Solve for the power series solution x(х — 2)у" — бу %3D0about x = 1

Name: Solve for the power series solution x(х — 2)у" — бу %3D0about x = 1
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Description: Solve for the power series solution x(х — 2)у" — бу %3D0about x = 1

Answered: x(х — 2)у" — бу %3D 0 about x =1

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Solve for the power series solution

Expert Solution

Step 1

Given differential equation $x (x - 2) y^{''} - 6 y = 0$

We have to find the power series solution of $x (x - 2) y^{''} - 6 y = 0$ about $x = 1$

Let $y = \sum_{n = 0}^{\infty} a_{n} {(x - 1)}^{n}$ be the solution.

Hence, $y^{'} = \sum_{n = 1}^{\infty} n a_{n} {(x - 1)}^{n - 1}$

And $y^{''} = \sum_{n = 2}^{\infty} n (n - 1) a_{n} {(x - 1)}^{n - 2}$

Substitute the values in $x (x - 2) y^{''} - 6 y = 0$

$\begin{array}{l} \Rightarrow x (x - 2) y^{''} - 6 y = 0 \\ \Rightarrow (x^{2} - 2 x) y^{''} - 6 y = 0 \\ \Rightarrow (x^{2} - 2 x) y^{''} + y^{''} - y^{''} - 6 y = 0 \\ \Rightarrow (x^{2} - 2 x + 1) y^{''} - y^{''} - 6 y = 0 \\ \Rightarrow {(x - 1)}^{2} y^{''} - y^{''} - 6 y = 0 \\ \Rightarrow {(x - 1)}^{2} \sum_{n = 2}^{\infty} n (n - 1) a_{n} {(x - 1)}^{n - 2} - \sum_{n = 2}^{\infty} n (n - 1) a_{n} {(x - 1)}^{n - 2} - 6 \sum_{n = 0}^{\infty} a_{n} {(x - 1)}^{n} = 0 \\ \Rightarrow \sum_{n = 2}^{\infty} n (n - 1) a_{n} {(x - 1)}^{n} - \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} {(x - 1)}^{n} - \sum_{n = 0}^{\infty} 6 a_{n} {(x - 1)}^{n} = 0 \\ \Rightarrow \sum_{n = 0}^{\infty} n (n - 1) a_{n} {(x - 1)}^{n} - \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} {(x - 1)}^{n} - \sum_{n = 0}^{\infty} 6 a_{n} {(x - 1)}^{n} = 0 \\ \Rightarrow \sum_{n = 0}^{\infty} [n (n - 1) a_{n} - (n + 2) (n + 1) a_{n + 2} - 6 a_{n}] {(x - 1)}^{n} = 0 \end{array}$

Step 2

Comparing both sides

$\begin{array}{l} \Rightarrow n (n - 1) a_{n} - (n + 2) (n + 1) a_{n + 2} - 6 a_{n} = 0 \\ \Rightarrow [n (n - 1) - 6] a_{n} - (n + 2) (n + 1) a_{n + 2} = 0 \\ \Rightarrow (n^{2} - n - 6) a_{n} - (n + 2) (n + 1) a_{n + 2} = 0 \\ \Rightarrow a_{n + 2} = \frac{(n^{2} - n - 6) a_{n}}{(n + 2) (n + 1)} \\ \Rightarrow a_{n + 2} = \frac{(n + 2) (n - 3) a_{n}}{(n + 2) (n + 1)} \\ \Rightarrow a_{n + 2} = \frac{(n - 3)}{(n + 1)} a_{n} \end{array}$

Substitute $n = 0$ , we get

$a_{2} = - 3 a_{0}$

Substitute $n = 1$ , we get

$a_{3} = - a_{1}$

Substitute $n = 2$ , we get

$a_{4} = a_{0}$

Substitute $n = 3$ , we get

$a_{5} = 0$

Substitute $n = 4$ , we get

$a_{6} = 0$

Similarly, $a_{7} = a_{8} = . . . . . . = 0$

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