{"id":1639,"date":"2021-07-23T16:36:59","date_gmt":"2021-07-23T16:36:59","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1639"},"modified":"2022-03-21T22:58:22","modified_gmt":"2022-03-21T22:58:22","slug":"first-order-differential-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/first-order-differential-equations\/","title":{"raw":"First-Order Differential Equations","rendered":"First-Order Differential Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Write a first-order linear differential equation in standard form<\/li>\r\n \t<li>Find an integrating factor and use it to solve a first-order linear differential equation<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1170572247743\">See the example on the introduction page for a first-order linear differential equation.<\/p>\r\n\r\n<div id=\"fs-id1170572225198\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170572222876\">A first-order differential equation is <span data-type=\"term\">linear<\/span> if it can be written in the form<\/p>\r\n\r\n<div id=\"fs-id1170572251046\" style=\"text-align: center;\" data-type=\"equation\">[latex]a\\left(x\\right){y}^{\\prime }+b\\left(x\\right)y=c\\left(x\\right)[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572168208\">where [latex]a\\left(x\\right),b\\left(x\\right)[\/latex], and [latex]c\\left(x\\right)[\/latex] are arbitrary functions of [latex]x[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170571696965\">Remember that the unknown function [latex]y[\/latex] depends on the variable [latex]x[\/latex]; that is, [latex]x[\/latex] is the independent variable and [latex]y[\/latex] is the dependent variable. Some examples of first-order linear differential equations are<\/p>\r\n\r\n<div id=\"fs-id1170572480509\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\left(3{x}^{2}-4\\right)y^{\\prime} +\\left(x - 3\\right)y&amp; =\\hfill &amp; \\sin{x}\\hfill \\\\ \\hfill \\left(\\sin{x}\\right)y^{\\prime} -\\left(\\cos{x}\\right)y&amp; =\\hfill &amp; \\cot{x}\\hfill \\\\ \\hfill 4xy^{\\prime} +\\left(3\\text{ln}x\\right)y&amp; =\\hfill &amp; {x}^{3}-4x.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572498765\">Examples of first-order nonlinear differential equations include<\/p>\r\n\r\n<div id=\"fs-id1170572208851\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\left(y^{\\prime} \\right)}^{4}-{\\left(y^{\\prime} \\right)}^{3}&amp; =\\hfill &amp; \\left(3x - 2\\right)\\left(y+4\\right)\\hfill \\\\ \\hfill 4y^{\\prime} +3{y}^{3}&amp; =\\hfill &amp; 4x - 5\\hfill \\\\ \\hfill {\\left(y^{\\prime} \\right)}^{2}&amp; =\\hfill &amp; \\sin{y}+\\cos{x}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572169162\">These equations are nonlinear because of terms like [latex]{\\left({y}^{\\prime }\\right)}^{4},{y}^{3}[\/latex], etc. Due to these terms, it is impossible to put these equations into the same form as the definition.<\/p>\r\n\r\n<section id=\"fs-id1170572228237\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Standard Form<\/h2>\r\n<p id=\"fs-id1170572224892\">Consider the differential equation<\/p>\r\n\r\n<div id=\"fs-id1170572228143\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left(3{x}^{2}-4\\right){y}^{\\prime }+\\left(x - 3\\right)y=\\sin{x}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572554740\">Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of [latex]{y}^{\\prime }[\/latex] be equal to [latex]1[\/latex]. To make this happen, we divide both sides by [latex]3{x}^{2}-4[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1170572210607\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }+\\left(\\frac{x - 3}{3{x}^{2}-4}\\right)y=\\frac{\\sin{x}}{3{x}^{2}-4}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572489215\">This is called the <span data-type=\"term\">standard form<\/span> of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to the definition, we can divide both sides of the equation by [latex]a\\left(x\\right)[\/latex]. This leads to the equation<\/p>\r\n\r\n<div id=\"fs-id1170572151745\" style=\"text-align: center;\" data-type=\"equation\">[latex]{y}^{\\prime }+\\frac{b\\left(x\\right)}{a\\left(x\\right)}y=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572363395\">Now define [latex]p\\left(x\\right)=\\frac{b\\left(x\\right)}{a\\left(x\\right)}[\/latex] and [latex]q\\left(x\\right)=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex]. Then the definition becomes<\/p>\r\n\r\n<div id=\"fs-id1170571712108\" style=\"text-align: center;\" data-type=\"equation\">[latex]{y}^{\\prime }+p\\left(x\\right)y=q\\left(x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572087796\">We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.<\/p>\r\n\r\n<div id=\"fs-id1170572447597\" data-type=\"example\">\r\n<div id=\"fs-id1170572141228\" data-type=\"exercise\">\r\n<div id=\"fs-id1170571623194\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing First-Order Linear Equations in Standard Form<\/h3>\r\n<div id=\"fs-id1170571623194\" data-type=\"problem\">\r\n<p id=\"fs-id1170571657281\">Put each of the following first-order linear differential equations into standard form. Identify [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex] for each equation.<\/p>\r\n\r\n<ol id=\"fs-id1170572099685\" type=\"a\">\r\n \t<li>[latex]y^{\\prime} =3x - 4y[\/latex]<\/li>\r\n \t<li>[latex]\\frac{3xy^{\\prime} }{4y - 3}=2[\/latex] (here [latex]x&gt;0[\/latex])<\/li>\r\n \t<li>[latex]y=3y^{\\prime} -4{x}^{2}+5[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1170572453259\" data-type=\"solution\">\r\n<ol id=\"fs-id1170572601312\" type=\"a\">\r\n \t<li>Add [latex]4y[\/latex] to both sides:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571602108\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} +4y=3x[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nIn this equation, [latex]p\\left(x\\right)=4[\/latex] and [latex]q\\left(x\\right)=3x[\/latex].<\/li>\r\n \t<li>Multiply both sides by [latex]4y - 3[\/latex], then subtract [latex]8y[\/latex] from each side:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571712896\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{3xy^{\\prime} }{4y - 3}&amp; =\\hfill &amp; 2\\hfill \\\\ \\hfill 3xy^{\\prime} &amp; =\\hfill &amp; 2\\left(4y - 3\\right)\\hfill \\\\ \\hfill 3xy^{\\prime} &amp; =\\hfill &amp; 8y - 6\\hfill \\\\ \\hfill 3xy^{\\prime} -8y&amp; =\\hfill &amp; -6.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nFinally, divide both sides by [latex]3x[\/latex] to make the coefficient of [latex]y^{\\prime} [\/latex] equal to [latex]1\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"eip-id1170800554716\" style=\"text-align: center;\" data-type=\"equation\">[latex]y^{\\prime} -\\frac{8}{3x}y=-\\frac{2}{3x}[\/latex].<\/div>\r\nThis is allowable because in the original statement of this problem we assumed that [latex]x&gt;0[\/latex]. (If [latex]x=0[\/latex] then the original equation becomes [latex]0=2[\/latex], which is clearly a false statement.)<span data-type=\"newline\">\r\n<\/span>\r\nIn this equation, [latex]p\\left(x\\right)=-\\frac{8}{3x}[\/latex] and [latex]q\\left(x\\right)=-\\frac{2}{3x}[\/latex].<\/li>\r\n \t<li>Subtract [latex]y[\/latex] from each side and add [latex]4{x}^{2}-5\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572373157\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]3y^{\\prime} -y=4{x}^{2}-5[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nNext divide both sides by [latex]3\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571637556\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} -\\frac{1}{3}y=\\frac{4}{3}{x}^{2}-\\frac{5}{3}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nIn this equation, [latex]p\\left(x\\right)=-\\frac{1}{3}[\/latex] and [latex]q\\left(x\\right)=\\frac{4}{3}{x}^{2}-\\frac{5}{3}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Writing First-Order Linear Equations in Standard Form.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6722828&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=SrNV1z_b_4o&amp;video_target=tpm-plugin-2sn3q61c-SrNV1z_b_4o\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.5.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for \"4.5.1\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1170572292285\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170572331245\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572331248\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170572331248\" data-type=\"problem\">\r\n\r\nPut the equation [latex]\\frac{\\left(x+3\\right)y^{\\prime} }{2x - 3y - 4}=5[\/latex] into standard form and identify [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex].\r\n\r\n[reveal-answer q=\"44558897\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558897\"]\r\n<div id=\"fs-id1170572241407\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170572484113\">Multiply both sides by the common denominator, then collect all terms involving [latex]y[\/latex] on one side.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1170572452404\" data-type=\"solution\">\r\n<p id=\"fs-id1170572169346\">[latex]y^{\\prime} +\\frac{15}{x+3}y=\\frac{10x - 20}{x+3};p\\left(x\\right)=\\frac{15}{x+3}[\/latex] and [latex]q\\left(x\\right)=\\frac{10x - 20}{x+3}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]169329[\/ohm_question]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1170572552065\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Integrating Factors<\/h2>\r\n<p id=\"fs-id1170572337196\">We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:<\/p>\r\n\r\n<div id=\"fs-id1170572130093\" style=\"text-align: center;\" data-type=\"equation\">[latex]y^{\\prime} +p\\left(x\\right)y=q\\left(x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572550947\">The first term on the left-hand side of [latex]{y}^{\\prime }+\\frac{b\\left(x\\right)}{a\\left(x\\right)}y=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex] is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the power rule from the Differentiation Rules section. If we multiply [latex]{y}^{\\prime }+p\\left(x\\right)y=q\\left(x\\right)[\/latex] by a yet-to-be-determined function [latex]\\mu \\left(x\\right)[\/latex], then the equation becomes<\/p>\r\n\r\n<div id=\"fs-id1170572449739\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\mu \\left(x\\right){y}^{\\prime }+\\mu \\left(x\\right)p\\left(x\\right)y=\\mu \\left(x\\right)q\\left(x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572168782\">The left-hand side of [latex]y^{\\prime} +p\\left(x\\right)y=q\\left(x\\right)[\/latex]\u00a0can be matched perfectly to the product rule:<\/p>\r\n\r\n<div id=\"fs-id1170571714376\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dx}\\left[f\\left(x\\right)g\\left(x\\right)\\right]={f}^{\\prime }\\left(x\\right)g\\left(x\\right)+f\\left(x\\right){g}^{\\prime }\\left(x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572482443\">Matching term by term gives [latex]y=f\\left(x\\right),g\\left(x\\right)=\\mu \\left(x\\right)[\/latex], and [latex]{g}^{\\prime }\\left(x\\right)=\\mu \\left(x\\right)p\\left(x\\right)[\/latex]. Taking the derivative of [latex]g\\left(x\\right)=\\mu \\left(x\\right)[\/latex] and setting it equal to the right-hand side of [latex]{g}^{\\prime }\\left(x\\right)=\\mu \\left(x\\right)p\\left(x\\right)[\/latex] leads to<\/p>\r\n\r\n<div id=\"fs-id1170571832251\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\mu }^{\\prime }\\left(x\\right)=\\mu \\left(x\\right)p\\left(x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572397269\">This is a first-order, separable differential equation for [latex]\\mu \\left(x\\right)[\/latex]. We know [latex]p\\left(x\\right)[\/latex] because it appears in the differential equation we are solving. Separating variables and integrating yields<\/p>\r\n\r\n<div id=\"fs-id1170571564354\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{{\\mu }^{\\prime }\\left(x\\right)}{\\mu \\left(x\\right)}&amp; =\\hfill &amp; p\\left(x\\right)\\hfill \\\\ \\hfill {\\displaystyle\\int \\frac{{\\mu }^{\\prime }\\left(x\\right)}{\\mu \\left(x\\right)}dx}&amp; =\\hfill &amp; {\\displaystyle\\int p\\left(x\\right)dx}\\hfill \\\\ \\hfill \\text{ln}|\\mu \\left(x\\right)|&amp; =\\hfill &amp; {\\displaystyle\\int p\\left(x\\right)dx+C}\\hfill \\\\ \\hfill {e}^{\\text{ln}|\\mu \\left(x\\right)|}&amp; =\\hfill &amp; {e}^{\\displaystyle\\int p\\left(x\\right)dx+C}\\hfill \\\\ \\hfill |\\mu \\left(x\\right)|&amp; =\\hfill &amp; {C}_{1}{e}^{\\displaystyle\\int p\\left(x\\right)dx}\\hfill \\\\ \\hfill \\mu \\left(x\\right)&amp; =\\hfill &amp; {C}_{2}{e}^{\\displaystyle\\int p\\left(x\\right)dx}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571593416\">Here [latex]{C}_{2}[\/latex] can be an arbitrary (positive or negative) constant. This leads to a general method for solving a first-order linear differential equation. We first multiply both sides of [latex]{y}^{\\prime }+p\\left(x\\right)y=q\\left(x\\right)[\/latex]\u00a0by the <span data-type=\"term\">integrating factor<\/span> [latex]\\mu \\left(x\\right)[\/latex]. This gives<\/p>\r\n\r\n<div id=\"fs-id1170571789027\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\mu \\left(x\\right){y}^{\\prime }+\\mu \\left(x\\right)p\\left(x\\right)y=\\mu \\left(x\\right)q\\left(x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571597633\">The left-hand side of the above equation can be rewritten as [latex]\\frac{d}{dx}\\left(\\mu \\left(x\\right)y\\right)[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1170572554001\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\frac{d}{dx}\\left(\\mu \\left(x\\right)y\\right)=\\mu \\left(x\\right)q\\left(x\\right)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571814541\">Next integrate both sides with respect to [latex]x[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1170572546914\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dx}\\left(\\mu \\left(x\\right)y\\right)dx}&amp; =\\hfill &amp; {\\displaystyle\\int \\mu \\left(x\\right)q\\left(x\\right)dx}\\hfill \\\\ \\hfill \\mu \\left(x\\right)y&amp; =\\hfill &amp; {\\displaystyle\\int \\mu \\left(x\\right)q\\left(x\\right)dx}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571699423\">Divide both sides\u00a0by [latex]\\mu \\left(x\\right)\\text{:}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1170571681325\" style=\"text-align: center;\" data-type=\"equation\">[latex]y=\\dfrac{1}{\\mu \\left(x\\right)}\\left[\\displaystyle\\int \\mu \\left(x\\right)q\\left(x\\right)dx+C\\right][\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170572290464\">Since [latex]\\mu \\left(x\\right)[\/latex] was previously calculated, we are now finished. An important note about the integrating constant [latex]C\\text{:}[\/latex] It may seem that we are inconsistent in the usage of the integrating constant. However, the integral involving [latex]p\\left(x\\right)[\/latex] is necessary in order to find an integrating factor for [latex]{y}^{\\prime }+\\frac{b\\left(x\\right)}{a\\left(x\\right)}y=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex]. Only one integrating factor is needed in order to solve the equation; therefore, it is safe to assign a value for [latex]C[\/latex] for this integral. We chose [latex]C=0[\/latex]. When calculating the integral inside the brackets in [latex]\\frac{d}{dx}\\left(\\mu \\left(x\\right)y\\right)=\\mu \\left(x\\right)q\\left(x\\right)[\/latex], it is necessary to keep our options open for the value of the integrating constant, because our goal is to find a general family of solutions to [latex]{y}^{\\prime }+\\frac{b\\left(x\\right)}{a\\left(x\\right)}y=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex]. This integrating factor guarantees just that.<\/p>\r\n\r\n<div id=\"fs-id1170572628487\" class=\"problem-solving\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3 data-type=\"title\">Problem-Solving Strategy: Solving a First-order Linear Differential Equation<\/h3>\r\n<ol id=\"fs-id1170571748988\" type=\"1\">\r\n \t<li>Put the equation into standard form and identify [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex].<\/li>\r\n \t<li>Calculate the integrating factor [latex]\\mu \\left(x\\right)={e}^{\\displaystyle\\int p\\left(x\\right)dx}[\/latex].<\/li>\r\n \t<li>Multiply both sides of the differential equation by [latex]\\mu \\left(x\\right)[\/latex].<\/li>\r\n \t<li>Integrate both sides of the equation obtained in step [latex]3[\/latex], and divide both sides by [latex]\\mu \\left(x\\right)[\/latex].<\/li>\r\n \t<li>If there is an initial condition, determine the value of [latex]C[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1170572539811\" data-type=\"example\">\r\n<div id=\"fs-id1170572321365\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572321367\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Solving a First-order Linear Equation<\/h3>\r\n<div id=\"fs-id1170572321367\" data-type=\"problem\">\r\n<p id=\"fs-id1170572572214\">Find a general solution for the differential equation [latex]xy^{\\prime} +3y=4{x}^{2}-3x[\/latex]. Assume [latex]x&gt;0[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1170572622201\" data-type=\"solution\">\r\n<ol id=\"fs-id1170572309310\" type=\"1\">\r\n \t<li>To put this differential equation into standard form, divide both sides by [latex]x\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572370908\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} +\\frac{3}{x}y=4x - 3[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore [latex]p\\left(x\\right)=\\frac{3}{x}[\/latex] and [latex]q\\left(x\\right)=4x - 3[\/latex].<\/li>\r\n \t<li>The integrating factor is [latex]\\mu \\left(x\\right)={e}^{\\displaystyle\\int \\left(\\frac{3}{x}\\right)dx}={e}^{3\\text{ln}x}={x}^{3}[\/latex].<\/li>\r\n \t<li>Multiplying both sides of the differential equation by [latex]\\mu \\left(x\\right)[\/latex] gives us<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572116162\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {x}^{3}{y}^{\\prime }+{x}^{3}\\left(\\frac{3}{x}\\right)y&amp; =\\hfill &amp; {x}^{3}\\left(4x - 3\\right)\\hfill \\\\ \\hfill {x}^{3}{y}^{\\prime }+3{x}^{2}y&amp; =\\hfill &amp; 4{x}^{4}-3{x}^{3}\\hfill \\\\ \\hfill \\frac{d}{dx}\\left({x}^{3}y\\right)&amp; =\\hfill &amp; 4{x}^{4}-3{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Integrate both sides of the equation.<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572369225\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dx}\\left({x}^{3}y\\right)dx}&amp; =\\hfill &amp; {\\displaystyle\\int 4{x}^{4}-3{x}^{3}dx}\\hfill \\\\ \\hfill {x}^{3}y&amp; =\\hfill &amp; \\frac{4{x}^{5}}{5}-\\frac{3{x}^{4}}{4}+C\\hfill \\\\ \\hfill y&amp; =\\hfill &amp; \\frac{4{x}^{2}}{5}-\\frac{3x}{4}+C{x}^{-3}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>There is no initial value, so the problem is complete.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"fs-id1170572585959\" data-type=\"commentary\">\r\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\r\n<p id=\"fs-id1170572169556\">You may have noticed the condition that was imposed on the differential equation; namely, [latex]x&gt;0[\/latex]. For any nonzero value of [latex]C[\/latex], the general solution is not defined at [latex]x=0[\/latex]. Furthermore, when [latex]x&lt;0[\/latex], the integrating factor changes. The integrating factor is given by [latex]\\mu \\left(x\\right){y}^{\\prime }+\\mu \\left(x\\right)p\\left(x\\right)y=\\mu \\left(x\\right)q\\left(x\\right)[\/latex]\u00a0as [latex]f\\left(x\\right)={e}^{\\displaystyle\\int p\\left(x\\right)dx}[\/latex]. For this [latex]p\\left(x\\right)[\/latex] we get<\/p>\r\n\r\n<div id=\"fs-id1170572505442\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{e}^{\\displaystyle\\int p\\left(x\\right)dx=}{e}^{\\displaystyle\\int \\left(\\frac{3}{x}\\right)dx}={e}^{3\\text{ln}|x|}={|x|}^{3}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170571600638\">since [latex]x&lt;0[\/latex]. The behavior of the general solution changes at [latex]x=0[\/latex] largely due to the fact that [latex]p\\left(x\\right)[\/latex] is not defined there.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example:\u00a0Solving a First-Order Linear Equation.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/doLN3RtE264?controls=0&amp;start=0&amp;end=300&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.5.2_0to300_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.5.2\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1170571598375\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170572332858\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572332860\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170572332860\" data-type=\"problem\">\r\n<p id=\"fs-id1170572332863\">Find the general solution to the differential equation [latex]\\left(x - 2\\right)y^{\\prime} +y=3{x}^{2}+2x[\/latex]. Assume [latex]x&gt;2[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558894\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44558894\"]\r\n<div id=\"fs-id1170571690904\" data-type=\"commentary\" data-element-type=\"hint\">\r\n<p id=\"fs-id1170571679832\">Use the method outlined in the problem-solving strategy for first-order linear differential equations.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"44558895\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558895\"]\r\n<div id=\"fs-id1170571611044\" data-type=\"solution\">\r\n<p id=\"fs-id1170571655144\">[latex]y=\\frac{{x}^{3}+{x}^{2}+C}{x - 2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<p id=\"fs-id1170572611124\">Now we use the same strategy to find the solution to an initial-value problem.<\/p>\r\n\r\n<div id=\"fs-id1170572611128\" data-type=\"example\">\r\n<div id=\"fs-id1170572611130\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572643199\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: A First-order Linear Initial-Value Problem<\/h3>\r\n<div id=\"fs-id1170572643199\" data-type=\"problem\">\r\n<p id=\"fs-id1170572643204\">Solve the initial-value problem<\/p>\r\n\r\n<div id=\"fs-id1170571609231\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }+3y=2x - 1,y\\left(0\\right)=3[\/latex].<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558893\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558893\"]\r\n<div id=\"fs-id1170571653907\" data-type=\"solution\">\r\n<ol id=\"fs-id1170571653909\" type=\"1\">\r\n \t<li>This differential equation is already in standard form with [latex]p\\left(x\\right)=3[\/latex] and [latex]q\\left(x\\right)=2x - 1[\/latex].<\/li>\r\n \t<li>The integrating factor is [latex]\\mu \\left(x\\right)={e}^{\\displaystyle\\int 3dx}={e}^{3x}[\/latex].<\/li>\r\n \t<li>Multiplying both sides of the differential equation by [latex]\\mu \\left(x\\right)[\/latex] gives<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572481493\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {e}^{3x}{y}^{\\prime }+3{e}^{3x}y&amp; =\\hfill &amp; \\left(2x - 1\\right){e}^{3x}\\hfill \\\\ \\hfill \\frac{d}{dx}\\left[y{e}^{3x}\\right]&amp; =\\hfill &amp; \\left(2x - 1\\right){e}^{3x}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nIntegrate both sides of the equation:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571593388\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dx}\\left[y{e}^{3x}\\right]dx}&amp; =\\hfill &amp; {\\displaystyle\\int \\left(2x - 1\\right){e}^{3x}dx}\\hfill \\\\ \\hfill y{e}^{3x}&amp; =\\hfill &amp; \\frac{{e}^{3x}}{3}\\left(2x - 1\\right)-{\\displaystyle\\int \\frac{2}{3}{e}^{3x}dx}\\hfill \\\\ \\hfill y{e}^{3x}&amp; =\\hfill &amp; \\frac{{e}^{3x}\\left(2x - 1\\right)}{3}-\\frac{2{e}^{3x}}{9}+C\\hfill \\\\ \\hfill y&amp; =\\hfill &amp; \\frac{2x - 1}{3}-\\frac{2}{9}+C{e}^{-3x}\\hfill \\\\ \\hfill y&amp; =\\hfill &amp; \\frac{2x}{3}-\\frac{5}{9}+C{e}^{-3x}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;<\/li>\r\n \t<li>Now substitute [latex]x=0[\/latex] and [latex]y=3[\/latex] into the general solution and solve for [latex]C\\text{:}[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170571595426\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill y&amp; =\\hfill &amp; \\frac{2}{3}x-\\frac{5}{9}+C{e}^{-3x}\\hfill \\\\ \\hfill 3&amp; =\\hfill &amp; \\frac{2}{3}\\left(0\\right)-\\frac{5}{9}+C{e}^{-3\\left(0\\right)}\\hfill \\\\ \\hfill 3&amp; =\\hfill &amp; -\\frac{5}{9}+C\\hfill \\\\ \\hfill C&amp; =\\hfill &amp; \\frac{32}{9}.\\hfill \\end{array}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nTherefore the solution to the initial-value problem is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1170572592117\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\frac{2}{3}x-\\frac{5}{9}+\\frac{32}{9}{e}^{-3x}[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n<\/ol>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: A First-Order Linear Initial-Value Problem.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/doLN3RtE264?controls=0&amp;start=301&amp;end=630&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.5.2_301to630_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.5.2\" here (opens in new window)<\/a>.\r\n<div id=\"fs-id1170571640637\" class=\"checkpoint\" data-type=\"note\">\r\n<div id=\"fs-id1170572505498\" data-type=\"exercise\">\r\n<div id=\"fs-id1170572505501\" data-type=\"problem\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<div id=\"fs-id1170572505501\" data-type=\"problem\">\r\n<p id=\"fs-id1170572505503\">Solve the initial-value problem [latex]y^{\\prime} -2y=4x+3, y\\left(0\\right)=-2[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558892\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558892\"]\r\n<div id=\"fs-id1170572449330\" data-type=\"solution\">\r\n<p id=\"fs-id1170572449332\">[latex]y=-2x - 4+2{e}^{2x}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]169340[\/ohm_question]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1170571593637\" data-depth=\"1\"><\/section>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Write a first-order linear differential equation in standard form<\/li>\n<li>Find an integrating factor and use it to solve a first-order linear differential equation<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1170572247743\">See the example on the introduction page for a first-order linear differential equation.<\/p>\n<div id=\"fs-id1170572225198\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170572222876\">A first-order differential equation is <span data-type=\"term\">linear<\/span> if it can be written in the form<\/p>\n<div id=\"fs-id1170572251046\" style=\"text-align: center;\" data-type=\"equation\">[latex]a\\left(x\\right){y}^{\\prime }+b\\left(x\\right)y=c\\left(x\\right)[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572168208\">where [latex]a\\left(x\\right),b\\left(x\\right)[\/latex], and [latex]c\\left(x\\right)[\/latex] are arbitrary functions of [latex]x[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170571696965\">Remember that the unknown function [latex]y[\/latex] depends on the variable [latex]x[\/latex]; that is, [latex]x[\/latex] is the independent variable and [latex]y[\/latex] is the dependent variable. Some examples of first-order linear differential equations are<\/p>\n<div id=\"fs-id1170572480509\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\left(3{x}^{2}-4\\right)y^{\\prime} +\\left(x - 3\\right)y& =\\hfill & \\sin{x}\\hfill \\\\ \\hfill \\left(\\sin{x}\\right)y^{\\prime} -\\left(\\cos{x}\\right)y& =\\hfill & \\cot{x}\\hfill \\\\ \\hfill 4xy^{\\prime} +\\left(3\\text{ln}x\\right)y& =\\hfill & {x}^{3}-4x.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572498765\">Examples of first-order nonlinear differential equations include<\/p>\n<div id=\"fs-id1170572208851\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\left(y^{\\prime} \\right)}^{4}-{\\left(y^{\\prime} \\right)}^{3}& =\\hfill & \\left(3x - 2\\right)\\left(y+4\\right)\\hfill \\\\ \\hfill 4y^{\\prime} +3{y}^{3}& =\\hfill & 4x - 5\\hfill \\\\ \\hfill {\\left(y^{\\prime} \\right)}^{2}& =\\hfill & \\sin{y}+\\cos{x}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572169162\">These equations are nonlinear because of terms like [latex]{\\left({y}^{\\prime }\\right)}^{4},{y}^{3}[\/latex], etc. Due to these terms, it is impossible to put these equations into the same form as the definition.<\/p>\n<section id=\"fs-id1170572228237\" data-depth=\"1\">\n<h2 data-type=\"title\">Standard Form<\/h2>\n<p id=\"fs-id1170572224892\">Consider the differential equation<\/p>\n<div id=\"fs-id1170572228143\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\left(3{x}^{2}-4\\right){y}^{\\prime }+\\left(x - 3\\right)y=\\sin{x}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572554740\">Our main goal in this section is to derive a solution method for equations of this form. It is useful to have the coefficient of [latex]{y}^{\\prime }[\/latex] be equal to [latex]1[\/latex]. To make this happen, we divide both sides by [latex]3{x}^{2}-4[\/latex].<\/p>\n<div id=\"fs-id1170572210607\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }+\\left(\\frac{x - 3}{3{x}^{2}-4}\\right)y=\\frac{\\sin{x}}{3{x}^{2}-4}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572489215\">This is called the <span data-type=\"term\">standard form<\/span> of the differential equation. We will use it later when finding the solution to a general first-order linear differential equation. Returning to the definition, we can divide both sides of the equation by [latex]a\\left(x\\right)[\/latex]. This leads to the equation<\/p>\n<div id=\"fs-id1170572151745\" style=\"text-align: center;\" data-type=\"equation\">[latex]{y}^{\\prime }+\\frac{b\\left(x\\right)}{a\\left(x\\right)}y=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572363395\">Now define [latex]p\\left(x\\right)=\\frac{b\\left(x\\right)}{a\\left(x\\right)}[\/latex] and [latex]q\\left(x\\right)=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex]. Then the definition becomes<\/p>\n<div id=\"fs-id1170571712108\" style=\"text-align: center;\" data-type=\"equation\">[latex]{y}^{\\prime }+p\\left(x\\right)y=q\\left(x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572087796\">We can write any first-order linear differential equation in this form, and this is referred to as the standard form for a first-order linear differential equation.<\/p>\n<div id=\"fs-id1170572447597\" data-type=\"example\">\n<div id=\"fs-id1170572141228\" data-type=\"exercise\">\n<div id=\"fs-id1170571623194\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: Writing First-Order Linear Equations in Standard Form<\/h3>\n<div id=\"fs-id1170571623194\" data-type=\"problem\">\n<p id=\"fs-id1170571657281\">Put each of the following first-order linear differential equations into standard form. Identify [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex] for each equation.<\/p>\n<ol id=\"fs-id1170572099685\" type=\"a\">\n<li>[latex]y^{\\prime} =3x - 4y[\/latex]<\/li>\n<li>[latex]\\frac{3xy^{\\prime} }{4y - 3}=2[\/latex] (here [latex]x>0[\/latex])<\/li>\n<li>[latex]y=3y^{\\prime} -4{x}^{2}+5[\/latex]<\/li>\n<\/ol>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558899\">Show Solution<\/span><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572453259\" data-type=\"solution\">\n<ol id=\"fs-id1170572601312\" type=\"a\">\n<li>Add [latex]4y[\/latex] to both sides:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571602108\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} +4y=3x[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIn this equation, [latex]p\\left(x\\right)=4[\/latex] and [latex]q\\left(x\\right)=3x[\/latex].<\/li>\n<li>Multiply both sides by [latex]4y - 3[\/latex], then subtract [latex]8y[\/latex] from each side:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571712896\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{3xy^{\\prime} }{4y - 3}& =\\hfill & 2\\hfill \\\\ \\hfill 3xy^{\\prime} & =\\hfill & 2\\left(4y - 3\\right)\\hfill \\\\ \\hfill 3xy^{\\prime} & =\\hfill & 8y - 6\\hfill \\\\ \\hfill 3xy^{\\prime} -8y& =\\hfill & -6.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nFinally, divide both sides by [latex]3x[\/latex] to make the coefficient of [latex]y^{\\prime}[\/latex] equal to [latex]1\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"eip-id1170800554716\" style=\"text-align: center;\" data-type=\"equation\">[latex]y^{\\prime} -\\frac{8}{3x}y=-\\frac{2}{3x}[\/latex].<\/div>\n<p>This is allowable because in the original statement of this problem we assumed that [latex]x>0[\/latex]. (If [latex]x=0[\/latex] then the original equation becomes [latex]0=2[\/latex], which is clearly a false statement.)<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIn this equation, [latex]p\\left(x\\right)=-\\frac{8}{3x}[\/latex] and [latex]q\\left(x\\right)=-\\frac{2}{3x}[\/latex].<\/li>\n<li>Subtract [latex]y[\/latex] from each side and add [latex]4{x}^{2}-5\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572373157\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]3y^{\\prime} -y=4{x}^{2}-5[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nNext divide both sides by [latex]3\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571637556\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} -\\frac{1}{3}y=\\frac{4}{3}{x}^{2}-\\frac{5}{3}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIn this equation, [latex]p\\left(x\\right)=-\\frac{1}{3}[\/latex] and [latex]q\\left(x\\right)=\\frac{4}{3}{x}^{2}-\\frac{5}{3}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Writing First-Order Linear Equations in Standard Form.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6722828&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=SrNV1z_b_4o&amp;video_target=tpm-plugin-2sn3q61c-SrNV1z_b_4o\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.5.1_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;4.5.1&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1170572292285\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170572331245\" data-type=\"exercise\">\n<div id=\"fs-id1170572331248\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170572331248\" data-type=\"problem\">\n<p>Put the equation [latex]\\frac{\\left(x+3\\right)y^{\\prime} }{2x - 3y - 4}=5[\/latex] into standard form and identify [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558897\">Hint<\/span><\/p>\n<div id=\"q44558897\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572241407\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170572484113\">Multiply both sides by the common denominator, then collect all terms involving [latex]y[\/latex] on one side.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558898\">Show Solution<\/span><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572452404\" data-type=\"solution\">\n<p id=\"fs-id1170572169346\">[latex]y^{\\prime} +\\frac{15}{x+3}y=\\frac{10x - 20}{x+3};p\\left(x\\right)=\\frac{15}{x+3}[\/latex] and [latex]q\\left(x\\right)=\\frac{10x - 20}{x+3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169329\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169329&theme=oea&iframe_resize_id=ohm169329&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/section>\n<section id=\"fs-id1170572552065\" data-depth=\"1\">\n<h2 data-type=\"title\">Integrating Factors<\/h2>\n<p id=\"fs-id1170572337196\">We now develop a solution technique for any first-order linear differential equation. We start with the standard form of a first-order linear differential equation:<\/p>\n<div id=\"fs-id1170572130093\" style=\"text-align: center;\" data-type=\"equation\">[latex]y^{\\prime} +p\\left(x\\right)y=q\\left(x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572550947\">The first term on the left-hand side of [latex]{y}^{\\prime }+\\frac{b\\left(x\\right)}{a\\left(x\\right)}y=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex] is the derivative of the unknown function, and the second term is the product of a known function with the unknown function. This is somewhat reminiscent of the power rule from the Differentiation Rules section. If we multiply [latex]{y}^{\\prime }+p\\left(x\\right)y=q\\left(x\\right)[\/latex] by a yet-to-be-determined function [latex]\\mu \\left(x\\right)[\/latex], then the equation becomes<\/p>\n<div id=\"fs-id1170572449739\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\mu \\left(x\\right){y}^{\\prime }+\\mu \\left(x\\right)p\\left(x\\right)y=\\mu \\left(x\\right)q\\left(x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572168782\">The left-hand side of [latex]y^{\\prime} +p\\left(x\\right)y=q\\left(x\\right)[\/latex]\u00a0can be matched perfectly to the product rule:<\/p>\n<div id=\"fs-id1170571714376\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{d}{dx}\\left[f\\left(x\\right)g\\left(x\\right)\\right]={f}^{\\prime }\\left(x\\right)g\\left(x\\right)+f\\left(x\\right){g}^{\\prime }\\left(x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572482443\">Matching term by term gives [latex]y=f\\left(x\\right),g\\left(x\\right)=\\mu \\left(x\\right)[\/latex], and [latex]{g}^{\\prime }\\left(x\\right)=\\mu \\left(x\\right)p\\left(x\\right)[\/latex]. Taking the derivative of [latex]g\\left(x\\right)=\\mu \\left(x\\right)[\/latex] and setting it equal to the right-hand side of [latex]{g}^{\\prime }\\left(x\\right)=\\mu \\left(x\\right)p\\left(x\\right)[\/latex] leads to<\/p>\n<div id=\"fs-id1170571832251\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\mu }^{\\prime }\\left(x\\right)=\\mu \\left(x\\right)p\\left(x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572397269\">This is a first-order, separable differential equation for [latex]\\mu \\left(x\\right)[\/latex]. We know [latex]p\\left(x\\right)[\/latex] because it appears in the differential equation we are solving. Separating variables and integrating yields<\/p>\n<div id=\"fs-id1170571564354\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill \\frac{{\\mu }^{\\prime }\\left(x\\right)}{\\mu \\left(x\\right)}& =\\hfill & p\\left(x\\right)\\hfill \\\\ \\hfill {\\displaystyle\\int \\frac{{\\mu }^{\\prime }\\left(x\\right)}{\\mu \\left(x\\right)}dx}& =\\hfill & {\\displaystyle\\int p\\left(x\\right)dx}\\hfill \\\\ \\hfill \\text{ln}|\\mu \\left(x\\right)|& =\\hfill & {\\displaystyle\\int p\\left(x\\right)dx+C}\\hfill \\\\ \\hfill {e}^{\\text{ln}|\\mu \\left(x\\right)|}& =\\hfill & {e}^{\\displaystyle\\int p\\left(x\\right)dx+C}\\hfill \\\\ \\hfill |\\mu \\left(x\\right)|& =\\hfill & {C}_{1}{e}^{\\displaystyle\\int p\\left(x\\right)dx}\\hfill \\\\ \\hfill \\mu \\left(x\\right)& =\\hfill & {C}_{2}{e}^{\\displaystyle\\int p\\left(x\\right)dx}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571593416\">Here [latex]{C}_{2}[\/latex] can be an arbitrary (positive or negative) constant. This leads to a general method for solving a first-order linear differential equation. We first multiply both sides of [latex]{y}^{\\prime }+p\\left(x\\right)y=q\\left(x\\right)[\/latex]\u00a0by the <span data-type=\"term\">integrating factor<\/span> [latex]\\mu \\left(x\\right)[\/latex]. This gives<\/p>\n<div id=\"fs-id1170571789027\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\mu \\left(x\\right){y}^{\\prime }+\\mu \\left(x\\right)p\\left(x\\right)y=\\mu \\left(x\\right)q\\left(x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571597633\">The left-hand side of the above equation can be rewritten as [latex]\\frac{d}{dx}\\left(\\mu \\left(x\\right)y\\right)[\/latex].<\/p>\n<div id=\"fs-id1170572554001\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\frac{d}{dx}\\left(\\mu \\left(x\\right)y\\right)=\\mu \\left(x\\right)q\\left(x\\right)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571814541\">Next integrate both sides with respect to [latex]x[\/latex].<\/p>\n<div id=\"fs-id1170572546914\" style=\"text-align: center;\" data-type=\"equation\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dx}\\left(\\mu \\left(x\\right)y\\right)dx}& =\\hfill & {\\displaystyle\\int \\mu \\left(x\\right)q\\left(x\\right)dx}\\hfill \\\\ \\hfill \\mu \\left(x\\right)y& =\\hfill & {\\displaystyle\\int \\mu \\left(x\\right)q\\left(x\\right)dx}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571699423\">Divide both sides\u00a0by [latex]\\mu \\left(x\\right)\\text{:}[\/latex]<\/p>\n<div id=\"fs-id1170571681325\" style=\"text-align: center;\" data-type=\"equation\">[latex]y=\\dfrac{1}{\\mu \\left(x\\right)}\\left[\\displaystyle\\int \\mu \\left(x\\right)q\\left(x\\right)dx+C\\right][\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170572290464\">Since [latex]\\mu \\left(x\\right)[\/latex] was previously calculated, we are now finished. An important note about the integrating constant [latex]C\\text{:}[\/latex] It may seem that we are inconsistent in the usage of the integrating constant. However, the integral involving [latex]p\\left(x\\right)[\/latex] is necessary in order to find an integrating factor for [latex]{y}^{\\prime }+\\frac{b\\left(x\\right)}{a\\left(x\\right)}y=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex]. Only one integrating factor is needed in order to solve the equation; therefore, it is safe to assign a value for [latex]C[\/latex] for this integral. We chose [latex]C=0[\/latex]. When calculating the integral inside the brackets in [latex]\\frac{d}{dx}\\left(\\mu \\left(x\\right)y\\right)=\\mu \\left(x\\right)q\\left(x\\right)[\/latex], it is necessary to keep our options open for the value of the integrating constant, because our goal is to find a general family of solutions to [latex]{y}^{\\prime }+\\frac{b\\left(x\\right)}{a\\left(x\\right)}y=\\frac{c\\left(x\\right)}{a\\left(x\\right)}[\/latex]. This integrating factor guarantees just that.<\/p>\n<div id=\"fs-id1170572628487\" class=\"problem-solving\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3 data-type=\"title\">Problem-Solving Strategy: Solving a First-order Linear Differential Equation<\/h3>\n<ol id=\"fs-id1170571748988\" type=\"1\">\n<li>Put the equation into standard form and identify [latex]p\\left(x\\right)[\/latex] and [latex]q\\left(x\\right)[\/latex].<\/li>\n<li>Calculate the integrating factor [latex]\\mu \\left(x\\right)={e}^{\\displaystyle\\int p\\left(x\\right)dx}[\/latex].<\/li>\n<li>Multiply both sides of the differential equation by [latex]\\mu \\left(x\\right)[\/latex].<\/li>\n<li>Integrate both sides of the equation obtained in step [latex]3[\/latex], and divide both sides by [latex]\\mu \\left(x\\right)[\/latex].<\/li>\n<li>If there is an initial condition, determine the value of [latex]C[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170572539811\" data-type=\"example\">\n<div id=\"fs-id1170572321365\" data-type=\"exercise\">\n<div id=\"fs-id1170572321367\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Solving a First-order Linear Equation<\/h3>\n<div id=\"fs-id1170572321367\" data-type=\"problem\">\n<p id=\"fs-id1170572572214\">Find a general solution for the differential equation [latex]xy^{\\prime} +3y=4{x}^{2}-3x[\/latex]. Assume [latex]x>0[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558896\">Show Solution<\/span><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572622201\" data-type=\"solution\">\n<ol id=\"fs-id1170572309310\" type=\"1\">\n<li>To put this differential equation into standard form, divide both sides by [latex]x\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572370908\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y^{\\prime} +\\frac{3}{x}y=4x - 3[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore [latex]p\\left(x\\right)=\\frac{3}{x}[\/latex] and [latex]q\\left(x\\right)=4x - 3[\/latex].<\/li>\n<li>The integrating factor is [latex]\\mu \\left(x\\right)={e}^{\\displaystyle\\int \\left(\\frac{3}{x}\\right)dx}={e}^{3\\text{ln}x}={x}^{3}[\/latex].<\/li>\n<li>Multiplying both sides of the differential equation by [latex]\\mu \\left(x\\right)[\/latex] gives us<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572116162\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {x}^{3}{y}^{\\prime }+{x}^{3}\\left(\\frac{3}{x}\\right)y& =\\hfill & {x}^{3}\\left(4x - 3\\right)\\hfill \\\\ \\hfill {x}^{3}{y}^{\\prime }+3{x}^{2}y& =\\hfill & 4{x}^{4}-3{x}^{3}\\hfill \\\\ \\hfill \\frac{d}{dx}\\left({x}^{3}y\\right)& =\\hfill & 4{x}^{4}-3{x}^{3}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>Integrate both sides of the equation.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572369225\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dx}\\left({x}^{3}y\\right)dx}& =\\hfill & {\\displaystyle\\int 4{x}^{4}-3{x}^{3}dx}\\hfill \\\\ \\hfill {x}^{3}y& =\\hfill & \\frac{4{x}^{5}}{5}-\\frac{3{x}^{4}}{4}+C\\hfill \\\\ \\hfill y& =\\hfill & \\frac{4{x}^{2}}{5}-\\frac{3x}{4}+C{x}^{-3}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>There is no initial value, so the problem is complete.<\/li>\n<\/ol>\n<\/div>\n<div id=\"fs-id1170572585959\" data-type=\"commentary\">\n<div data-type=\"title\"><strong>Analysis<\/strong><\/div>\n<p id=\"fs-id1170572169556\">You may have noticed the condition that was imposed on the differential equation; namely, [latex]x>0[\/latex]. For any nonzero value of [latex]C[\/latex], the general solution is not defined at [latex]x=0[\/latex]. Furthermore, when [latex]x<0[\/latex], the integrating factor changes. The integrating factor is given by [latex]\\mu \\left(x\\right){y}^{\\prime }+\\mu \\left(x\\right)p\\left(x\\right)y=\\mu \\left(x\\right)q\\left(x\\right)[\/latex]\u00a0as [latex]f\\left(x\\right)={e}^{\\displaystyle\\int p\\left(x\\right)dx}[\/latex]. For this [latex]p\\left(x\\right)[\/latex] we get<\/p>\n<div id=\"fs-id1170572505442\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{e}^{\\displaystyle\\int p\\left(x\\right)dx=}{e}^{\\displaystyle\\int \\left(\\frac{3}{x}\\right)dx}={e}^{3\\text{ln}|x|}={|x|}^{3}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170571600638\">since [latex]x<0[\/latex]. The behavior of the general solution changes at [latex]x=0[\/latex] largely due to the fact that [latex]p\\left(x\\right)[\/latex] is not defined there.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example:\u00a0Solving a First-Order Linear Equation.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/doLN3RtE264?controls=0&amp;start=0&amp;end=300&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.5.2_0to300_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.5.2&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1170571598375\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170572332858\" data-type=\"exercise\">\n<div id=\"fs-id1170572332860\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170572332860\" data-type=\"problem\">\n<p id=\"fs-id1170572332863\">Find the general solution to the differential equation [latex]\\left(x - 2\\right)y^{\\prime} +y=3{x}^{2}+2x[\/latex]. Assume [latex]x>2[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558894\">Hint<\/span><\/p>\n<div id=\"q44558894\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571690904\" data-type=\"commentary\" data-element-type=\"hint\">\n<p id=\"fs-id1170571679832\">Use the method outlined in the problem-solving strategy for first-order linear differential equations.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558895\">Show Solution<\/span><\/p>\n<div id=\"q44558895\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571611044\" data-type=\"solution\">\n<p id=\"fs-id1170571655144\">[latex]y=\\frac{{x}^{3}+{x}^{2}+C}{x - 2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1170572611124\">Now we use the same strategy to find the solution to an initial-value problem.<\/p>\n<div id=\"fs-id1170572611128\" data-type=\"example\">\n<div id=\"fs-id1170572611130\" data-type=\"exercise\">\n<div id=\"fs-id1170572643199\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example: A First-order Linear Initial-Value Problem<\/h3>\n<div id=\"fs-id1170572643199\" data-type=\"problem\">\n<p id=\"fs-id1170572643204\">Solve the initial-value problem<\/p>\n<div id=\"fs-id1170571609231\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{y}^{\\prime }+3y=2x - 1,y\\left(0\\right)=3[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558893\">Show Solution<\/span><\/p>\n<div id=\"q44558893\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170571653907\" data-type=\"solution\">\n<ol id=\"fs-id1170571653909\" type=\"1\">\n<li>This differential equation is already in standard form with [latex]p\\left(x\\right)=3[\/latex] and [latex]q\\left(x\\right)=2x - 1[\/latex].<\/li>\n<li>The integrating factor is [latex]\\mu \\left(x\\right)={e}^{\\displaystyle\\int 3dx}={e}^{3x}[\/latex].<\/li>\n<li>Multiplying both sides of the differential equation by [latex]\\mu \\left(x\\right)[\/latex] gives<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572481493\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {e}^{3x}{y}^{\\prime }+3{e}^{3x}y& =\\hfill & \\left(2x - 1\\right){e}^{3x}\\hfill \\\\ \\hfill \\frac{d}{dx}\\left[y{e}^{3x}\\right]& =\\hfill & \\left(2x - 1\\right){e}^{3x}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIntegrate both sides of the equation:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571593388\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill {\\displaystyle\\int \\frac{d}{dx}\\left[y{e}^{3x}\\right]dx}& =\\hfill & {\\displaystyle\\int \\left(2x - 1\\right){e}^{3x}dx}\\hfill \\\\ \\hfill y{e}^{3x}& =\\hfill & \\frac{{e}^{3x}}{3}\\left(2x - 1\\right)-{\\displaystyle\\int \\frac{2}{3}{e}^{3x}dx}\\hfill \\\\ \\hfill y{e}^{3x}& =\\hfill & \\frac{{e}^{3x}\\left(2x - 1\\right)}{3}-\\frac{2{e}^{3x}}{9}+C\\hfill \\\\ \\hfill y& =\\hfill & \\frac{2x - 1}{3}-\\frac{2}{9}+C{e}^{-3x}\\hfill \\\\ \\hfill y& =\\hfill & \\frac{2x}{3}-\\frac{5}{9}+C{e}^{-3x}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/li>\n<li>Now substitute [latex]x=0[\/latex] and [latex]y=3[\/latex] into the general solution and solve for [latex]C\\text{:}[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170571595426\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccc}\\hfill y& =\\hfill & \\frac{2}{3}x-\\frac{5}{9}+C{e}^{-3x}\\hfill \\\\ \\hfill 3& =\\hfill & \\frac{2}{3}\\left(0\\right)-\\frac{5}{9}+C{e}^{-3\\left(0\\right)}\\hfill \\\\ \\hfill 3& =\\hfill & -\\frac{5}{9}+C\\hfill \\\\ \\hfill C& =\\hfill & \\frac{32}{9}.\\hfill \\end{array}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nTherefore the solution to the initial-value problem is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1170572592117\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]y=\\frac{2}{3}x-\\frac{5}{9}+\\frac{32}{9}{e}^{-3x}[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: A First-Order Linear Initial-Value Problem.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/doLN3RtE264?controls=0&amp;start=301&amp;end=630&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/4.5.2_301to630_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.5.2&#8221; here (opens in new window)<\/a>.<\/p>\n<div id=\"fs-id1170571640637\" class=\"checkpoint\" data-type=\"note\">\n<div id=\"fs-id1170572505498\" data-type=\"exercise\">\n<div id=\"fs-id1170572505501\" data-type=\"problem\">\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<div id=\"fs-id1170572505501\" data-type=\"problem\">\n<p id=\"fs-id1170572505503\">Solve the initial-value problem [latex]y^{\\prime} -2y=4x+3, y\\left(0\\right)=-2[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558892\">Show Solution<\/span><\/p>\n<div id=\"q44558892\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1170572449330\" data-type=\"solution\">\n<p id=\"fs-id1170572449332\">[latex]y=-2x - 4+2{e}^{2x}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169340\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169340&theme=oea&iframe_resize_id=ohm169340&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1170571593637\" data-depth=\"1\"><\/section>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1639\">\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>4.5.1. <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>4.5.2. <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 2. <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-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/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-2\/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":17533,"menu_order":18,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.5.1\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"4.5.2\",\"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-1639","chapter","type-chapter","status-publish","hentry"],"part":159,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1639","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1639\/revisions"}],"predecessor-version":[{"id":2209,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1639\/revisions\/2209"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/159"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1639\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1639"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1639"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1639"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1639"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}