How do I do Question 5. Show every step with detail. 5. We will find the Green Function G(r, r') satisfyingI(1+1)d² Gdr²-G = 8(r — r'),where a a > 0, G(a, r') = G(b, r')=G(r, a) = G(r, b) = 0,and 10 is a constant.(a) Find the general solutions to the homogeneous equation by assuming asolution of the form rm.(b) Determine G(r, r') from the homogeneous solutions, the boundary condi-tions, and behavior at r = r².6. The Hypergeometric differential equation isx(x − 1)y” + [(1 + a+b)x — c)]y' + aby = 0, where a, b, and c are constants.(a) Find the singular points and classify them. Include in your considerationthe point x→∞.(b) Find the Frobenius series around x =Q Search0 corresponding to s = 0 root of@4x

5) Here consider associated homogeneous ODE Noted that its solution is of the form ,then it…

Answered: 5. We will find the Green Function G(r,…

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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How do I do Question 5. Show every step with detail.

5. We will find the Green Function G(r, r') satisfying
I
(1+1)
d² G
dr²
-G = 8(r — r'),
where a <r<b, a ≤ r' ≤ b, b> a > 0, G(a, r') = G(b, r')=G(r, a) = G(r, b) = 0,
and 10 is a constant.
(a) Find the general solutions to the homogeneous equation by assuming a
solution of the form rm.
(b) Determine G(r, r') from the homogeneous solutions, the boundary condi-
tions, and behavior at r = r².
6. The Hypergeometric differential equation is
x(x − 1)y” + [(1 + a+b)x — c)]y' + aby = 0, where a, b, and c are constants.
(a) Find the singular points and classify them. Include in your consideration
the point x→∞.
(b) Find the Frobenius series around x =
Q Search
0 corresponding to s = 0 root of
@4x

Expert Solution

Step 1: Green's function find out

5) Here consider associated homogeneous ODE $y apostrophe apostrophe minus fraction numerator l open parentheses l plus 1 close parentheses over denominator r squared end fraction y equals 0 comma y left parenthesis a right parenthesis equals 0 comma y left parenthesis b right parenthesis equals 0$

Noted that its solution is of the form $y open parentheses r close parentheses equals r to the power of m$ ,then it satisfies given ODE ,so we get

$m open parentheses m minus 1 close parentheses minus l open parentheses l plus 1 close parentheses equals 0 rightwards double arrow m equals l plus 1 comma negative l$

Hence general solution become $y left parenthesis r right parenthesis equals A. r to the power of l plus 1 end exponent plus B. r to the power of negative l end exponent comma w h e r e space A comma B space a r e space a r b i t r a r y space c o n s tan t s$ .

Now $y left parenthesis a right parenthesis equals 0 rightwards double arrow A. a to the power of l plus 1 end exponent plus B. a to the power of negative l end exponent equals 0 space a n d space y left parenthesis b right parenthesis equals 0 rightwards double arrow A. b to the power of l plus 1 end exponent plus B. b to the power of negative l end exponent equals 0$

solving we get $A equals B equals 0$

So we get trivial solution here under given boundary conditions.

Hence Green's function exists.