{"id":1334,"date":"2021-11-24T06:41:35","date_gmt":"2021-11-24T06:41:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1334"},"modified":"2022-11-01T23:07:05","modified_gmt":"2022-11-01T23:07:05","slug":"series-solutions-of-differential-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/series-solutions-of-differential-equations\/","title":{"raw":"Series Solutions of Differential Equations","rendered":"Series Solutions of Differential 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\">Use power series to solve first-order and second-order differential equations.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1170571376919\" class=\"problem-solving ui-has-child-title\" 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: finding power series solutions to differential equations<\/h3>\r\n<ol id=\"fs-id1170571248526\" type=\"1\">\r\n \t<li>Assume the differential equation has a solution of the form [latex]y(x)=\\displaystyle\\sum_{n=0}^\\infty a_nx^n[\/latex].<\/li>\r\n \t<li>Differentiate the power series term by term to get [latex]y^\\prime(x)=\\displaystyle\\sum_{n=0}^\\infty na_nx^{n-1}[\/latex] and\u00a0[latex]y^{\\prime\\prime}(x)=\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}[\/latex].<\/li>\r\n \t<li>Substitute the power series expressions into the differential equation.<\/li>\r\n \t<li>Re-index sums as necessary to combine terms and simplify the expression.<\/li>\r\n \t<li>Equate coefficients of like powers of [latex]x[\/latex] to determine values for the coefficients [latex]a_n[\/latex] in the power series.<\/li>\r\n \t<li>Substitute the coefficients back into the power series and write the solution.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: series solutions to differential equations<\/h3>\r\n<p id=\"fs-id1170573434701\">Find a power series solution for the following differential equations.<\/p>\r\n\r\n<ol id=\"fs-id1170573437980\" type=\"a\">\r\n \t<li>[latex]y^{\\prime\\prime}-y=0[\/latex]<\/li>\r\n \t<li>[latex](x^2-1)y^{\\prime\\prime}+6xy^\\prime+4y=-4[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"226627144\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"226627144\"]\r\n<ol type=\"a\">\r\n \t<li style=\"list-style-type: none;\">\r\n<ol type=\"a\">\r\n \t<li>Assume [latex]y(x)=\\displaystyle\\sum_{n=0}^\\infty a_nx^n[\/latex] (step 1). Then, [latex]y^\\prime(x)=\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}[\/latex] and [latex]y^{\\prime\\prime}(x)=\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}[\/latex] (step 2). We want to find values for the coefficients [latex]a_n[\/latex] such that<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny^{\\prime\\prime}-y&amp;=0 \\\\\r\n\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}-\\displaystyle\\sum_{n=0}^\\infty a_nx^n&amp;=0\r\n\\end{aligned}[\/latex] (step 3).<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with [latex]n=0[\/latex].\u00a0<span data-type=\"newline\">\r\n<\/span>To re-index the first term, replace [latex]n[\/latex] with [latex]n+2[\/latex] inside the sum, and change the lower summation limit to [latex]n=0[\/latex]. We get<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}=\\displaystyle\\sum_{n=0}^\\infty(n+2)(n+1)a_{n+2}x^n[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>This gives<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\sum_{n=0}^\\infty(n+2)(n+1)a_{n+2}x^n-\\displaystyle\\sum_{n=0}^\\infty a_nx^n&amp;=0 \\\\\r\n\\displaystyle\\sum_{n=0}^\\infty[(n+2)(n+1)a_{n+2}-a_n]x^n&amp;=0 \\text{ (step 4)}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of [latex]x[\/latex] are zero. So we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\large{(n+2)(n+1)a_{n+2}-a_n=0\\text{ for }n=0,1,2,\\ldots}[\/latex]<\/p>\r\nThis recurrence relationship allows us to express each coefficient [latex]a_n[\/latex] in terms of the coefficient two terms earlier. This yields one expression for even values of [latex]n[\/latex] and another expression for odd values of [latex]n[\/latex]. Looking first at the equations involving even values of [latex]n[\/latex], we see that<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\na_2&amp;=\\frac{a_0}2 \\\\\r\na_4&amp;=\\frac{a_2}{4\\cdot3}=\\frac{a_0}{4!} \\\\\r\na_6&amp;=\\frac{a_4}{6\\cdot5}=\\frac{a_0}{6!} \\\\\r\n&amp;\\vdots\r\n\\end{aligned}[\/latex]<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Thus, in general, when [latex]n[\/latex] is even, [latex]a_n=\\frac{a_0}{n!}[\/latex] (step 5).<span data-type=\"newline\">\r\n<\/span>For the equations involving odd values of [latex]n[\/latex], we see that<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\na_3&amp;=\\frac{a_1}{3\\cdot2}=\\frac{a_1}{3!} \\\\\r\na_5&amp;=\\frac{a_3}{5\\cdot4}=\\frac{a_1}{5!} \\\\\r\na_7&amp;=\\frac{a_5}{7\\cdot6}=\\frac{a_1}{7!} \\\\\r\n&amp;\\vdots\r\n\\end{aligned}[\/latex]<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Therefore, in general, when [latex]n[\/latex] is odd, [latex]a_n=\\frac{a_1}{n!}[\/latex] (step 5 continued).<span data-type=\"newline\">\r\n<\/span>Putting this together, we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny(x)&amp;=\\displaystyle\\sum_{n=0}^\\infty a_nx^n \\\\\r\n&amp;=a_0+a_1x+\\frac{a_0}2x^2+\\frac{a_1}{3!}x^3+\\frac{a_0}{4!}x^4+\\frac{a_1}{5!}x^5+\\cdots\r\n\\end{aligned}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Re-indexing the sums to account for the even and odd values of [latex]n[\/latex] separately, we obtain<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]y(x)=a_0\\displaystyle\\sum_{k=0}^\\infty\\frac{1}{(2k)!}x^{2k}+a_1\\displaystyle\\sum_{k=0}^\\infty\\frac{1}{(2k+1)!}x^{2k+1}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span><strong data-effect=\"bold\">Analysis for part a.<\/strong><span data-type=\"newline\">\r\n<\/span>As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since [latex]y(x)=c_1e^x+c_2e^{-x}[\/latex] is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.<span data-type=\"newline\">\r\n<\/span>Fortunately, after writing the power series representations of [latex]e^x[\/latex] and [latex]e^{-x}[\/latex], and doing some algebra, we find that if we choose<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\large{c_0=\\frac{(a_0+a_1)}2, \\ c_1=\\frac{(a_0-a_1)}2}[\/latex],<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>we then have [latex]a_0=c_0+c_1[\/latex] and [latex]a_1=c_0-c_1[\/latex], and\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\ny(x)&amp;=a_0+a_1x+\\frac{a_0}2x^2+\\frac{a_1}{3!}x^3+\\frac{a_0}{4!}x^4+\\frac{a_0}{5!}x^5+\\ldots \\\\\r\n&amp;=(c_0+c_1)+(c_0-c_1)x+\\frac{(c_0+c_1)}2x^2+\\frac{(c_0-c_1)}{3!}x^3+\\frac{(c_0+c_1)}{4!}x^4+\\frac{(c_0-c_1)}{5!}x^5+\\ldots \\\\\r\n&amp;=c_0\\displaystyle\\sum_{n=0}^\\infty\\frac{x^n}{n!}+c_1\\displaystyle\\sum_{n=0}^\\infty\\frac{(-x)^n}{n!} \\\\\r\n&amp;=c_0e^x+c_1e^{-x}\r\n\\end{aligned}[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1170571232470\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<div class=\"MathJax_Display\"><\/div>\r\n<\/div>\r\nSo we have, in fact, found the same general solution. Note that this choice of [latex]c_1[\/latex] and\u00a0[latex]c_2[\/latex] is not obvious. This is a case when we know what the answer should be, and have essentially \u201creverse-engineered\u201d our choice of coefficients.\r\n<ul>\r\n \t<li>Assume [latex]y(x)=\\displaystyle\\sum_{n=0}^\\infty a_nx^n[\/latex] (step 1). Then, [latex]y^\\prime(x)=\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}[\/latex] and [latex]y^{\\prime\\prime}=\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}[\/latex] (step 2). We want to find values for the coefficients [latex]a_n[\/latex] such that<\/li>\r\n<\/ul>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n(x^2-1)y^{\\prime\\prime}+6xy^\\prime+4y&amp;=-4 \\\\\r\n(x^2-1)\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}+6x\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}+4\\displaystyle\\sum_{n=0}^\\infty a_nx^n&amp;=-4 \\\\\r\nx^2\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}-\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}+6x\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}+4\\displaystyle\\sum_{n=0}^\\infty a_nx^n&amp;=-4\r\n\\end{aligned}[\/latex].<\/p>\r\nTaking the external factors inside the summations, we get\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n}-\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}+\\displaystyle\\sum_{n=1}^\\infty 6na_nx^n+\\displaystyle\\sum_{n=0}^\\infty 4a_nx^n=-4[\/latex] (step 3).<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Now, in the first summation, we see that when [latex]n=0[\/latex] or[latex]n=1[\/latex], the term evaluates to zero, so we can add these terms back into our sum to get\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n}=\\displaystyle\\sum_{n=0}^\\infty n(n-1)a_nx^{n}[\/latex].<\/p>\r\nSimilarly, in the third term, we see that when [latex]n=0[\/latex], the expression evaluates to zero, so we can add that term back in as well. We have\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=1}^\\infty 6na_nx^n=\\displaystyle\\sum_{n=0}^\\infty 6na_nx^n[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Then, we need only shift the indices in our second term. We get<span data-type=\"newline\">\r\n<\/span>\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}=\\displaystyle\\sum_{n=0}^\\infty(n+2)(n+1)a_{n+2}x^n[\/latex]<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Thus, we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\displaystyle\\sum_{n=0}^\\infty n(n-1)a_nx^{n}-\\displaystyle\\sum_{n=0}^\\infty(n+2)(n+1)a_{n+2}x^n+\\displaystyle\\sum_{n=0}^\\infty6na_nx^n+\\displaystyle\\sum_{n=0}^\\infty4a_nx^n&amp;=-4\\text{ (step 4).} \\\\\r\n\\displaystyle\\sum_{n=0}^\\infty[n(n-1)a_n-(n+2)(n+1)a_{n+2}6na_n+4a_n]x^n&amp;=-4 \\\\\r\n\\displaystyle\\sum_{n=0}^\\infty[(n^2-n)a_n+6na_n+4a_n-(n+2)(n+1)a_{n+2}]x^n&amp;=-4 \\\\\r\n\\displaystyle\\sum_{n=0}^\\infty[n^2a_n+5na_n+4a_n-(n+2)(n+1)a_{n+2}]x^n&amp;=-4 \\\\\r\n\\displaystyle\\sum_{n=0}^\\infty[(n^2+5n+4)a_n-(n+2)(n+1)a_{n+2}]x^n&amp;=-4 \\\\\r\n\\displaystyle\\sum_{n=0}^\\infty[(n+4)(n+1)a_n-(n+2)(n+1)a_{n+2}]x^n&amp;=-4\r\n\\end{aligned}[\/latex]<\/p>\r\nLooking at the coefficients of each power of [latex]x[\/latex], we see that the constant term must be equal to [latex]-4[\/latex] and the coefficients of all other powers of [latex]x[\/latex] must be zero. Then, looking first at the constant term,<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n4a_0-2a_2&amp;=-4 \\\\\r\na_2&amp;=2a_0+2\r\n\\end{aligned}[\/latex] (step 3).<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>For [latex]n\\geq1[\/latex], we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n(n+4)(n+1)a_n-(n+2)(n+1)a_{a+2}&amp;=0 \\\\\r\n(n+1)[(n+4)a_n-(n+2)a_{n+2}]&amp;=0\r\n\\end{aligned}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>Since [latex]n\\geq1[\/latex], [latex]n+1\\ne0[\/latex], we see that<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex](n+4)a_n-(n+2)a_{n+2}=0[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>and thus<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]a_{n+2}=\\frac{n+4}{n+2}a_n[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>For even values of [latex]n[\/latex], we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\na_4&amp;=\\frac64(2a_0+2)=3a_0+3 \\\\\r\na_6&amp;=\\frac86(3a_0+3)=4a_0+4 \\\\\r\n&amp;\\vdots\r\n\\end{aligned}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>In general, [latex]a_{2k}=(k+1)(a_0+1)[\/latex] (step 5).<span data-type=\"newline\">\r\n<\/span>For odd values of [latex]n[\/latex], we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\na_3&amp;=\\frac53a_1 \\\\\r\na_5&amp;=\\frac75a_3=\\frac73a_1 \\\\\r\na_7&amp;=\\frac97a_5=\\frac93a_1=3a_1 \\\\\r\n&amp;\\vdots\r\n\\end{aligned}[\/latex].<\/p>\r\n<span data-type=\"newline\">\r\n<\/span>In general, [latex]a_{2k+1}=\\frac{2k+3}3a_1[\/latex] (step 5 continued).<span data-type=\"newline\">\r\n<\/span>Putting this together, we have<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]y(x)=\\displaystyle\\sum_{k=0}^\\infty (k+1)(a_0+1)x^{2k}+\\displaystyle\\sum_{k=0}^\\infty\\left(\\frac{2k+3}3\\right)a_1x^{2k+1}[\/latex] (step 6).<\/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-id1170571291319\">Find a power series solution for the following differential equations.<\/p>\r\n\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]y^{\\prime}+2xy=0[\/latex]<\/li>\r\n \t<li>[latex](x+1)y^\\prime=3y[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"652781369\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"652781369\"]\r\n\r\na. [latex]y(x)=a_0\\displaystyle\\sum_{n=0}^\\infty\\frac{(-1)^n}{n!}x^{2n}=a_0e^{-x^2}[\/latex]\r\n\r\nb. [latex]y(x)=a_0(x+1)^3[\/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=8250347&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=d5VG6PM48vQ&amp;video_target=tpm-plugin-7w3p47zd-d5VG6PM48vQ\" 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\/CP7.22a_transcript.html\">transcript for \u201cCP 7.22a\u201d here (opens in new window).<\/a><\/center><center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8250348&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=2ukZj1dveVs&amp;video_target=tpm-plugin-lphrhkde-2ukZj1dveVs\" 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\/CP7.22b_transcript.html\">transcript for \u201cCP 7.22b\u201d here (opens in new window).<\/a><\/center>\r\n<p id=\"fs-id1170573391386\">We close this section with a brief introduction to\u00a0<span id=\"1f5ff083-3081-490b-b429-61fc1959738c_term323\" class=\"no-emphasis\" data-type=\"term\">Bessel functions<\/span>. Complete treatment of Bessel functions is well beyond the scope of this course, but we get a little taste of the topic here so we can see how series solutions to differential equations are used in real-world applications. The Bessel equation of order [latex]n[\/latex] is given by<\/p>\r\n<p style=\"text-align: center;\">[latex]x^2y^{\\prime\\prime}+xy^\\prime+(x^2-n^2)y=0[\/latex].<\/p>\r\n<p id=\"fs-id1170571370989\">This equation arises in many physical applications, particularly those involving cylindrical coordinates, such as the vibration of a circular drum head and transient heating or cooling of a cylinder. In the next example, we find a power series solution to the Bessel equation of order 0.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: power series solution to the bessel equation<\/h3>\r\nFind a power series solution to the Bessel equation of order [latex]0[\/latex] and graph the solution.\r\n\r\n[reveal-answer q=\"314159296\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"314159296\"]\r\n<p id=\"fs-id1170573506395\">The Bessel equation of order\u00a0[latex]0[\/latex] is given by<\/p>\r\n<p style=\"text-align: center;\">[latex]x^2y^{\\prime\\prime}xy^\\prime+x^2y=0[\/latex].<\/p>\r\n<p id=\"fs-id1170573519636\">We assume a solution of the form [latex]y=\\displaystyle\\sum_{n=0}^\\infty a_nx^n[\/latex]. Then [latex]y^\\prime(x)=\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}[\/latex] and [latex]y^{\\prime\\prime}(x)=\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}[\/latex]. Substituting this into the differential equation, we get<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx^2\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}+x\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}+x^2\\displaystyle\\sum_{n=0}^\\infty a_nx^n&amp;=0 &amp;\\quad &amp;\\text{Substitution} \\\\\r\n\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^n+\\displaystyle\\sum_{n=1}^\\infty na_nx^n+\\displaystyle\\sum_{n=0}^\\infty a_nx^{n+2} &amp;=0 &amp;\\quad &amp;\\text{Bring external factors within sums} \\\\\r\n\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^n+\\displaystyle\\sum_{n=1}^\\infty na_nx^n+\\displaystyle\\sum_{n=2}^\\infty a_{n-2}x^n&amp;=0 &amp;\\quad &amp;\\text{Re-index third sum.} \\\\\r\n\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^n+a_1x+\\displaystyle\\sum_{n=2}^\\infty na_nx^n+\\displaystyle\\sum_{n=2}^\\infty a_{n-2}x^n&amp;=0 &amp;\\quad &amp;\\text{Separate }n=1\\text{ term from second sum} \\\\\r\na_1x+\\displaystyle\\sum_{n=2}^\\infty [n(n-1)a_n+na_n+a_{n-2}]x^n&amp;=0 &amp;\\quad &amp;\\text{Collect summation terms.} \\\\\r\na_1x+\\displaystyle\\sum_{n=2}^\\infty[(n^2-n)a_n+na_n+a_{n-2}]x^n&amp;=0 &amp;\\quad &amp;\\text{Multiply through first term.} \\\\\r\na_1x+\\displaystyle\\sum_{n=2}^\\infty [n^2a_n+a_{n-2}]x^n&amp;=0 &amp;\\quad &amp;\\text{Simplify.}\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1170570998410\">Then [latex]a_1=0[\/latex], and for [latex]n\\geq2[\/latex],<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nn^2a_n+a_{n-2}&amp;=0 \\\\\r\na_n=-\\frac{1}{n^2}a_{n-2}\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170573608117\">Because [latex]a_1=0[\/latex], all odd terms are zero. Then, for even values of [latex]n[\/latex], we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\na_2&amp;=-\\frac{1}{2^2}a_0 \\\\\r\na_4&amp;=-\\frac{1}{4^2}a_2=\\frac{1}{4^2\\cdot2^2}a_0 \\\\\r\na_6&amp;=-\\frac{1}{6^2}a_4=-\\frac{1}{6^2\\cdot4^2\\cdot2^2}a_0\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1170573551067\">In general,<\/p>\r\n<p style=\"text-align: center;\">[latex]a_{2k}=\\frac{(-1)^k}{(2)^{2k}(k!)^2}a_0[\/latex].<\/p>\r\n<p id=\"fs-id1170571284290\">Thus, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]y(x)=a_0\\displaystyle\\sum_{k=0}^\\infty\\frac{(-1)^k}{(2)^{2k}(k!)^2}x^{2k}[\/latex].<\/p>\r\n<p id=\"fs-id1170573525687\">The graph appears below.<\/p>\r\n\r\n\r\n[caption id=\"attachment_3296\" align=\"alignnone\" width=\"487\"]<img class=\"wp-image-3296 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18023459\/7-4-1.jpeg\" alt=\"This figure is the graph of a function. The graph is oscillating with the highest amplitude above the origin. The horizontal axis is labeled in increments of 2.5. The vertical axis is labeled in increments of 0.2.\" width=\"487\" height=\"316\" \/> Figure 1.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nVerify that the expression found in\u00a0Example \"Power Series Solution to the Bessel Equation\"\u00a0is a solution to the Bessel equation of order 0.\r\n\r\n<\/div>\r\n&nbsp;\r\n\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\">Use power series to solve first-order and second-order differential equations.<\/span><\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1170571376919\" class=\"problem-solving ui-has-child-title\" 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: finding power series solutions to differential equations<\/h3>\n<ol id=\"fs-id1170571248526\" type=\"1\">\n<li>Assume the differential equation has a solution of the form [latex]y(x)=\\displaystyle\\sum_{n=0}^\\infty a_nx^n[\/latex].<\/li>\n<li>Differentiate the power series term by term to get [latex]y^\\prime(x)=\\displaystyle\\sum_{n=0}^\\infty na_nx^{n-1}[\/latex] and\u00a0[latex]y^{\\prime\\prime}(x)=\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}[\/latex].<\/li>\n<li>Substitute the power series expressions into the differential equation.<\/li>\n<li>Re-index sums as necessary to combine terms and simplify the expression.<\/li>\n<li>Equate coefficients of like powers of [latex]x[\/latex] to determine values for the coefficients [latex]a_n[\/latex] in the power series.<\/li>\n<li>Substitute the coefficients back into the power series and write the solution.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: series solutions to differential equations<\/h3>\n<p id=\"fs-id1170573434701\">Find a power series solution for the following differential equations.<\/p>\n<ol id=\"fs-id1170573437980\" type=\"a\">\n<li>[latex]y^{\\prime\\prime}-y=0[\/latex]<\/li>\n<li>[latex](x^2-1)y^{\\prime\\prime}+6xy^\\prime+4y=-4[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q226627144\">Show Solution<\/span><\/p>\n<div id=\"q226627144\" class=\"hidden-answer\" style=\"display: none\">\n<ol type=\"a\">\n<li style=\"list-style-type: none;\">\n<ol type=\"a\">\n<li>Assume [latex]y(x)=\\displaystyle\\sum_{n=0}^\\infty a_nx^n[\/latex] (step 1). Then, [latex]y^\\prime(x)=\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}[\/latex] and [latex]y^{\\prime\\prime}(x)=\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}[\/latex] (step 2). We want to find values for the coefficients [latex]a_n[\/latex] such that<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y^{\\prime\\prime}-y&=0 \\\\  \\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}-\\displaystyle\\sum_{n=0}^\\infty a_nx^n&=0  \\end{aligned}[\/latex] (step 3).<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with [latex]n=0[\/latex].\u00a0<span data-type=\"newline\"><br \/>\n<\/span>To re-index the first term, replace [latex]n[\/latex] with [latex]n+2[\/latex] inside the sum, and change the lower summation limit to [latex]n=0[\/latex]. We get<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}=\\displaystyle\\sum_{n=0}^\\infty(n+2)(n+1)a_{n+2}x^n[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>This gives<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\sum_{n=0}^\\infty(n+2)(n+1)a_{n+2}x^n-\\displaystyle\\sum_{n=0}^\\infty a_nx^n&=0 \\\\  \\displaystyle\\sum_{n=0}^\\infty[(n+2)(n+1)a_{n+2}-a_n]x^n&=0 \\text{ (step 4)}.  \\end{aligned}[\/latex]<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of [latex]x[\/latex] are zero. So we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{(n+2)(n+1)a_{n+2}-a_n=0\\text{ for }n=0,1,2,\\ldots}[\/latex]<\/p>\n<p>This recurrence relationship allows us to express each coefficient [latex]a_n[\/latex] in terms of the coefficient two terms earlier. This yields one expression for even values of [latex]n[\/latex] and another expression for odd values of [latex]n[\/latex]. Looking first at the equations involving even values of [latex]n[\/latex], we see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  a_2&=\\frac{a_0}2 \\\\  a_4&=\\frac{a_2}{4\\cdot3}=\\frac{a_0}{4!} \\\\  a_6&=\\frac{a_4}{6\\cdot5}=\\frac{a_0}{6!} \\\\  &\\vdots  \\end{aligned}[\/latex]<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Thus, in general, when [latex]n[\/latex] is even, [latex]a_n=\\frac{a_0}{n!}[\/latex] (step 5).<span data-type=\"newline\"><br \/>\n<\/span>For the equations involving odd values of [latex]n[\/latex], we see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  a_3&=\\frac{a_1}{3\\cdot2}=\\frac{a_1}{3!} \\\\  a_5&=\\frac{a_3}{5\\cdot4}=\\frac{a_1}{5!} \\\\  a_7&=\\frac{a_5}{7\\cdot6}=\\frac{a_1}{7!} \\\\  &\\vdots  \\end{aligned}[\/latex]<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Therefore, in general, when [latex]n[\/latex] is odd, [latex]a_n=\\frac{a_1}{n!}[\/latex] (step 5 continued).<span data-type=\"newline\"><br \/>\n<\/span>Putting this together, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y(x)&=\\displaystyle\\sum_{n=0}^\\infty a_nx^n \\\\  &=a_0+a_1x+\\frac{a_0}2x^2+\\frac{a_1}{3!}x^3+\\frac{a_0}{4!}x^4+\\frac{a_1}{5!}x^5+\\cdots  \\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Re-indexing the sums to account for the even and odd values of [latex]n[\/latex] separately, we obtain<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]y(x)=a_0\\displaystyle\\sum_{k=0}^\\infty\\frac{1}{(2k)!}x^{2k}+a_1\\displaystyle\\sum_{k=0}^\\infty\\frac{1}{(2k+1)!}x^{2k+1}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span><strong data-effect=\"bold\">Analysis for part a.<\/strong><span data-type=\"newline\"><br \/>\n<\/span>As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since [latex]y(x)=c_1e^x+c_2e^{-x}[\/latex] is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.<span data-type=\"newline\"><br \/>\n<\/span>Fortunately, after writing the power series representations of [latex]e^x[\/latex] and [latex]e^{-x}[\/latex], and doing some algebra, we find that if we choose<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{c_0=\\frac{(a_0+a_1)}2, \\ c_1=\\frac{(a_0-a_1)}2}[\/latex],<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>we then have [latex]a_0=c_0+c_1[\/latex] and [latex]a_1=c_0-c_1[\/latex], and<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  y(x)&=a_0+a_1x+\\frac{a_0}2x^2+\\frac{a_1}{3!}x^3+\\frac{a_0}{4!}x^4+\\frac{a_0}{5!}x^5+\\ldots \\\\  &=(c_0+c_1)+(c_0-c_1)x+\\frac{(c_0+c_1)}2x^2+\\frac{(c_0-c_1)}{3!}x^3+\\frac{(c_0+c_1)}{4!}x^4+\\frac{(c_0-c_1)}{5!}x^5+\\ldots \\\\  &=c_0\\displaystyle\\sum_{n=0}^\\infty\\frac{x^n}{n!}+c_1\\displaystyle\\sum_{n=0}^\\infty\\frac{(-x)^n}{n!} \\\\  &=c_0e^x+c_1e^{-x}  \\end{aligned}[\/latex].<\/p>\n<div id=\"fs-id1170571232470\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div class=\"MathJax_Display\"><\/div>\n<\/div>\n<p>So we have, in fact, found the same general solution. Note that this choice of [latex]c_1[\/latex] and\u00a0[latex]c_2[\/latex] is not obvious. This is a case when we know what the answer should be, and have essentially \u201creverse-engineered\u201d our choice of coefficients.<\/p>\n<ul>\n<li>Assume [latex]y(x)=\\displaystyle\\sum_{n=0}^\\infty a_nx^n[\/latex] (step 1). Then, [latex]y^\\prime(x)=\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}[\/latex] and [latex]y^{\\prime\\prime}=\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}[\/latex] (step 2). We want to find values for the coefficients [latex]a_n[\/latex] such that<\/li>\n<\/ul>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  (x^2-1)y^{\\prime\\prime}+6xy^\\prime+4y&=-4 \\\\  (x^2-1)\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}+6x\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}+4\\displaystyle\\sum_{n=0}^\\infty a_nx^n&=-4 \\\\  x^2\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}-\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}+6x\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}+4\\displaystyle\\sum_{n=0}^\\infty a_nx^n&=-4  \\end{aligned}[\/latex].<\/p>\n<p>Taking the external factors inside the summations, we get<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n}-\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}+\\displaystyle\\sum_{n=1}^\\infty 6na_nx^n+\\displaystyle\\sum_{n=0}^\\infty 4a_nx^n=-4[\/latex] (step 3).<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Now, in the first summation, we see that when [latex]n=0[\/latex] or[latex]n=1[\/latex], the term evaluates to zero, so we can add these terms back into our sum to get<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n}=\\displaystyle\\sum_{n=0}^\\infty n(n-1)a_nx^{n}[\/latex].<\/p>\n<p>Similarly, in the third term, we see that when [latex]n=0[\/latex], the expression evaluates to zero, so we can add that term back in as well. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=1}^\\infty 6na_nx^n=\\displaystyle\\sum_{n=0}^\\infty 6na_nx^n[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Then, we need only shift the indices in our second term. We get<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}=\\displaystyle\\sum_{n=0}^\\infty(n+2)(n+1)a_{n+2}x^n[\/latex]<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Thus, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\displaystyle\\sum_{n=0}^\\infty n(n-1)a_nx^{n}-\\displaystyle\\sum_{n=0}^\\infty(n+2)(n+1)a_{n+2}x^n+\\displaystyle\\sum_{n=0}^\\infty6na_nx^n+\\displaystyle\\sum_{n=0}^\\infty4a_nx^n&=-4\\text{ (step 4).} \\\\  \\displaystyle\\sum_{n=0}^\\infty[n(n-1)a_n-(n+2)(n+1)a_{n+2}6na_n+4a_n]x^n&=-4 \\\\  \\displaystyle\\sum_{n=0}^\\infty[(n^2-n)a_n+6na_n+4a_n-(n+2)(n+1)a_{n+2}]x^n&=-4 \\\\  \\displaystyle\\sum_{n=0}^\\infty[n^2a_n+5na_n+4a_n-(n+2)(n+1)a_{n+2}]x^n&=-4 \\\\  \\displaystyle\\sum_{n=0}^\\infty[(n^2+5n+4)a_n-(n+2)(n+1)a_{n+2}]x^n&=-4 \\\\  \\displaystyle\\sum_{n=0}^\\infty[(n+4)(n+1)a_n-(n+2)(n+1)a_{n+2}]x^n&=-4  \\end{aligned}[\/latex]<\/p>\n<p>Looking at the coefficients of each power of [latex]x[\/latex], we see that the constant term must be equal to [latex]-4[\/latex] and the coefficients of all other powers of [latex]x[\/latex] must be zero. Then, looking first at the constant term,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  4a_0-2a_2&=-4 \\\\  a_2&=2a_0+2  \\end{aligned}[\/latex] (step 3).<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>For [latex]n\\geq1[\/latex], we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  (n+4)(n+1)a_n-(n+2)(n+1)a_{a+2}&=0 \\\\  (n+1)[(n+4)a_n-(n+2)a_{n+2}]&=0  \\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Since [latex]n\\geq1[\/latex], [latex]n+1\\ne0[\/latex], we see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex](n+4)a_n-(n+2)a_{n+2}=0[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>and thus<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]a_{n+2}=\\frac{n+4}{n+2}a_n[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>For even values of [latex]n[\/latex], we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  a_4&=\\frac64(2a_0+2)=3a_0+3 \\\\  a_6&=\\frac86(3a_0+3)=4a_0+4 \\\\  &\\vdots  \\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>In general, [latex]a_{2k}=(k+1)(a_0+1)[\/latex] (step 5).<span data-type=\"newline\"><br \/>\n<\/span>For odd values of [latex]n[\/latex], we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  a_3&=\\frac53a_1 \\\\  a_5&=\\frac75a_3=\\frac73a_1 \\\\  a_7&=\\frac97a_5=\\frac93a_1=3a_1 \\\\  &\\vdots  \\end{aligned}[\/latex].<\/p>\n<p><span data-type=\"newline\"><br \/>\n<\/span>In general, [latex]a_{2k+1}=\\frac{2k+3}3a_1[\/latex] (step 5 continued).<span data-type=\"newline\"><br \/>\n<\/span>Putting this together, we have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]y(x)=\\displaystyle\\sum_{k=0}^\\infty (k+1)(a_0+1)x^{2k}+\\displaystyle\\sum_{k=0}^\\infty\\left(\\frac{2k+3}3\\right)a_1x^{2k+1}[\/latex] (step 6).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-id1170571291319\">Find a power series solution for the following differential equations.<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]y^{\\prime}+2xy=0[\/latex]<\/li>\n<li>[latex](x+1)y^\\prime=3y[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q652781369\">Show Solution<\/span><\/p>\n<div id=\"q652781369\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. [latex]y(x)=a_0\\displaystyle\\sum_{n=0}^\\infty\\frac{(-1)^n}{n!}x^{2n}=a_0e^{-x^2}[\/latex]<\/p>\n<p>b. [latex]y(x)=a_0(x+1)^3[\/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=8250347&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=d5VG6PM48vQ&amp;video_target=tpm-plugin-7w3p47zd-d5VG6PM48vQ\" 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\/CP7.22a_transcript.html\">transcript for \u201cCP 7.22a\u201d here (opens in new window).<\/a><\/div>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8250348&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=2ukZj1dveVs&amp;video_target=tpm-plugin-lphrhkde-2ukZj1dveVs\" 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\/CP7.22b_transcript.html\">transcript for \u201cCP 7.22b\u201d here (opens in new window).<\/a><\/div>\n<p id=\"fs-id1170573391386\">We close this section with a brief introduction to\u00a0<span id=\"1f5ff083-3081-490b-b429-61fc1959738c_term323\" class=\"no-emphasis\" data-type=\"term\">Bessel functions<\/span>. Complete treatment of Bessel functions is well beyond the scope of this course, but we get a little taste of the topic here so we can see how series solutions to differential equations are used in real-world applications. The Bessel equation of order [latex]n[\/latex] is given by<\/p>\n<p style=\"text-align: center;\">[latex]x^2y^{\\prime\\prime}+xy^\\prime+(x^2-n^2)y=0[\/latex].<\/p>\n<p id=\"fs-id1170571370989\">This equation arises in many physical applications, particularly those involving cylindrical coordinates, such as the vibration of a circular drum head and transient heating or cooling of a cylinder. In the next example, we find a power series solution to the Bessel equation of order 0.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: power series solution to the bessel equation<\/h3>\n<p>Find a power series solution to the Bessel equation of order [latex]0[\/latex] and graph the solution.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q314159296\">Show Solution<\/span><\/p>\n<div id=\"q314159296\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1170573506395\">The Bessel equation of order\u00a0[latex]0[\/latex] is given by<\/p>\n<p style=\"text-align: center;\">[latex]x^2y^{\\prime\\prime}xy^\\prime+x^2y=0[\/latex].<\/p>\n<p id=\"fs-id1170573519636\">We assume a solution of the form [latex]y=\\displaystyle\\sum_{n=0}^\\infty a_nx^n[\/latex]. Then [latex]y^\\prime(x)=\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}[\/latex] and [latex]y^{\\prime\\prime}(x)=\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}[\/latex]. Substituting this into the differential equation, we get<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x^2\\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^{n-2}+x\\displaystyle\\sum_{n=1}^\\infty na_nx^{n-1}+x^2\\displaystyle\\sum_{n=0}^\\infty a_nx^n&=0 &\\quad &\\text{Substitution} \\\\  \\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^n+\\displaystyle\\sum_{n=1}^\\infty na_nx^n+\\displaystyle\\sum_{n=0}^\\infty a_nx^{n+2} &=0 &\\quad &\\text{Bring external factors within sums} \\\\  \\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^n+\\displaystyle\\sum_{n=1}^\\infty na_nx^n+\\displaystyle\\sum_{n=2}^\\infty a_{n-2}x^n&=0 &\\quad &\\text{Re-index third sum.} \\\\  \\displaystyle\\sum_{n=2}^\\infty n(n-1)a_nx^n+a_1x+\\displaystyle\\sum_{n=2}^\\infty na_nx^n+\\displaystyle\\sum_{n=2}^\\infty a_{n-2}x^n&=0 &\\quad &\\text{Separate }n=1\\text{ term from second sum} \\\\  a_1x+\\displaystyle\\sum_{n=2}^\\infty [n(n-1)a_n+na_n+a_{n-2}]x^n&=0 &\\quad &\\text{Collect summation terms.} \\\\  a_1x+\\displaystyle\\sum_{n=2}^\\infty[(n^2-n)a_n+na_n+a_{n-2}]x^n&=0 &\\quad &\\text{Multiply through first term.} \\\\  a_1x+\\displaystyle\\sum_{n=2}^\\infty [n^2a_n+a_{n-2}]x^n&=0 &\\quad &\\text{Simplify.}  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1170570998410\">Then [latex]a_1=0[\/latex], and for [latex]n\\geq2[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  n^2a_n+a_{n-2}&=0 \\\\  a_n=-\\frac{1}{n^2}a_{n-2}  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170573608117\">Because [latex]a_1=0[\/latex], all odd terms are zero. Then, for even values of [latex]n[\/latex], we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  a_2&=-\\frac{1}{2^2}a_0 \\\\  a_4&=-\\frac{1}{4^2}a_2=\\frac{1}{4^2\\cdot2^2}a_0 \\\\  a_6&=-\\frac{1}{6^2}a_4=-\\frac{1}{6^2\\cdot4^2\\cdot2^2}a_0  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1170573551067\">In general,<\/p>\n<p style=\"text-align: center;\">[latex]a_{2k}=\\frac{(-1)^k}{(2)^{2k}(k!)^2}a_0[\/latex].<\/p>\n<p id=\"fs-id1170571284290\">Thus, we have<\/p>\n<p style=\"text-align: center;\">[latex]y(x)=a_0\\displaystyle\\sum_{k=0}^\\infty\\frac{(-1)^k}{(2)^{2k}(k!)^2}x^{2k}[\/latex].<\/p>\n<p id=\"fs-id1170573525687\">The graph appears below.<\/p>\n<div id=\"attachment_3296\" style=\"width: 497px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-3296\" class=\"wp-image-3296 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/18023459\/7-4-1.jpeg\" alt=\"This figure is the graph of a function. The graph is oscillating with the highest amplitude above the origin. The horizontal axis is labeled in increments of 2.5. The vertical axis is labeled in increments of 0.2.\" width=\"487\" height=\"316\" \/><\/p>\n<p id=\"caption-attachment-3296\" class=\"wp-caption-text\">Figure 1.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Verify that the expression found in\u00a0Example &#8220;Power Series Solution to the Bessel Equation&#8221;\u00a0is a solution to the Bessel equation of order 0.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\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-1334\">\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 7.22a. <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 7.22b. <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":15,"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 7.22a\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 7.22b\",\"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-1334","chapter","type-chapter","status-publish","hentry"],"part":25,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1334","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":40,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1334\/revisions"}],"predecessor-version":[{"id":6402,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1334\/revisions\/6402"}],"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\/1334\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1334"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1334"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1334"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1334"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}