Determine two linearly independent solutions to x2y′′+xy′−(4+x)y =0, x > 0.

Answered: Determine two linearly independent…

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 two linearly independent solutions to x²y′′+xy′−(4+x)y =0, x > 0.

Expert Solution

Step 1

Differential equations are used to describe a system. It's state at every different value of time for different initial values can be solved using a single differential equation. Power series is a method to solve differential equation where the solution is obtained using a recurrence relation.

Given: $x^{2} y'' + x y' - (4 + x) y = 0$

Step 2

Assuming power series solution:

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

Putting this in given equation

$\begin{array}{rcl} x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - (4 + x) y & = & 0 \\ x^{2} \frac{d^{2}}{d x^{2}} \sum_{n = 0}^{\infty} c_{n} x^{n + k} + x \frac{d}{d x} \sum_{n = 0}^{\infty} c_{n} x^{n + k} - (4 + x) \sum_{n = 0}^{\infty} c_{n} x^{n + k} & = & 0 \\ x^{2} \sum_{n = 2}^{\infty} (n + k) (n + k - 1) c_{n} x^{n + k - 2} + x \sum_{n = 1}^{\infty} (n + k) c_{n} x^{n + k - 1} - (4 + x) \sum_{n = 0}^{\infty} c_{n} x^{n + k} & = & 0 \\ \sum_{n = 2}^{\infty} (n + k) (n + k - 1) c_{n} x^{n + k} + \sum_{n = 1}^{\infty} (n + k) c_{n} x^{n + k} - \sum_{n = 0}^{\infty} 4 c_{n} x^{n + k} - \sum_{n = 0}^{\infty} c_{n} x^{n + k + 1} & = & 0 \end{array}$

Putting coefficient of lowest power equal to zero:

$\begin{array}{rcl} (0 + k) (0 + k - 1) c_{0} + (0 + k) c_{0} - 4 c_{0} & = & 0 \\ k^{2} - k + k - 4 & = & 0 \\ k^{2} & = & 4 \\ k & = & \pm 2 \end{array}$

here k can only be positive so taking k=2 series becomes.

$\begin{array}{rcl} \sum_{n = 0}^{\infty} (n + 2) (n + 2 - 1) c_{n} x^{n + 2} + \sum_{n = 0}^{\infty} (n + 2) c_{n} x^{n + 2} - \sum_{n = 0}^{\infty} 4 c_{n} x^{n + 2} - \sum_{n = 0}^{\infty} c_{n} x^{n + 2 + 1} & = & 0 \\ \sum_{n = 0}^{\infty} (n + 2) (n + 1) c_{n} x^{n + 2} + \sum_{n = 0}^{\infty} (n + 2) c_{n} x^{n + 2} - \sum_{n = 0}^{\infty} 4 c_{n} x^{n + 2} - \sum_{n = 0}^{\infty} c_{n} x^{n + 3} & = & 0 \end{array}$

Similarly all other coefficients will be also zero so looking at xⁿ⁺² we get recurrence relation :

$\begin{array}{rcl} (n + 2) (n + 1) c_{n} + (n + 2) c_{n} - 4 c_{n} - c_{n - 1} & = & 0 \\ (n^{2} + 3 n + 2 + n + 2 - 4) c_{n} & = & c_{n - 1} \\ c_{n} & = & \frac{1}{n^{2} + 4 n} c_{n - 1} \\ c_{n} & = & \frac{1}{n (n + 4)} c_{n - 1} \end{array}$

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