{"id":833,"date":"2025-06-20T17:16:30","date_gmt":"2025-06-20T17:16:30","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=833"},"modified":"2025-08-15T16:51:55","modified_gmt":"2025-08-15T16:51:55","slug":"first-order-linear-equations-and-applications-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/first-order-linear-equations-and-applications-learn-it-1\/","title":{"raw":"First-Order Linear Equations and Applications: Learn It 1","rendered":"First-Order Linear Equations and Applications: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Write first-order linear differential equations in their standard form<\/li>\r\n \t<li>Find and use integrating factors to solve first-order linear equations<\/li>\r\n \t<li>Understand how carrying capacity affects population growth in the logistic model<\/li>\r\n \t<li>Work with logistic equations and interpret what their solutions mean<\/li>\r\n \t<li>Solve real-world problems using first-order linear differential equations<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>First-Order Differential Equations<\/h2>\r\n<p id=\"fs-id1170572444259\">Let's build on what you already know about objects in motion. Previously, we looked at a ball thrown upward and modeled its velocity using the differential equation:<\/p>\r\n\r\n<div id=\"fs-id1170572387982\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{dv}{dt}=-32,v\\left(0\\right)={v}_{0}[\/latex].<\/div>\r\n<p class=\"whitespace-normal break-words\">This assumed only gravity affected the ball. But what happens when we add air resistance to our model?<\/p>\r\n<p class=\"whitespace-normal break-words\">Air resistance always opposes motion. When an object is rising, air resistance acts in a downward direction. If the object is falling, air resistance acts in an upward direction (Figure 1). There's no exact relationship between velocity and air resistance, but for small objects, we can approximate that air resistance is proportional to velocity.<\/p>\r\n<p class=\"whitespace-normal break-words\">For small objects, air resistance is proportional to velocity. If [latex]k &gt; 0[\/latex] is a constant, then the force due to air resistance is [latex]F_A = -kv[\/latex]. The negative sign ensures air resistance opposes the direction of motion.<\/p>\r\nNow we can identify the total forces acting on our object. The gravitational force is [latex]-mg[\/latex] (acting downward), and the air resistance is [latex]-kv[\/latex] (opposing motion).\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234222\/CNX_Calc_Figure_08_05_001.jpg\" alt=\"A diagram of a baseball with an arrow above it pointing up and an arrow below it pointing down. The upper arrow is labeled \" width=\"325\" height=\"352\" data-media-type=\"image\/jpeg\" \/> Figure 1. Forces acting on a moving baseball: gravity acts in a downward direction and air resistance acts in a direction opposite to the direction of motion.[\/caption]\r\n\r\nUsing Newton's second law, the sum of these forces equals mass times acceleration:\r\n<p style=\"text-align: center;\">[latex]m\\frac{dv}{dt} = -kv - mg[\/latex]<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\u00a0<strong>Newton's Second Law<\/strong>: The sum of forces acting on an object equals mass times acceleration: [latex]F = ma[\/latex]<\/section>\r\n<p id=\"fs-id1170572142332\">Adding an initial condition [latex]v(0) = v_0[\/latex] gives us the complete model for motion with air resistance:<\/p>\r\n\r\n<div id=\"fs-id1170572593954\" style=\"text-align: center;\" data-type=\"equation\">[latex]m\\frac{dv}{dt}=\\text{-}kv-mg,v\\left(0\\right)={v}_{0}[\/latex].<\/div>\r\n<div data-type=\"equation\">Where [latex]m[\/latex] is mass, [latex]k[\/latex] is the air resistance coefficient, [latex]g[\/latex] is acceleration due to gravity, and [latex]v_0[\/latex] is initial velocity.<\/div>\r\n<div data-type=\"equation\"><\/div>\r\n<div data-type=\"equation\">The equation [latex]m\\frac{dv}{dt} = -kv - mg[\/latex] is a <strong>first-order linear differential equation<\/strong> because the highest-order derivative is first-order ([latex]\\frac{dv}{dt}[\/latex]) and the equation is linear in the unknown function and its derivative.<\/div>\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>first-order linear differential equation<\/h3>\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<\/section>Remember that the unknown function [latex]y[\/latex] depends on the variable [latex]x[\/latex]. Here [latex]x[\/latex] is the independent variable and [latex]y[\/latex] is the dependent variable.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Some examples of first-order linear differential equations are:\r\n<p style=\"text-align: center;\">[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]<\/p>\r\n<p class=\"whitespace-normal break-words\">In contrast, examples of first-order <strong>nonlinear<\/strong> differential equations include:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[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]<\/p>\r\nThese 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.\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">The key to identifying linear equations is that [latex]y[\/latex] and [latex]y'[\/latex] appear only to the first power and are not multiplied together. To identify if an equation is linear, look for powers of [latex]y[\/latex] or [latex]y'[\/latex] other than [latex]1[\/latex], or products involving [latex]y[\/latex] and [latex]y'[\/latex] together. If you find any, the equation is nonlinear.<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Write first-order linear differential equations in their standard form<\/li>\n<li>Find and use integrating factors to solve first-order linear equations<\/li>\n<li>Understand how carrying capacity affects population growth in the logistic model<\/li>\n<li>Work with logistic equations and interpret what their solutions mean<\/li>\n<li>Solve real-world problems using first-order linear differential equations<\/li>\n<\/ul>\n<\/section>\n<h2>First-Order Differential Equations<\/h2>\n<p id=\"fs-id1170572444259\">Let&#8217;s build on what you already know about objects in motion. Previously, we looked at a ball thrown upward and modeled its velocity using the differential equation:<\/p>\n<div id=\"fs-id1170572387982\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\frac{dv}{dt}=-32,v\\left(0\\right)={v}_{0}[\/latex].<\/div>\n<p class=\"whitespace-normal break-words\">This assumed only gravity affected the ball. But what happens when we add air resistance to our model?<\/p>\n<p class=\"whitespace-normal break-words\">Air resistance always opposes motion. When an object is rising, air resistance acts in a downward direction. If the object is falling, air resistance acts in an upward direction (Figure 1). There&#8217;s no exact relationship between velocity and air resistance, but for small objects, we can approximate that air resistance is proportional to velocity.<\/p>\n<p class=\"whitespace-normal break-words\">For small objects, air resistance is proportional to velocity. If [latex]k > 0[\/latex] is a constant, then the force due to air resistance is [latex]F_A = -kv[\/latex]. The negative sign ensures air resistance opposes the direction of motion.<\/p>\n<p>Now we can identify the total forces acting on our object. The gravitational force is [latex]-mg[\/latex] (acting downward), and the air resistance is [latex]-kv[\/latex] (opposing motion).<\/p>\n<figure style=\"width: 325px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11234222\/CNX_Calc_Figure_08_05_001.jpg\" alt=\"A diagram of a baseball with an arrow above it pointing up and an arrow below it pointing down. The upper arrow is labeled\" width=\"325\" height=\"352\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 1. Forces acting on a moving baseball: gravity acts in a downward direction and air resistance acts in a direction opposite to the direction of motion.<\/figcaption><\/figure>\n<p>Using Newton&#8217;s second law, the sum of these forces equals mass times acceleration:<\/p>\n<p style=\"text-align: center;\">[latex]m\\frac{dv}{dt} = -kv - mg[\/latex]<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\u00a0<strong>Newton&#8217;s Second Law<\/strong>: The sum of forces acting on an object equals mass times acceleration: [latex]F = ma[\/latex]<\/section>\n<p id=\"fs-id1170572142332\">Adding an initial condition [latex]v(0) = v_0[\/latex] gives us the complete model for motion with air resistance:<\/p>\n<div id=\"fs-id1170572593954\" style=\"text-align: center;\" data-type=\"equation\">[latex]m\\frac{dv}{dt}=\\text{-}kv-mg,v\\left(0\\right)={v}_{0}[\/latex].<\/div>\n<div data-type=\"equation\">Where [latex]m[\/latex] is mass, [latex]k[\/latex] is the air resistance coefficient, [latex]g[\/latex] is acceleration due to gravity, and [latex]v_0[\/latex] is initial velocity.<\/div>\n<div data-type=\"equation\"><\/div>\n<div data-type=\"equation\">The equation [latex]m\\frac{dv}{dt} = -kv - mg[\/latex] is a <strong>first-order linear differential equation<\/strong> because the highest-order derivative is first-order ([latex]\\frac{dv}{dt}[\/latex]) and the equation is linear in the unknown function and its derivative.<\/div>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>first-order linear differential equation<\/h3>\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<\/section>\n<p>Remember that the unknown function [latex]y[\/latex] depends on the variable [latex]x[\/latex]. Here [latex]x[\/latex] is the independent variable and [latex]y[\/latex] is the dependent variable.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">Some examples of first-order linear differential equations are:<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p class=\"whitespace-normal break-words\">In contrast, examples of first-order <strong>nonlinear<\/strong> differential equations include:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[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]<\/p>\n<p>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>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">The key to identifying linear equations is that [latex]y[\/latex] and [latex]y'[\/latex] appear only to the first power and are not multiplied together. To identify if an equation is linear, look for powers of [latex]y[\/latex] or [latex]y'[\/latex] other than [latex]1[\/latex], or products involving [latex]y[\/latex] and [latex]y'[\/latex] together. If you find any, the equation is nonlinear.<\/section>\n","protected":false},"author":15,"menu_order":22,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":669,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/833"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/833\/revisions"}],"predecessor-version":[{"id":1840,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/833\/revisions\/1840"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/669"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/833\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=833"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=833"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=833"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=833"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}