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Expert-verified Found in: Page 449 ### 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 # Find at least the first four non-zero terms in a power series expansion about x0 for a general solution to the given differential equation with the value for x0.$\left({x}^{2}-2x\right){\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}{\mathbf{;}}{\mathbf{}}{{\mathbf{x}}}_{{\mathbf{0}}}{\mathbf{=}}{\mathbf{1}}$

The first four nonzero terms in a power series expansion about x0 for a general solution:

$\mathrm{y}={\mathrm{a}}_{0}\left(1+{\left(\mathrm{x}+1\right)}^{2}+\frac{1}{3}{\left(\mathrm{x}-1\right)}^{4}+...\right)+{\mathrm{a}}_{1}\left(\left(\mathrm{x}-1\right)+\frac{1}{3}{\left(\mathrm{x}-1\right)}^{3}+...\right)$

See the step by step solution

## Step 1: Power series expansion

A power series expansion of can be obtained simply by expanding the exponential and integrating term-by-term. This series converges for all, but the convergence becomes extremely slow if significantly exceeds unity.

## Step 2: To determine the first four nonzero terms in a power series expansion about x0 for a general solution.

$\begin{array}{l}\left({\mathrm{x}}^{2}-2\mathrm{x}\right)\mathrm{y}\text{'}\text{'}+2\mathrm{y}=0\\ \mathrm{y}\text{'}\text{'}+\frac{2}{{\mathrm{x}}^{2}-2\mathrm{x}}\mathrm{y}=0\end{array}$

We get the value of $\mathrm{q}\left(\mathrm{x}\right)=\frac{2}{{\mathrm{x}}^{2}-2\mathrm{x}}$.

$\begin{array}{l}\mathrm{y}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}-1\right)}^{\mathrm{n}}\\ \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right){\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}-1\right)}^{\mathrm{n}-2}+2\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}-1\right)}^{\mathrm{n}}=0\end{array}$

Let, $\mathrm{x}-1=\mathrm{t}$:

role="math" localid="1664089919710" $\begin{array}{l}\left(\mathrm{t}+1\right)\left(\mathrm{t}-1\right)\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right){\mathrm{a}}_{\mathrm{n}}\left(\mathrm{t}{\right)}^{\mathrm{n}-2}+2\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0\\ \sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right){\mathrm{a}}_{\mathrm{n}}\left(\mathrm{t}{\right)}^{\mathrm{n}}-\sum _{\mathrm{n}=2}^{\infty }\mathrm{n}\left(\mathrm{n}-1\right){\mathrm{a}}_{\mathrm{n}}\left(\mathrm{t}{\right)}^{\mathrm{n}-2}+2\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{t}\right)}^{\mathrm{n}}=0\\ 2{\mathrm{a}}_{2}{\mathrm{t}}^{2}+6{\mathrm{a}}_{3}{\mathrm{t}}^{3}+12{\mathrm{a}}_{4}{\mathrm{t}}^{4}+20{\mathrm{a}}_{5}{\mathrm{t}}^{5}+...\\ 2{\mathrm{a}}_{0}+2{\mathrm{a}}_{1}\mathrm{t}+2{\mathrm{a}}_{2}{\mathrm{t}}^{2}+2{\mathrm{a}}_{3}{\mathrm{t}}^{3}+2{\mathrm{a}}_{4}{\mathrm{t}}^{4}+...=0\\ \left(2{\mathrm{a}}_{0}-2{\mathrm{a}}_{2}\right)+\left(2{\mathrm{a}}_{1}-6{\mathrm{a}}_{3}\right)\left(\mathrm{x}-1\right)+\left(2{\mathrm{a}}_{2}-12{\mathrm{a}}_{4}+2{\mathrm{a}}_{2}\right)\left(\mathrm{x}-1{\right)}^{2}+\left(2{\mathrm{a}}_{3}-20{\mathrm{a}}_{5}\right)\left(\mathrm{x}-1{\right)}^{3}+....=0\\ 2{\mathrm{a}}_{0}-2{\mathrm{a}}_{2}=0⇒{\mathrm{a}}_{2}={\mathrm{a}}_{0}\\ 2{\mathrm{a}}_{1}-6{\mathrm{a}}_{3}=0⇒{\mathrm{a}}_{3}=\frac{1}{3}{\mathrm{a}}_{1}\\ 2{\mathrm{a}}_{2}-12{\mathrm{a}}_{4}+2{\mathrm{a}}_{2}=0⇒{\mathrm{a}}_{4}=\frac{1}{3}{\mathrm{a}}_{0}\end{array}$

In the end,

$\begin{array}{l}\mathrm{y}=\sum _{\mathrm{n}=0}^{\infty }{\mathrm{a}}_{\mathrm{n}}{\left(\mathrm{x}-1\right)}^{\mathrm{n}}={\mathrm{a}}_{0}+{\mathrm{a}}_{1}\left(\mathrm{x}-1\right)+{\mathrm{a}}_{2}\left(\mathrm{x}-1{\right)}^{2}+{\mathrm{a}}_{3}\left(\mathrm{x}-1{\right)}_{3}+...\\ \mathrm{y}={\mathrm{a}}_{0}\left(1+{\left(\mathrm{x}+1\right)}^{2}+\frac{1}{3}{\left(\mathrm{x}-1\right)}^{4}+...\right)+{\mathrm{a}}_{1}\left(\left(\mathrm{x}-1\right)+\frac{1}{3}{\left(\mathrm{x}-1\right)}^{3}+...\right)\end{array}$

Hence, the final answer is $\mathrm{y}={\mathrm{a}}_{0}\left(1+{\left(\mathrm{x}+1\right)}^{2}+\frac{1}{3}{\left(\mathrm{x}-1\right)}^{4}+...\right)+{\mathrm{a}}_{1}\left(\left(\mathrm{x}-1\right)+\frac{1}{3}{\left(\mathrm{x}-1\right)}^{3}+...\right)$ ### Want to see more solutions like these? 