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Q18 E

Expert-verified
Found in: Page 450

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

In Problems 13-19, find at least the first four nonzero terms in a power series expansion of the solution to the given initial value problem.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{\left(}}{\mathbf{cosx}}{\mathbf{\right)}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}\phantom{\rule{0ex}{0ex}}{\mathbf{y}}{\mathbf{\left(}}{\mathbf{\pi }}{\mathbf{/}}{\mathbf{2}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}{\mathbf{,}}{\mathbf{}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\left(}}{\mathbf{\pi }}{\mathbf{/}}{\mathbf{2}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}$

The first four nonzero terms in the power series expansion of the given initial value problem $\mathrm{y}\text{'}\text{'}-\left(\mathrm{cosx}\right)\mathrm{y}\text{'}-\mathrm{y}=0$ is $\mathrm{Y}\left(\mathrm{x}\right)=1+\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)+\frac{1}{2}{\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}^{2}+\frac{1}{3}{\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}^{3}+\cdots$

See the step by step solution

Step 1: Define power series expansion.

The power series approach is used in mathematics to find a power series solution to certain differential equations. In general, such a solution starts with an unknown power series and then plugs that solution into the differential equation to obtain a coefficient recurrence relation.

A differential equation's power series solution is a function with an infinite number of terms, each holding a different power of the dependent variable. It is generally given by the formula,

${\mathbf{y}}{\mathbf{\left(}}{\mathbf{x}}{\mathbf{\right)}}{\mathbf{=}}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{{\mathbf{a}}}_{{\mathbf{n}}}{{\mathbf{x}}}^{{\mathbf{n}}}$

Step 2: Find the relation.

Given,

$\mathrm{y}\text{'}\text{'}-\left(\mathrm{cosx}\right)\mathrm{y}\text{'}-\mathrm{y}=0\phantom{\rule{0ex}{0ex}}\mathrm{y}\left(\mathrm{\pi }/2\right)=1,\mathrm{y}\text{'}\left(\mathrm{\pi }/2\right)=1$

Apply a substitution and transform the equation,

$\mathrm{y}\text{'}\text{'}-\mathrm{cos}\left(\mathrm{t}+\frac{\mathrm{\pi }}{2}\right)·\mathrm{y}\text{'}-\mathrm{y}=0\phantom{\rule{0ex}{0ex}}\mathrm{y}\text{'}\text{'}-\mathrm{sint}·\mathrm{y}\text{'}-\mathrm{y}=0$

Use the formula

$\mathrm{Y}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}\phantom{\rule{0ex}{0ex}}\mathrm{Y}\text{'}\left(\mathrm{t}\right)=\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}\phantom{\rule{0ex}{0ex}}\mathrm{Y}\text{'}\text{'}\left(\mathrm{t}\right)=\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2}$

Substitute it in the above equation we get,

$\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2}-\left(\mathrm{t}-\frac{{\mathrm{t}}^{3}}{3!}+\frac{{\mathrm{t}}^{5}}{5!}-\frac{{\mathrm{t}}^{7}}{7!}+\cdots \right)\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}-\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0$

Hence we get the relation:

$\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right)·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-2}-\left(\mathrm{t}-\frac{{\mathrm{t}}^{3}}{3!}+\frac{{\mathrm{t}}^{5}}{5!}-\frac{{\mathrm{t}}^{7}}{7!}+\cdots \right)\sum _{\mathrm{n}=1}^{\infty }\mathrm{n}·{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}-1}-\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0$.

Step 3: Find the expression after expansion.

The series expansion for the function is

$\left(2{\mathrm{a}}_{2}+6{\mathrm{a}}_{3}\mathrm{t}+12{\mathrm{a}}_{4}{\mathrm{t}}^{2}+20{\mathrm{a}}_{5}{\mathrm{t}}^{3}+\cdots \right)-\left({\mathrm{a}}_{1}\mathrm{t}+2{\mathrm{a}}_{2}{\mathrm{t}}^{2}+3{\mathrm{a}}_{3}{\mathrm{t}}^{3}+4{\mathrm{a}}_{4}{\mathrm{t}}^{4}+\cdots \right)+\left({\mathrm{a}}_{1}\frac{{\mathrm{t}}^{3}}{3!}+2{\mathrm{a}}_{2}\frac{{\mathrm{t}}^{4}}{3!}+3{\mathrm{a}}_{3}\frac{{\mathrm{t}}^{5}}{3!}+4{\mathrm{a}}_{4}\frac{{\mathrm{t}}^{6}}{3!}+\cdots \right)\phantom{\rule{0ex}{0ex}}-\left({\mathrm{a}}_{1}\frac{{\mathrm{t}}^{5}}{5!}+2{\mathrm{a}}_{2}\frac{{\mathrm{t}}^{6}}{5!}+3{\mathrm{a}}_{3}\frac{{\mathrm{t}}^{7}}{5!}+4{\mathrm{a}}_{4}\frac{{\mathrm{t}}^{8}}{5!}+\cdots \right)+\cdots -\left({\mathrm{a}}_{0}+{\mathrm{a}}_{1}\mathrm{t}+{\mathrm{a}}_{2}{\mathrm{t}}^{2}+{\mathrm{a}}_{3}{\mathrm{t}}^{3}+{\mathrm{a}}_{4}{\mathrm{t}}^{4}+\cdots \right)=0$

Taking coefficients and exponents of the same power.

$\left(2{\mathrm{a}}_{2}-{\mathrm{a}}_{0}\right)+\left(6{\mathrm{a}}_{3}-{\mathrm{a}}_{1}-{\mathrm{a}}_{1}\right)\mathrm{t}+\left(12{\mathrm{a}}_{4}-2{\mathrm{a}}_{2}-{\mathrm{a}}_{2}\right){\mathrm{t}}^{2}+\left(20{\mathrm{a}}_{5}-3{\mathrm{a}}_{3}+\frac{{\mathrm{a}}_{1}}{6}-{\mathrm{a}}_{3}\right){\mathrm{t}}^{3}+\cdots =0$

Simplify the expression:

$\left(2{\mathrm{a}}_{2}-{\mathrm{a}}_{0}\right)+\left(6{\mathrm{a}}_{3}-{\mathrm{a}}_{1}-{\mathrm{a}}_{1}\right)\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)+\left(12{\mathrm{a}}_{4}-2{\mathrm{a}}_{2}-{\mathrm{a}}_{2}\right){\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}^{2}+\left(20{\mathrm{a}}_{5}-3{\mathrm{a}}_{3}+\frac{{\mathrm{a}}_{1}}{6}-{\mathrm{a}}_{3}\right){\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}^{3}+\cdots =0$

Hence, the expression after the expansion is:

$\left(2{\mathrm{a}}_{2}-{\mathrm{a}}_{0}\right)+\left(6{\mathrm{a}}_{3}-{\mathrm{a}}_{1}-{\mathrm{a}}_{1}\right)\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)+\left(12{\mathrm{a}}_{4}-2{\mathrm{a}}_{2}-{\mathrm{a}}_{2}\right){\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}^{2}+\left(20{\mathrm{a}}_{5}-3{\mathrm{a}}_{3}+\frac{{\mathrm{a}}_{1}}{6}-{\mathrm{a}}_{3}\right){\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}^{3}+\cdots =0$

Step 4: Find the first four nonzero terms.

By equating the coefficients, we get,

$2{\mathrm{a}}_{2}-{\mathrm{a}}_{0}=0\to {\mathrm{a}}_{2}=\frac{{\mathrm{a}}_{0}}{2}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}6{\mathrm{a}}_{3}-{\mathrm{a}}_{1}-{\mathrm{a}}_{1}\to {\mathrm{a}}_{3}=\frac{{\mathrm{a}}_{1}}{3}=\frac{1}{3}$

The general solution was

$\mathrm{Y}\left(\mathrm{t}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}={\mathrm{a}}_{0}+{\mathrm{a}}_{1}\mathrm{t}+{\mathrm{a}}_{2}{\mathrm{t}}^{2}+{\mathrm{a}}_{3}{\mathrm{t}}^{3}+\cdots$

Apply the initial condition and substitute the coefficient.

$\mathrm{Y}\left(\mathrm{x}\right)=1+\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)+\frac{1}{2}{\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}^{2}+\frac{1}{3}{\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}^{3}+\cdots$

Hence, the first four nonzero terms are:

$\mathrm{Y}\left(\mathrm{x}\right)=1+\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)+\frac{1}{2}{\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}^{2}+\frac{1}{3}{\left(\mathrm{x}-\frac{\mathrm{\pi }}{2}\right)}^{3}+\cdots$

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