By means of a power series (centred at the ordinary point xotwo linearly independent solutions (y1 (x) and y2(x)) of the differential equation:10.0), find they" + 4 x² y = 0

Answered: By means of a power series (centred at…

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

By means of a power series (centred at the ordinary point xo
two linearly independent solutions (y1 (x) and y2(x)) of the differential equation:
10.
0), find the
y" + 4 x² y = 0

Expert Solution

Step 1

Assume the power series solution for the given differential equation to be as follows:

$y = \sum_{k = 0}^{\infty} a_{k} {(x - x_{0})}^{k}$

In our problem, it is given that $x_{0} = 0$ .

$\{y = \sum_{k = 0}^{\infty} a_{k} x^{k} y' = \sum_{k = 1}^{\infty} k a_{k} x^{k - 1} y'' = \sum_{k = 2}^{\infty} k (k - 1) a_{k} x^{k - 2}\}$ ..........................................(1)

Substitute equation (1) in the differential equation.

$y'' + 4 x^{2} y = 0 \sum_{k = 2}^{\infty} k (k - 1) a_{k} x^{k - 2} + 4 x^{2} \sum_{k = 0}^{\infty} a_{k} x^{k} = 0$

$\sum_{k = 2}^{\infty} k (k - 1) a_{k} x^{k - 2} + 4 \sum_{k = 0}^{\infty} a_{k} x^{k + 2} = 0$ ...........................(2)

Step 2

Substitute $k - 2 = n$ for the first term and $k + 2 = n$ for the 2nd term in equation (1).

$\sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} + 4 \sum_{n = 2}^{\infty} a_{n - 2} x^{n} = 0$

Taking the first two coefficients of the 1^st summation and equating it to zero, we get,

$2 a_{2} + 6 a_{3} + \sum_{n = 2}^{\infty} [(n + 2) (n + 1) a_{n + 2} + 4 a_{n - 2}] x^{n} = 0 \Rightarrow a_{2} = 0, a_{3} = 0, (n + 2) (n + 1) a_{n + 2} + 4 a_{n - 2} = 0$

$\Rightarrow a_{n + 2} = - \frac{4 a_{n - 2}}{(n + 2) (n + 1)}$ ..............................................(3)

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