Answered: Determine the ﬁrst ﬁve nonzero terms in…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

Determine the ﬁrst ﬁve nonzero terms in each of two linearly independent Frobenius series solutions to 3x²y′′+x(1+3x²)y′−2xy=0, x > 0.

Expert Solution

Step 1

Frobenius series or Frobenius method is a method to solve second-order differential equations, this method gives an infinite power series solution whose function is expressed in terms of a polynomial equation. This method finds power series around the singular point of the differential equation.

Step 2

Given the differential equation:

$3 x^{2} y'' + x (1 + 3 x^{2}) y' - 2 x y = 0$

Let's assume the standard solution:

$y = \sum_{k = 0}^{\infty} a_{k} \cdot x^{k + r}$

Putting this in the differential equation we get:

$\begin{array}{rcl} 3 x^{2} \frac{d^{2}}{d x^{2}} \sum_{k = 0}^{\infty} [a_{k} \cdot x^{k + r}] + x (1 + 3 x^{2}) \frac{d}{d x} \sum_{k = 0}^{\infty} [a_{k} \cdot x^{k + r}] - 2 x \sum_{k = 0}^{\infty} [a_{k} \cdot x^{k + r}] & = & 0 \\ 3 x^{2} \sum_{k = 0}^{\infty} [(k + r) (k + r - 1) a_{k} \cdot x^{k + r - 2}] + x (1 + 3 x^{2}) \sum_{k = 0}^{\infty} [(k + r) a_{k} \cdot x^{k + r - 1}] - 2 x \sum_{k = 0}^{\infty} [a_{k} \cdot x^{k + r}] & = & 0 \\ \sum_{k = 0}^{\infty} [3 (k + r) (k + r - 1) a_{k} \cdot x^{k + r}] + \sum_{k = 0}^{\infty} [(k + r) a_{k} \cdot x^{k + r}] + \sum_{k = 0}^{\infty} [3 (k + r) a_{k} \cdot x^{k + r + 2}] - \sum_{k = 0}^{\infty} [2 a_{k} \cdot x^{k + r + 1}] & = & 0 \end{array}$

Shifting the value of k in the third sum $(k \to k - 2)$ and fourth sum $(k \to k - 1)$ to get the same powers of x:

$\begin{array}{rcl} \sum_{k = 0}^{\infty} 3 (k + r) (k + r - 1) a_{k} \cdot x^{k + r} + \sum_{k = 0}^{\infty} (k + r) a_{k} \cdot x^{k + r} + \sum_{k = 2}^{\infty} 3 (k - 2 + r) a_{k - 2} \cdot x^{k + r} - \sum_{k = 1}^{\infty} 2 a_{k - 1} \cdot x^{k + r} & = & 0 \end{array}$

Putting the coefficient of lowest power equal to zero i.e. x^r (which can be obtained by putting k=0 in the first two sums):

$\begin{array}{rcl} 3 (0 + r) (0 + r - 1) a_{0} x^{r} + (0 + r) a_{0} x^{r} & = & 0 \\ 3 r (r - 1) + r & = & 0 \\ 3 r^{2} - 3 r + r & = & 0 \\ 3 r^{2} & = & 2 r \\ r & = & \frac{2}{3} \\ o r r & = & 0 \end{array}$

Now put the coefficient of x^k+r equal to zero to get a recurrence relation:

$\begin{array}{rcl} 3 (k + r) (k + r - 1) a_{k} + (k + r) a_{k} + 3 (k + r - 2) a_{k - 2} - 2 a_{k - 1} & = & 0 \\ \{3 (k + r - 1) + 1\} (k + r) a_{k} & = & 2 a_{k - 1} - 3 (k + r - 2) a_{k - 2} \\ (3 k + 3 r - 2) (k + r) a_{k} & = & 2 a_{k - 1} - 3 (k + r - 2) a_{k - 2} \\ a_{k} & = & \frac{2}{(3 k + 3 r - 2) (k + r)} a_{k - 1} - \frac{3 (k + r - 2)}{(3 k + 3 r - 2) (k + r)} a_{k - 2} \end{array}$

Step by step

Solved in 4 steps

Knowledge Booster

$Laplace Transformation$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

Similar questions