ondiere Consider the diferential equation where O is a constant. iscifferential equation has an irreqular singular POint @ =D. Show that the substitution t yields the different ial equation d'y 2 dy + t dt + Ay=o which now has a reqularsinqular poit at t-o. dt2 b) Find two lincarly ihdependent Frobenius series sulutions 0f this new equation. C) Express each series solution af the original equatioh in terms of elementary functions 2. Consider tthe equat ion indicial equation has roots 4-420 ahose a) Derive the FrobeniuS scries solution n! (n-) 12u b) Substitute 2:Cy, nxt n in the equation xy"-y0 to derive e recurrence rtlation nni b= 2nt C. (nti)in! C) Conclude frorm uhis result that a seuond solution is n! (n-i! where H

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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1. Parts a,b,& c 

ondiere
Consider the diferential equation
where
O is
a constant.
iscifferential equation has an irreqular singular
POint @ =D. Show that the substitution
t yields the different ial equation
d'y
2 dy
+
t dt
+ Ay=o
which now has a reqularsinqular poit at t-o.
dt2
b) Find two lincarly ihdependent Frobenius series sulutions
0f this new equation.
C) Express each series solution af the original equatioh
in terms of elementary functions
2.
Consider tthe equat ion
indicial equation has roots
4-420 ahose
a) Derive the FrobeniuS scries solution
n! (n-)
12u
b) Substitute 2:Cy, nxt
n
in the equation xy"-y0 to derive e recurrence
rtlation
nni b=
2nt
C.
(nti)in!
C) Conclude frorm uhis result that a seuond solution is
n! (n-i!
where H
Transcribed Image Text:ondiere Consider the diferential equation where O is a constant. iscifferential equation has an irreqular singular POint @ =D. Show that the substitution t yields the different ial equation d'y 2 dy + t dt + Ay=o which now has a reqularsinqular poit at t-o. dt2 b) Find two lincarly ihdependent Frobenius series sulutions 0f this new equation. C) Express each series solution af the original equatioh in terms of elementary functions 2. Consider tthe equat ion indicial equation has roots 4-420 ahose a) Derive the FrobeniuS scries solution n! (n-) 12u b) Substitute 2:Cy, nxt n in the equation xy"-y0 to derive e recurrence rtlation nni b= 2nt C. (nti)in! C) Conclude frorm uhis result that a seuond solution is n! (n-i! where H
Expert Solution
Step 1

1 a,b,c) To solve the first differential equation by suitable coordinate transformation and power series technique

(1)
(x> 0, A,a cons tan t
(consider: y"
0;
0, as x->0
As
x0 is anirreular sin gular point
Step 2

Solution to a)

a)x'y"Ay 0, (> 0)
(1)
dy dy dt
dr dt dx
dt
so
dac
1
-:y'=
=-i'y')
d-f y'(t)
-21x(-f)y'f)=ř y*(O-?)
y"(x)=
dx
='y"()+2r'y)
So, (l) fransforms to
(y")2y')+åy()=0,(t>0),
y"(f)+y*)+ Ay)= 0, (t > 0),
(2)
2
as required (astx=-2, as t-0
the equation (2) is regular sin gular at t 0
Step 3

b) First step in Frobenius method is to solve for the indices . here r =0 , -1. The difference is an integer. By the theory , Frobenius solution exists for r =0 (the larger root); Frobenius solution for r=-1 remains to be investigated.

b): y")+y)+y(0)=0, (t > 0), (2)
(may assume 1, by a linear change of variables)
Re write, y'()+2ty(t)+y() 0, (>0), (3)
Compare with the standard form
fy"(t)+tp(y(t)+ q)y(t)= 0, (t > 0) (4)
then, p(t)= 2, q(t) = t;s0,the indicial equation is
r(r-)p(0)r+q(0) = 0
r(r-1)2r 0r 0,r =-1
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