equation, so this is a valid guess (step 3 again). We now want to find values for [latex]A[\/latex] and [latex]B[\/latex], so we substitute [latex]y_p[\/latex] into the differential equation. We have [latex]y_p^\\prime(t)=2At+B[\/latex] and [latex]y_p^{\\prime\\prime}(t)=2A[\/latex], so we want to find values of [latex]A[\/latex] and [latex]B[\/latex] such that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y^{\\prime\\prime}-3y^\\prime&=-12t \\\\  2A-3(2At+B)=-12t \\\\  -6At+(2A-3B)&=-12t  \\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Therefore,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  -6A&=-12 \\\\  2A-3B&=0  \\end{aligned}[\/latex].<\/p>\n<p style=\"text-align: center;\"><span data-type=\"newline\"><br \/>\n<\/span>This gives [latex]A=2[\/latex] and [latex]b=4\/3[\/latex], so [latex]y_p(t)=2t^2+(4\/3)t[\/latex] (step 4). Putting everything together, we have the general solution<br \/>\n[latex]y(t)=c_1e^{3t}+c_2+2t^2+\\frac43t[\/latex].<\/p>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-id1170571715424\">Find the general solution to the following differential equations.<\/p>\n<ol id=\"fs-id1170571715427\" type=\"a\">\n<li>[latex]y^{\\prime\\prime}-5y^\\prime+4y=3e^x[\/latex]<\/li>\n<li>[latex]y^{\\prime\\prime}+y^\\prime-6y=52\\cos2t[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q597293813\">Show Solution<\/span><\/p>\n<div id=\"q597293813\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0[latex]y(x)=c_1e^{4x}+c_2e^x-xe^x[\/latex]<\/p>\n<p>b.\u00a0[latex]y(t)=c_1e^t+c_2te^t+te^t\\ln|t|[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Variation of Parameters<\/h2>\n<p id=\"fs-id1170572593832\">Sometimes, [latex]r(x)[\/latex] is not a combination of polynomials, exponentials, or sines and cosines. When this is the case, the method of undetermined coefficients does not work, and we have to use another approach to find a particular solution to the differential equation. We use an approach called the\u00a0<span id=\"e2cb0fd1-6cb0-486c-93d7-2edd081caff6_term311\" data-type=\"term\">method of variation of parameters<\/span>.<\/p>\n<p id=\"fs-id1170572593856\">To simplify our calculations a little, we are going to divide the differential equation through by [latex]a[\/latex], so we have a leading coefficient of 1. Then the differential equation has the form<\/p>\n<div id=\"fs-id1170572604296\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div style=\"text-align: center;\">[latex]y^{\\prime\\prime}+py^\\prime+qy=r(x)[\/latex],<\/div>\n<div><\/div>\n<p>where [latex]p[\/latex] and [latex]q[\/latex] are constants.<\/p>\n<\/div>\n<p id=\"fs-id1170572604354\">If the general solution to the complementary equation is given by [latex]c_1y_1(x)+c_2y_2(x)[\/latex], we are going to look for a particular solution of the form [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex]. In this case, we use the two linearly independent solutions to the complementary equation to form our particular solution. However, we are assuming the coefficients are functions of [latex]x[\/latex], rather than constants. We want to find functions [latex]u(x)[\/latex] and [latex]v(x)[\/latex] such that [latex]y_p(x)[\/latex] satisfies the differential equation. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y_p&=uy_1+vy_2 \\\\  y_p^\\prime&=u^\\prime+uy_1^\\prime+v^\\prime y_2+vy_2^\\prime \\\\  y_p^{\\prime\\prime}&=(u^\\prime y_1+v^\\prime y_2)^\\prime+u^\\prime y_1^\\prime+uy_1^{\\prime\\prime}+v^\\prime y_2^\\prime+vy_2^{\\prime\\prime}  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170571579727\">Substituting into the differential equation, we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y_p^{\\prime\\prime}+py_p^\\prime+qy_p&=[(u^\\prime y_1+v^\\prime y_2)+u^\\prime y_1^\\prime+uy_1^{\\prime\\prime}+v^\\prime y_2^\\prime+vy_2^{\\prime\\prime}]+p[u^\\prime y_1+uy_1^\\prime+v^\\prime y_2+vy_2^\\prime]+q[uy_1+vy_2] \\\\  &=u[y_1^{\\prime\\prime}+py_1^\\prime+qy_1]+v[y_2^{\\prime\\prime}+py_2^\\prime+qy_2]+(u^\\prime y_1+v^\\prime y_2)^\\prime+p(u^\\prime y_1+v^\\prime y_2)+(u^\\prime y_1^\\prime+v^\\prime y_2^\\prime)  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170572185030\">Note that [latex]y_1[\/latex] and [latex]y_2[\/latex] are solutions to the complementary equation, so the first two terms are zero. Thus, we have<\/p>\n<p style=\"text-align: center;\">[latex](u^\\prime y_1+v^\\prime y_2)^\\prime+p(u^\\prime y_1+v^\\prime y_2)+(u^\\prime y_1^\\prime+v^\\prime y_2^\\prime)=r(x)[\/latex].<\/p>\n<p id=\"fs-id1170571670359\">If we simplify this equation by imposing the additional condition [latex]u^\\prime y_1+v^\\prime y_2=0[\/latex], the first two terms are zero, and this reduces to [latex]u^\\prime y_1^\\prime+v^\\prime y_2=r(x)[\/latex]. So, with this additional condition, we have a system of two equations in two unknowns:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  u^\\prime y_1+v^\\prime y_2&=0 \\\\  u^\\prime y_1^\\prime+v^\\prime y_2^\\prime&=r(x)  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170571613606\">Solving this system gives us [latex]u'[\/latex] and [latex]v'[\/latex], which we can integrate to find [latex]u[\/latex] and\u00a0[latex]v[\/latex].<\/p>\n<p id=\"fs-id1170571613638\">Then, [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex] is a particular solution to the differential equation. Solving this system of equations is sometimes challenging, so let\u2019s take this opportunity to review Cramer\u2019s rule, which allows us to solve the system of equations using determinants.<\/p>\n<div id=\"fs-id1170572548183\" class=\"ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">rule: cramer&#8217;s rule<\/h3>\n<hr \/>\n<p id=\"fs-id1170572548189\">The system of equations<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  a_1z_1+b_1z_2=r_1 \\\\  a_2z_1+b_2z_2&=r_2  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1170572387640\">has a unique solution if and only if the determinant of the coefficients is not zero. In this case, the solution is given by<\/p>\n<p style=\"text-align: center;\">[latex]z_1=\\frac{\\begin{vmatrix}r_1&b_1 \\\\ r_2&b_2\\end{vmatrix}}{\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}}[\/latex] and\u00a0[latex]z_2=\\frac{\\begin{vmatrix}a_1&r_1 \\\\ a_2&r_2\\end{vmatrix}}{\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: using cramer&#8217;s rule<\/h3>\n<p id=\"fs-id1170572504528\">Use Cramer\u2019s rule to solve the following system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x^2z_1+2xz_2&=0 \\\\  z_1-3x^2z_2&=2x  \\end{aligned}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q923921637\">Show Solution<\/span><\/p>\n<div id=\"q923921637\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170572116382\">We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  a_1(x)&=x^2 \\\\  a_2(x)&=1 \\\\  b_1(x)&=2x \\\\  b_2(x)&=-3x^2 \\\\  r_1(x)&=0 \\\\  r_2(x)&=2x  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170571748971\">Then,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}=\\begin{vmatrix}x^2&2x \\\\ 1&-3x^2\\end{vmatrix}=-3x^4-2x[\/latex]<\/p>\n<p id=\"fs-id1170572088607\">and<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{vmatrix}r_1&b_1 \\\\ r_2&b_2\\end{vmatrix}=\\begin{vmatrix}0&2x \\\\ 2x&-3x^2\\end{vmatrix}=0-4x^2=-4x^2[\/latex].<\/p>\n<p id=\"fs-id1170571688223\">Thus,<\/p>\n<p style=\"text-align: center;\">[latex]z_1=\\frac{\\begin{vmatrix}r_1&b_1 \\\\ r_2&b_2\\end{vmatrix}}{\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}}=\\frac{-4x^2}{-3x^4-2x}=\\frac{4x}{3x^3+2}[\/latex].<\/p>\n<p id=\"fs-id1170571611310\">In addition,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{vmatrix}a_1&r_1 \\\\ a_2&r_2\\end{vmatrix}=\\begin{vmatrix}x^2&0 \\\\1&2x\\end{vmatrix}=2x^3-0=2x^3[\/latex].<\/p>\n<p id=\"fs-id1170572330219\">Thus,<\/p>\n<p style=\"text-align: center;\">[latex]z_2=\\frac{\\begin{vmatrix}a_1&r_1 \\\\ a_2&r_2\\end{vmatrix}}{\\begin{vmatrix}a_1&b_1 \\\\ a_2&b_2\\end{vmatrix}}=\\frac{2x^3}{-3x^4-2x}=\\frac{-2x^2}{3x^3+2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-id1170571553660\">Use Cramer\u2019s rule to solve the following system of equations.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  2xz_1-3z_2&=0 \\\\  x^2z_1+4xz_2&=x+1  \\end{aligned}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q734571894\">Show Solution<\/span><\/p>\n<div id=\"q734571894\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]z_1=\\frac{3x+3}{11x^2}, \\ z_2\\frac{2x+2}{11x}[\/latex]<\/p>\n<\/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: method of variation of parameters<\/h3>\n<ol id=\"fs-id1170571789088\" type=\"1\">\n<li>Solve the complementary equation and write down the general solution<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571789098\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]c_1y_1(x)+c_2y_2(x)[\/latex].<\/p>\n<\/div>\n<\/li>\n<li>Use Cramer\u2019s rule or another suitable technique to find functions [latex]u^\\prime(x)[\/latex] and [latex]v^\\prime(x)[\/latex] satisfying<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  u^\\prime y_1+v^\\prime y_2&=0 \\\\  u^\\prime y_1^\\prime+v^\\prime y_2^\\prime&=r(x)  \\end{aligned}[\/latex]<\/p>\n<\/li>\n<li>Integrate [latex]u^\\prime[\/latex] and [latex]v^\\prime[\/latex] to find [latex]u(x)[\/latex] and [latex]v(x)[\/latex]. Then, [latex]y_p(x)=u(x)y_1(x)+v(x)y_2(x)[\/latex] is a particular solution to the equation.<\/li>\n<li>Add the general solution to the complementary equation and the particular solution found in step 3 to obtain the general solution to the nonhomogeneous equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>example: using the method of variation of parameters<\/h3>\n<p id=\"fs-id1170571823253\">Find the general solution to the following differential equations.<\/p>\n<ol id=\"fs-id1170571823257\" type=\"a\">\n<li>[latex]y^{\\prime\\prime}-2y^\\prime+y=\\frac{e^t}{t^2}[\/latex]<\/li>\n<li>[latex]y^{\\prime\\prime}+y=3\\sin^2x[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q982348083\">Show Solution<\/span><\/p>\n<div id=\"q982348083\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170572589220\" type=\"a\">\n<li>The complementary equation is [latex]y^{\\prime\\prime}-2y^\\prime+y=0[\/latex] with associated general solution [latex]c_1e^t+c_2te^t[\/latex]. Therefore, [latex]y_1(t)=e^t[\/latex] and [latex]y_2(t)=te^t[\/latex]. Calculating the derivatives, we get [latex]y_1^\\prime(t)=e^t[\/latex] and [latex]y_2^\\prime(t)=e^t+te^t[\/latex] (step 1). Then, we want to find functions [latex]u'(t)[\/latex] and [latex]v'(t)[\/latex] so that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  u^\\prime e^t+v^\\prime te^t&=0 \\\\  u^\\prime e^t+v^\\prime(e^t+te^t)=\\frac{e^t}{t^2}  \\end{aligned}[\/latex]<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Applying Cramer\u2019s rule, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]u^\\prime=\\frac{\\begin{vmatrix}0&te^t \\\\ \\frac{e^t}{t^2}&e^t+te^t\\end{vmatrix}}{\\begin{vmatrix}e^t&te^t \\\\e^t&e^t+te^t\\end{vmatrix}}=\\frac{0-te^t\\left(\\frac{e^t}{t^2}\\right)}{e^t(e^t+te^t)-e^tte^t}=\\frac{-\\frac{e^{2t}}{t}}{e^{2t}}=-\\frac1t[\/latex]<\/p>\n<p>and<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572296671\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]v^\\prime=\\frac{\\begin{vmatrix}e^t&0\\\\e^t&\\frac{e^t}{t^2}\\end{vmatrix}}{\\begin{vmatrix}e^t&te^t\\\\e^t&e^t+te^t\\end{vmatrix}}=\\frac{e^t\\left(\\frac{e^t}{t^2}\\right)}{e^{2t}}=\\frac{1}{t^2}[\/latex] (step 2).<\/p>\n<\/div>\n<p>Integrating, we get<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  u&=-\\displaystyle\\int\\frac1t \\ dt=-\\ln|t| \\\\  v&=\\displaystyle\\int\\frac{1}{t^2}dt=-\\frac{1}{t}\\text{(step 3)}.  \\end{aligned}[\/latex]<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Then we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y_p&=-e^t\\ln|t|-\\frac{1}{t}te^t \\\\  &=-e^t\\ln|t|-e^t \\text{(step 4)}.  \\end{aligned}[\/latex]<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>The [latex]e^{t}[\/latex] term is a solution to the complementary equation, so we don\u2019t need to carry that term into our general solution explicitly. The general solution is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]y(t)=c_1e^t+c_2te^t-e^t\\ln|t|[\/latex] (step 5).<\/p>\n<\/li>\n<li>The complementary equation is [latex]y^{\\prime\\prime}+y+0[\/latex] with associated general solution [latex]c_1\\cos x+c_2\\sin x[\/latex]. So, [latex]y_1(x)=\\cos x[\/latex] and [latex]y_2(x)=\\sin x[\/latex] (step 1). Then, we want to find functions [latex]u'(x)[\/latex] and [latex]v'(x)[\/latex] such that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}<\/p>\n<p>  u^\\prime\\cos x+v^\\prime\\sin x&=0 \\\\  -y^\\prime\\sin x+v^\\prime\\cos x&=3\\sin^2x  \\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Applying Cramer\u2019s rule, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]u^\\prime=\\frac{\\begin{vmatrix}0&\\sin x \\\\3\\sin^2x&\\cos x\\end{vmatrix}}{\\begin{vmatrix}\\cos x&\\sin x \\\\ -\\sin x&\\cos x\\end{vmatrix}}=\\frac{0-3\\sin^3x}{\\cos^2x+\\sin^2x}=-3\\sin^3x[\/latex]<\/p>\n<p>and<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]v^\\prime=\\frac{\\begin{vmatrix}\\cos x&0\\\\ -\\sin x&3\\sin^2x\\end{vmatrix}}{\\begin{vmatrix}\\cos x&\\sin x \\\\-\\sin x&\\cos x\\end{vmatrix}}=\\frac{3\\sin^2x\\cos x}{1}=3\\sin^2x\\cos x[\/latex] (step 2).<\/p>\n<p>Integrating first to find [latex]u[\/latex], we get<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]u=\\displaystyle\\int-3\\sin^3xdx=-3\\left[-\\frac13\\sin^2x\\cos x+\\frac23\\displaystyle\\int\\sin xdx\\right]=\\sin^2x\\cos x+2\\cos x[\/latex] (step 3).<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Now, we integrate to find [latex]v[\/latex]. Using substitution (with [latex]w=\\sin x[\/latex]), we get<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]v=\\displaystyle\\int3\\sin^2x\\cos xdx=\\displaystyle\\int3w^2dw=w^3=\\sin^3x[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Then,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y_p&=(\\sin^2x\\cos x+2\\cos x)\\cos x+(\\sin^3x)\\sin x \\\\  &=\\sin^2x\\cos^2x+2\\cos^2x+\\sin^4x \\\\  &=2\\cos^2x+\\sin^2x(\\cos^2x+\\sin^2x) \\ \\ \\ \\ (step 4). \\\\  &=2\\cos^2x+\\sin^2 \\\\  &=\\cos^2x+1  \\end{aligned}[\/latex]<\/p>\n<p>The general solution is<\/p>\n<p style=\"text-align: center;\">[latex]y(x)=c_1\\cos x+c_2\\sin x+1+\\cos^2x[\/latex] (step 5).<\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-id1170572554234\">Find the general solution to the following differential equations.<\/p>\n<ol id=\"fs-id1170572554237\" type=\"a\">\n<li>[latex]y^{\\prime\\prime}+y=\\sec x[\/latex]<\/li>\n<li>[latex]x^{\\prime\\prime}-2x^\\prime+x=\\frac{e^t}t[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q687247112\">Show Solution<\/span><\/p>\n<div id=\"q687247112\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0[latex]y(x)=c_1\\cos x+c_2\\sin x+\\cos x\\ln|\\cos x|+x\\sin x[\/latex]<\/p>\n<p>b.\u00a0[latex]x(t)=c_1e^t+c_2te^t+te^t\\ln|t|[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following videos 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=8250341&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=8N2A8lhcxPI&amp;video_target=tpm-plugin-y9gjmauh-8N2A8lhcxPI\" 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\/CP7.14a_transcript.html\">transcript for \u201cCP 7.14a\u201d here (opens in new window).<\/a><\/div>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250340&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=HCMJodmpYKg&amp;video_target=tpm-plugin-4khktbip-HCMJodmpYKg\" 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\/CP7.14b_transcript.html\">transcript for \u201cCP 7.14b\u201d here (opens in new window).<\/a><\/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-1330\">\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 7.11. <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><li>CP 7.14a. <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><li>CP 7.14b. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":428269,"menu_order":8,"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 7.11\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 7.14a\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 7.14b\",\"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-1330","chapter","type-chapter","status-publish","hentry"],"part":25,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1330","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/428269"}],"version-history":[{"count":61,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1330\/revisions"}],"predecessor-version":[{"id":6403,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1330\/revisions\/6403"}],"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\/1330\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1330"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1330"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1330"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1330"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}