Answered: determine whether x = 0 is an ordinary…

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

determine whether x = 0 is an ordinary point or a regular singular point of the given differential equation. Then obtain two linearly independent solutions to the differential equation and state the maximum interval on which your solutions are valid.

Q. xy′′+2y′+xy=0

Expert Solution

Step 1

Consider the differential equation:

$x y'' + 2 y' + x y = 0$

Divide by x throughout.

$y'' + \frac{2}{x} y' + y = 0$

Compare with,

$y'' + P (x) y' + Q (x) y = 0$

So,

$P (x) = \frac{2}{x} Q (x) = 1$

Since, $P (x) \to \infty$ at $x = 0$ , therefore $x = 0$ is a singular point.

Now,

$\begin{array}{rcl} x P (x) & = & 2 \\ x^{2} Q (x) & = & x^{2} \end{array}$

Since both are finite at $x = 0$ , therefore $x = 0$ is a regular singular point.

Step 2

Suppose, the power series solution is:

$y = \sum_{n = 0}^{\infty} c_{n} x^{n + r}$

Differentiate with respect to x.

$\begin{array}{rcl} y' & = & \sum_{n = 0}^{\infty} (n + r) c_{n} x^{n + r - 1} \\ y'' & = & \sum_{n = 0}^{\infty} (n + r) (n + r - 1) c_{n} x^{n + r - 2} \end{array}$

Substitute into the differential equation.

$\begin{array}{rcl} x \sum_{n = 0}^{\infty} (n + r) (n + r - 1) c_{n} x^{n + r - 2} + 2 \sum_{n = 0}^{\infty} (n + r) c_{n} x^{n + r - 1} + x \sum_{n = 0}^{\infty} c_{n} x^{n + r} & = & 0 \\ \sum_{n = 0}^{\infty} (n + r) (n + r - 1) c_{n} x^{n + r - 1} + \sum_{n = 0}^{\infty} 2 (n + r) c_{n} x^{n + r - 1} + \sum_{n = 0}^{\infty} c_{n} x^{n + r + 1} & = & 0 \\ \sum_{n = 0}^{\infty} (n + r) (n + r - 1) c_{n} x^{n + r - 1} + \sum_{n = 0}^{\infty} 2 (n + r) c_{n} x^{n + r - 1} + \sum_{n = 2}^{\infty} c_{n - 2} x^{n + r - 1} & = & 0 \\ [r (r - 1) c_{0} + 2 r c_{0}] x^{r - 1} + [r (r + 1) c_{1} + 2 (r + 1) c_{1}] x^{r} + \sum_{n = 2}^{\infty} [(n + r) (n + r - 1) c_{n} + 2 (n + r) c_{n} + c_{n - 2}] x^{n + r - 1} & = & 0 \end{array}$

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