{"id":959,"date":"2025-06-20T17:25:03","date_gmt":"2025-06-20T17:25:03","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&p=959"},"modified":"2025-08-27T12:47:22","modified_gmt":"2025-08-27T12:47:22","slug":"applications-of-series-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/applications-of-series-learn-it-3\/","title":{"raw":"Applications of Series: Learn It 3","rendered":"Applications of Series: Learn It 3"},"content":{"raw":"

Solving Differential Equations with Power Series<\/h2>\r\n

Consider the differential equation<\/p>\r\n\r\n

[latex]{y}^{\\prime }\\left(x\\right)=y[\/latex].<\/div>\r\nThis is a first-order separable equation with solution [latex]y=C{e}^{x}[\/latex]. We can solve this using techniques from earlier chapters.\r\n\r\nFor most differential equations, however, we do not yet have analytical tools to solve them. Power series are an extremely useful tool for solving many types of differential equations. In this technique, we look for a solution of the form [latex]y=\\displaystyle\\sum_{n=0}^{\\infty}{c}_{n}{x}^{n}[\/latex] and determine what the coefficients need to be.\r\n\r\nIn the next example, we consider an initial-value problem involving [latex]{y}^{\\prime}=y[\/latex] to illustrate the technique.\r\n\r\n
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Use power series to solve the initial-value problem<\/p>\r\n\r\n

[latex]{y}^{\\prime }=y,y\\left(0\\right)=3[\/latex].<\/div>\r\n \r\n\r\n<\/div>\r\n[reveal-answer q=\"44558890\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558890\"]\r\n
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Suppose that there exists a power series solution<\/p>\r\n\r\n

[latex]y\\left(x\\right)=\\displaystyle\\sum _{n=0}^{\\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\\cdots [\/latex].<\/div>\r\n \r\n

Differentiating this series term by term, we obtain<\/p>\r\n\r\n

[latex]{y}^{\\prime }={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{4}{x}^{3}+\\cdots [\/latex].<\/div>\r\n \r\n

If y<\/em> satisfies the differential equation, then<\/p>\r\n\r\n

[latex]{c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\\cdots ={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+4{c}_{3}{x}^{3}+\\cdots [\/latex].<\/div>\r\n \r\n

Using [link] on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,<\/p>\r\n\r\n

[latex]\\begin{array}{c}{c}_{0}={c}_{1},\\hfill \\\\ {c}_{1}=2{c}_{2},\\hfill \\\\ {c}_{2}=3{c}_{3},\\hfill \\\\ {c}_{3}=4{c}_{4},\\hfill \\\\ \\hfill \\vdots.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

Using the initial condition [latex]y\\left(0\\right)=3[\/latex] combined with the power series representation<\/p>\r\n\r\n

[latex]y\\left(x\\right)={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\\cdots [\/latex],<\/div>\r\n \r\n

we find that [latex]{c}_{0}=3[\/latex]. We are now ready to solve for the rest of the coefficients. Using the fact that [latex]{c}_{0}=3[\/latex], we have<\/p>\r\n\r\n

[latex]\\begin{array}{}\\\\ \\\\ {c}_{1}={c}_{0}=3=\\frac{3}{1\\text{!}},\\hfill \\\\ {c}_{2}=\\frac{{c}_{1}}{2}=\\frac{3}{2}=\\frac{3}{2\\text{!}},\\hfill \\\\ {c}_{3}=\\frac{{c}_{2}}{3}=\\frac{3}{3\\cdot 2}=\\frac{3}{3\\text{!}},\\hfill \\\\ {c}_{4}=\\frac{{c}_{3}}{4}=\\frac{3}{4\\cdot 3\\cdot 2}=\\frac{3}{4\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

Therefore,<\/p>\r\n\r\n

[latex]\\begin{array}{cc}\\hfill y& =3\\left[1+\\frac{1}{1\\text{!}}x+\\frac{1}{2\\text{!}}{x}^{2}+\\frac{1}{3\\text{!}}{x}^{3}\\frac{1}{4\\text{!}}{x}^{4}+\\cdots \\right]\\hfill \\\\ & =3{\\displaystyle\\sum _{n=0}^{\\infty}}\\frac{{x}^{n}}{n\\text{!}}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n

You might recognize<\/p>\r\n\r\n

[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{{x}^{n}}{n\\text{!}}[\/latex]<\/div>\r\n \r\n

as the Taylor series for [latex]{e}^{x}[\/latex]. Therefore, the solution is [latex]y=3{e}^{x}[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Watch the following video to see the worked solution to the example above.