Solutions to Differential Equations with Boundary Conditions

Q1i)
Given Chebyshev’s equation
(1 – x2)y’’ – xy’ + α2y = 0
A point x0 is said to be a regular singular point of DE y’’ + p(x)y’ + Q(x)y = 0. If the functions p
(x) = (x – x0)P(x) and q(x) = (x – x0)2 Q(x) are both analytic at x0’’
y’’ - x
(1−x2 ) y’ + 2 α
(1−x2 ) y = 0
y’’ - x
(x−1)(1+ x) y’ + 2 α
(x−1)(1+ x) y = 0
here,
P(x) = x
(x−1)(1+ x)
Q(x) = 2 α
(x−1)(1+ x)
y’’ –P(x)y’ + Q(x)y = 0
Q1ii)
The Chebyshev’s polynomial o degree n ≥ 0 is defined as
Tn(x) = cos(ncos-1(x)), x ∈ [-1, 1]
Tn(x) = cos(nθ) , where x = cosθ, θ ε [0 , π ]
∫
−1
1
( 2 x )
−1
2 ∗T m(x) * Tn(x)dx
= -∫
π
0
cos ( nθ ) ∗cos ( mθ ) ∗dθ [ since , dθ= −dx
√1−x2 ]

= ∫
0
π
1
2 [cos {(n+m)θ }¿+cos {(n−m)θ }]dθ ¿
= 1
2 [ sin ( n+m ) θ
n+ m + sin ( n−m ) θ
n−m ]π
0
= 1
2 [ { sin ( n+m ) π
n+m +sin ( n−m ) π
n−m }−{ sin 0
n+m + sin 0
n−m }] , m ≠ n
= 1
2 [ ( 0+ 0 )− ( 0+ 0 ) ]
= 0
Hence ,
∫
−1
1
( 1−x2 )
1
2∗T m(x)*Tn(x)dx = 0, when m ≠ n
Q2)
u’’ + λu = 0
u’(0) = u’(1) = 0
so r2 + λ = 0
r = √ λ− ¿
+¿ ¿ ¿ = i √λ− ¿
+¿ ¿ ¿
so u(n) = c1cos(√ λn) + c2sin(√ λn)
using the first boundary condition u(0) = 0
therefore 0 = c1
So,
u(n) = c2sin(√ λn)
u’(n) = c2√ λ * cos( √ λ n)
using the second boundary condition u’(1) = 0

c2√ λ * cos(1 √ λ)
1 √ λ=−π
2 , 3 π
2
2n+ 1
2 π , n is an integer
So,
√ λ=2 n+1
2∗1 π
Therefore, λn = ( 2n+ 1 )2
4 π 2, n is integer
So, u(n) = c2sin( 2n+ 1
2 πn¿
For λ = 0, u’’ = 0, u(0) = 0, u’(1) = 0
u(n) = c1 + c2n
u(0) = 0, c1 = 0
u(n) = c2(n)
u’(n) = c2 => u’(1) = 0 => 0 = c2
so, λ = 0, can not be an eigenvalue for this problem
now λ < 0
r = √ λ− ¿
+¿ ¿ ¿ ,
so, u(n) = c1cosh(√ λ n ¿ + c2sinh(√ λ n ¿
u(0) = 0
0 = c1cosh(0 ¿ + c2sinh(0 ¿
0 = c1(1) + c2(0)
c1 = 0
so, u(n) = c2sinh(√−λ n ¿