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Physics100AAsslgnment
Solutions".r.r.rl-2;
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Curtis Asplund
December2,2007
For solutions to the iirst group of problcuis, seethe Midtenn Solutions. The following are from
the ,.econdgr ciup of problems.
o Boas Ch.12, section 1, p. 564
\\,c solve the followirrg ODE fist using the power series n.rethod and then using a morc
clernentarl' method to verify our solution:
a "- 2 a ' + a : 0 '
(1)
The starting I'oint is assuming the solution takes the form of a power series:
{.2)
Plugging this into (1) we obtain
\-","-r,".,,-2
/-"'"
\-r""""'r\-,
m=2
i:2
",
-.
13
i=0
+ L(a,,+,(m+ l)Qn+z) - 2a^a1(m
t 1 )+ a " ) r - : 0 .
(4)
m-0
81. the linear ldependenceof the powers of r (in thc space of polynomials of r), the above
equality implies tha,t for each rn, > 0,
o ^ 1 2 ( m t 1 ) ( m + 2 ) - 2 a * . , 1 ( m* 1 ) + a * : 0
a*
^ -2o* t l
)(rn | 2)'
m-2
(m
t
a!8
(5)
f6l
The above recurrencerelation determinesthe solution for y(z), in terms of the initial choices
o0 and a1. Since no boundary conditions are specified in the problem, we have gone as far
a,swe ca.nrvith the power seriessohrtion.
Next we solve the ODE using the elementary method outlined in Boas Ch. 8, section 5. \\'e
write (1) a^s(D2 2D + 1.)y: 0 + (D - 1)(, - i)y : O. tt" generalsolution for such an
equation is given bv Bon^sequation (5.15):
(.7)
aa"*(z):(Ar+B)e'.
1b vr:r'if1'tha,t this is consistent with our series solution, rve check that the power scries for
thc abol.c (ana,lr.tic)solution does indeed saiisly the recurrence relation (6).
,!J"t.-\.r)
- (A.r+ B)f
-
(8)
I7 1 !
. . , l Lt ' - . I BZ't Int
n,.
lt u ,
:B+i
"
-t ) :
\m -
(e)
B+Bt+
4.
( 1 0)
ml
(, A,,,+4)'-.
'
3,\
n,
. rl
(11)
rrt'.
f
So ri'e sec lhat for rn > 0, tlLe m.tnterm in the powcr series for y"1" (r) ir
,3,
Now rvc check thai this satisfiesthe recurrence relatiol:
+ #. :t b^.
(m+I)(m,+2)
( 1 2)
m:1
o-
BB
+ r-( m . + 2 ) ( m + 1 ) ! (m,+ 2)l
A
m,A
"r,-rAr+U2ntA + 2A - rn,A
-
B
( m , + 2 ) ( 1 n , + I ) l( o * 2 ) l
AB
( n + 1 ) ! ( m+ 2 ) l
(m + 2)l
(13)
(14)
(15)
(16)
- bm+2,
r4rich is what we were looking for. So we have vcrified our recurrence rela,tion that constil u t - , 1 o U r ! , u \ r a l5 r .ira 5" o l u r i o n .
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