{"id":830,"date":"2025-10-09T15:42:05","date_gmt":"2025-10-09T15:42:05","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=830"},"modified":"2025-11-11T17:46:08","modified_gmt":"2025-11-11T17:46:08","slug":"series-solutions-of-differential-equations-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/series-solutions-of-differential-equations-learn-it-1\/","title":{"raw":"Series Solutions of Differential Equations: Learn It 1","rendered":"Series Solutions of Differential Equations: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Solve differential equations using power series<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 id=\"fs-id1170571376919\" class=\"problem-solving ui-has-child-title\" data-type=\"note\">Series Solutions of Differential Equations<\/h2>\r\nFunctions can be represented as power series, [latex]y(x)=\\displaystyle\\sum_{n=0}^{\\infty} a_{n}x^{n}[\/latex]. We can find series representations of the derivatives of such functions by differentiating the power series term by term. This gives [latex]y^{\\prime}(x)=\\displaystyle\\sum_{n=1}^{\\infty} na_{n}x^{n-1}[\/latex] and [latex]y^{\\prime\\prime}(x)=\\displaystyle\\sum_{n=2}^{\\infty} n(n-1)a_{n}x^{n-2}[\/latex]. In some cases, these power series representations can be used to find solutions to differential equations.\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">This text provides only a brief treatment of power series solutions. Most introductory differential equations textbooks include an entire module on this topic. Several important issues are not addressed here, particularly issues related to existence of solutions. The examples and exercises were chosen for which power solutions exist. However, it is not always the case that power solutions exist.<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Find Power Series Solutions to Differential Equations<\/strong>\r\n<ol id=\"fs-id1170571248526\" type=\"1\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Assume a power series solution<\/strong>: [latex]y(x)=\\displaystyle\\sum_{n=0}^\\infty a_nx^n[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Differentiate term by term<\/strong>: Get [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].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Substitute into the differential equation<\/strong>: Replace [latex]y[\/latex], [latex]y^\\prime[\/latex], and [latex]y^{\\prime\\prime}[\/latex] with their power series expressions.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Re-index and combine terms<\/strong>: Re-index sums as necessary to combine terms and simplify the expression.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Equate coefficients<\/strong>: Set coefficients of like powers of [latex]x[\/latex] equal to determine values for the coefficients [latex]a_n[\/latex].<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Write the solution<\/strong>: Substitute the coefficients back into the power series.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\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>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\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.<\/li>\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\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<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3>Bessel Functions<\/h3>\r\n<p class=\"whitespace-normal break-words\">We close with a brief introduction to Bessel functions. Complete treatment of Bessel functions is well beyond the scope of this course, but we get a taste of the topic here so we can see how series solutions to differential equations are used in real-world applications.<\/p>\r\n<p class=\"whitespace-normal break-words\">The <strong>Bessel equation of order [latex]n[\/latex]<\/strong> 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\nThis 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.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">Find a power series solution to the Bessel equation of order [latex]0[\/latex] and graph the solution.[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=\"aligncenter\" 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<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Solve differential equations using power series<\/li>\n<\/ul>\n<\/section>\n<h2 id=\"fs-id1170571376919\" class=\"problem-solving ui-has-child-title\" data-type=\"note\">Series Solutions of Differential Equations<\/h2>\n<p>Functions can be represented as power series, [latex]y(x)=\\displaystyle\\sum_{n=0}^{\\infty} a_{n}x^{n}[\/latex]. We can find series representations of the derivatives of such functions by differentiating the power series term by term. This gives [latex]y^{\\prime}(x)=\\displaystyle\\sum_{n=1}^{\\infty} na_{n}x^{n-1}[\/latex] and [latex]y^{\\prime\\prime}(x)=\\displaystyle\\sum_{n=2}^{\\infty} n(n-1)a_{n}x^{n-2}[\/latex]. In some cases, these power series representations can be used to find solutions to differential equations.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">This text provides only a brief treatment of power series solutions. Most introductory differential equations textbooks include an entire module on this topic. Several important issues are not addressed here, particularly issues related to existence of solutions. The examples and exercises were chosen for which power solutions exist. However, it is not always the case that power solutions exist.<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How to: Find Power Series Solutions to Differential Equations<\/strong><\/p>\n<ol id=\"fs-id1170571248526\" type=\"1\">\n<li class=\"whitespace-normal break-words\"><strong>Assume a power series solution<\/strong>: [latex]y(x)=\\displaystyle\\sum_{n=0}^\\infty a_nx^n[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Differentiate term by term<\/strong>: Get [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].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Substitute into the differential equation<\/strong>: Replace [latex]y[\/latex], [latex]y^\\prime[\/latex], and [latex]y^{\\prime\\prime}[\/latex] with their power series expressions.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Re-index and combine terms<\/strong>: Re-index sums as necessary to combine terms and simplify the expression.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Equate coefficients<\/strong>: Set coefficients of like powers of [latex]x[\/latex] equal to determine values for the coefficients [latex]a_n[\/latex].<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Write the solution<\/strong>: Substitute the coefficients back into the power series.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q226627144\">Show Solution<\/button><\/p>\n<div id=\"q226627144\" class=\"hidden-answer\" style=\"display: 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\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.<\/li>\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\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<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<h3>Bessel Functions<\/h3>\n<p class=\"whitespace-normal break-words\">We close with a brief introduction to Bessel functions. Complete treatment of Bessel functions is well beyond the scope of this course, but we get a taste of the topic here so we can see how series solutions to differential equations are used in real-world applications.<\/p>\n<p class=\"whitespace-normal break-words\">The <strong>Bessel equation of order [latex]n[\/latex]<\/strong> 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>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.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q314159296\">Show Solution<\/button><\/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<figure id=\"attachment_3296\" aria-describedby=\"caption-attachment-3296\" style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption id=\"caption-attachment-3296\" class=\"wp-caption-text\">Figure 1.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":23,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-830","chapter","type-chapter","status-publish","hentry"],"part":440,"module-header":"learn_it","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\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/830","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/830\/revisions"}],"predecessor-version":[{"id":2161,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/830\/revisions\/2161"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/440"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/830\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=830"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=830"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=830"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=830"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}