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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:

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