Engineering Mathematics II (2M03)Tutorial 10
Marina ChugunovaDepartment of Math. & Stat., office: HH403
e-mail: [email protected]
office hours: Math Help Centre, Thursday 1:30 - 3:30
November 15-16, 2007
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 14)Determine the number of power series solutions at the point x = 0 forthe ODE:
xy′′ + y′ + 10y = 0.
Solution:
2
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 14)Determine the number of power series solutions at the point x = 0 forthe ODE:
xy′′ + y′ + 10y = 0.
Solution:x = 0 is a regular singular point. (Check !) Frobenius method:
3
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 14)Determine the number of power series solutions at the point x = 0 forthe ODE:
xy′′ + y′ + 10y = 0.
Solution:x = 0 is a regular singular point. (Check !) Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
4
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 14)Determine the number of power series solutions at the point x = 0 forthe ODE:
xy′′ + y′ + 10y = 0.
Solution:x = 0 is a regular singular point. (Check !) Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 +
∞∑n=0
(n + r)cnxn+r−1 + 10
∞∑n=0
cnxn+r = 0
5
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 14)Determine the number of power series solutions at the point x = 0 forthe ODE:
xy′′ + y′ + 10y = 0.
Solution:x = 0 is a regular singular point. (Check !) Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 +
∞∑n=0
(n + r)cnxn+r−1 + 10
∞∑n=0
cnxn+r = 0
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 +
∞∑n=0
(n + r)cnxn+r−1 + 10
∞∑n=1
cn−1xn+r−1 = 0
6
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 14)Determine the number of power series solutions at the point x = 0 forthe ODE:
xy′′ + y′ + 10y = 0.
Solution:x = 0 is a regular singular point. (Check !) Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 +
∞∑n=0
(n + r)cnxn+r−1 + 10
∞∑n=0
cnxn+r = 0
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 +
∞∑n=0
(n + r)cnxn+r−1 + 10
∞∑n=1
cn−1xn+r−1 = 0
n = 0 : r(r − 1)c0 + rc0 = 0, r(r − 1) + r = 0, r2 = 0
7
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 14)Determine the number of power series solutions at the point x = 0 forthe ODE:
xy′′ + y′ + 10y = 0.
Solution:x = 0 is a regular singular point. (Check !) Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 +
∞∑n=0
(n + r)cnxn+r−1 + 10
∞∑n=0
cnxn+r = 0
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 +
∞∑n=0
(n + r)cnxn+r−1 + 10
∞∑n=1
cn−1xn+r−1 = 0
n = 0 : r(r − 1)c0 + rc0 = 0, r(r − 1) + r = 0, r2 = 0
Repeated root = only one power series solution.
8
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 16)Find two power series solutions at the point x = 0 for the ODE:
2xy′′ + 5y′ + xy = 0.
Solution:
9
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 16)Find two power series solutions at the point x = 0 for the ODE:
2xy′′ + 5y′ + xy = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
10
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 16)Find two power series solutions at the point x = 0 for the ODE:
2xy′′ + 5y′ + xy = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
11
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 16)Find two power series solutions at the point x = 0 for the ODE:
2xy′′ + 5y′ + xy = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=0
cnxn+r+1 = 0
12
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 16)Find two power series solutions at the point x = 0 for the ODE:
2xy′′ + 5y′ + xy = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=0
cnxn+r+1 = 0
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
13
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 16)Find two power series solutions at the point x = 0 for the ODE:
2xy′′ + 5y′ + xy = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=0
cnxn+r+1 = 0
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
n = 0 : 2r(r − 1)c0 + 5rc0 = 0, 2r2 + 3r = 0, r1 = 0, r2 = −3
2
14
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 16)Find two power series solutions at the point x = 0 for the ODE:
2xy′′ + 5y′ + xy = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=0
cnxn+r+1 = 0
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
n = 0 : 2r(r − 1)c0 + 5rc0 = 0, 2r2 + 3r = 0, r1 = 0, r2 = −3
2
r1 − r2 =3
2it’s not an integer number (2 power series solutions)
15
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r1 = 0:
2
∞∑n=0
n(n− 1)cnxn−1 + 5
∞∑n=0
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
16
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r1 = 0:
2
∞∑n=0
n(n− 1)cnxn−1 + 5
∞∑n=0
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
2
∞∑n=2
n(n− 1)cnxn−1 + 5
∞∑n=1
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
17
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r1 = 0:
2
∞∑n=0
n(n− 1)cnxn−1 + 5
∞∑n=0
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
2
∞∑n=2
n(n− 1)cnxn−1 + 5
∞∑n=1
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
5c1 +
∞∑n=2
[(2n(n− 1) + 5n)cn + cn−2]xn−1 = 0
18
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r1 = 0:
2
∞∑n=0
n(n− 1)cnxn−1 + 5
∞∑n=0
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
2
∞∑n=2
n(n− 1)cnxn−1 + 5
∞∑n=1
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
5c1 +
∞∑n=2
[(2n(n− 1) + 5n)cn + cn−2]xn−1 = 0
5c1 +
∞∑n=2
[(2n2 + 3n)cn + cn−2]xn−1 = 0
19
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r1 = 0:
2
∞∑n=0
n(n− 1)cnxn−1 + 5
∞∑n=0
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
2
∞∑n=2
n(n− 1)cnxn−1 + 5
∞∑n=1
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
5c1 +
∞∑n=2
[(2n(n− 1) + 5n)cn + cn−2]xn−1 = 0
5c1 +
∞∑n=2
[(2n2 + 3n)cn + cn−2]xn−1 = 0
c1 = 0, cn = − 1
2n2 + 3ncn−2, n = 2, 3, 4... c2 = − 1
14c0, c3 = 0
20
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r1 = 0:
2
∞∑n=0
n(n− 1)cnxn−1 + 5
∞∑n=0
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
2
∞∑n=2
n(n− 1)cnxn−1 + 5
∞∑n=1
ncnxn−1 +
∞∑n=2
cn−2xn−1 = 0
5c1 +
∞∑n=2
[(2n(n− 1) + 5n)cn + cn−2]xn−1 = 0
5c1 +
∞∑n=2
[(2n2 + 3n)cn + cn−2]xn−1 = 0
c1 = 0, cn = − 1
2n2 + 3ncn−2, n = 2, 3, 4... c2 = − 1
14c0, c3 = 0
y1(x) = c0(1−1
14x2...)
21
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r2 = −32:
22
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r2 = −32:
2
∞∑n=0
(n− 3/2)(n− 5/2)cnxn−5/2 + 5
∞∑n=0
(n− 3/2)cnxn−5/2 +
∞∑n=2
cn−2xn−5/2 = 0
23
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r2 = −32:
2
∞∑n=0
(n− 3/2)(n− 5/2)cnxn−5/2 + 5
∞∑n=0
(n− 3/2)cnxn−5/2 +
∞∑n=2
cn−2xn−5/2 = 0
−c1x−3/2 +
∞∑n=2
[(2(n− 3/2)(n− 5/2) + 5(n− 3/2))cn + cn−2]xn−5/2 = 0
24
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r2 = −32:
2
∞∑n=0
(n− 3/2)(n− 5/2)cnxn−5/2 + 5
∞∑n=0
(n− 3/2)cnxn−5/2 +
∞∑n=2
cn−2xn−5/2 = 0
−c1x−3/2 +
∞∑n=2
[(2(n− 3/2)(n− 5/2) + 5(n− 3/2))cn + cn−2]xn−5/2 = 0
c1 = 0, cn = − 1
(2n− 3)ncn−2, n = 2, 3, 4... c2 = −1
2c0...
25
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r2 = −32:
2
∞∑n=0
(n− 3/2)(n− 5/2)cnxn−5/2 + 5
∞∑n=0
(n− 3/2)cnxn−5/2 +
∞∑n=2
cn−2xn−5/2 = 0
−c1x−3/2 +
∞∑n=2
[(2(n− 3/2)(n− 5/2) + 5(n− 3/2))cn + cn−2]xn−5/2 = 0
c1 = 0, cn = − 1
(2n− 3)ncn−2, n = 2, 3, 4... c2 = −1
2c0...
y2(x) = c0x−3/2(1− 1
2x2...)
26
The Series Solutions of ODE (5.2 Solutions about Singular Points )
2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 5
∞∑n=0
(n + r)cnxn+r−1 +
∞∑n=2
cn−2xn+r−1 = 0
Solve for r2 = −32:
2
∞∑n=0
(n− 3/2)(n− 5/2)cnxn−5/2 + 5
∞∑n=0
(n− 3/2)cnxn−5/2 +
∞∑n=2
cn−2xn−5/2 = 0
−c1x−3/2 +
∞∑n=2
[(2(n− 3/2)(n− 5/2) + 5(n− 3/2))cn + cn−2]xn−5/2 = 0
c1 = 0, cn = − 1
(2n− 3)ncn−2, n = 2, 3, 4... c2 = −1
2c0...
y2(x) = c0x−3/2(1− 1
2x2...)
General solution:
y(x) = c1y1 + c2y2 = c1(1−1
14x2...) + c2x
−3/2(1− 1
2x2...)
27
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:
28
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
29
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
30
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=0
cnxn+r = 0
31
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=0
cnxn+r = 0
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=2
cn−2xn+r−2 = 0
32
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=0
cnxn+r = 0
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=2
cn−2xn+r−2 = 0
n = 0 : r(r − 1)c0 + 3rc0 = 0, r2 + 2r = 0, r1 = 0, r2 = −2
33
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=0
cnxn+r = 0
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=2
cn−2xn+r−2 = 0
n = 0 : r(r − 1)c0 + 3rc0 = 0, r2 + 2r = 0, r1 = 0, r2 = −2
r2−r1 = 2 it’s an integer number (in general only 1 power series solution)
34
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=2
cn−2xn+r−2 = 0
Solve for r1 = 0:
35
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=2
cn−2xn+r−2 = 0
Solve for r1 = 0:∞∑
n=0
n(n− 1)cnxn−2 + 3
∞∑n=0
ncnxn−2 − 2
∞∑n=2
cn−2xn−2 = 0
36
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=2
cn−2xn+r−2 = 0
Solve for r1 = 0:∞∑
n=0
n(n− 1)cnxn−2 + 3
∞∑n=0
ncnxn−2 − 2
∞∑n=2
cn−2xn−2 = 0
∞∑n=2
n(n− 1)cnxn−2 + 3
∞∑n=1
ncnxn−2 − 2
∞∑n=2
cn−2xn−2 = 0
37
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 28)Find solutions at the point x = 0 for the ODE:
y′′ +3
xy′ − 2y = 0.
Solution:
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=2
cn−2xn+r−2 = 0
Solve for r1 = 0:∞∑
n=0
n(n− 1)cnxn−2 + 3
∞∑n=0
ncnxn−2 − 2
∞∑n=2
cn−2xn−2 = 0
∞∑n=2
n(n− 1)cnxn−2 + 3
∞∑n=1
ncnxn−2 − 2
∞∑n=2
cn−2xn−2 = 0
c1 = 0, cn =2
n2 + 2ncn−2, c2 =
1
4c0, c3 = 0...
38
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Solution:
∞∑n=0
(n + r)(n + r − 1)cnxn+r−2 + 3
∞∑n=0
(n + r)cnxn+r−2 − 2
∞∑n=2
cn−2xn+r−2 = 0
Solve for r1 = 0:∞∑
n=0
n(n− 1)cnxn−2 + 3
∞∑n=0
ncnxn−2 − 2
∞∑n=2
cn−2xn−2 = 0
∞∑n=2
n(n− 1)cnxn−2 + 3
∞∑n=1
ncnxn−2 − 2
∞∑n=2
cn−2xn−2 = 0
c1 = 0, cn =2
n2 + 2ncn−2, c2 =
1
4c0, c3 = 0...
y1(x) = c0 +1
4c0x
2... = c0(1 +1
4x2..)
Use numerical computations to find y2(x).
39
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 32)Find solutions at the point x = 0 for the ODE:
x(x− 1)y′′ + 3y′ − 2y = 0.
Solution:
40
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 32)Find solutions at the point x = 0 for the ODE:
x(x− 1)y′′ + 3y′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
41
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 32)Find solutions at the point x = 0 for the ODE:
x(x− 1)y′′ + 3y′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
42
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 32)Find solutions at the point x = 0 for the ODE:
x(x− 1)y′′ + 3y′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r− 1)cnxn+r−
∞∑n=0
(n + r)(n + r− 1)cnxn+r−1 + 3
∞∑n=0
(n + r)cnxn+r−1−
2∞∑
n=0cnx
n+r = 0
43
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 32)Find solutions at the point x = 0 for the ODE:
x(x− 1)y′′ + 3y′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r− 1)cnxn+r−
∞∑n=0
(n + r)(n + r− 1)cnxn+r−1 + 3
∞∑n=0
(n + r)cnxn+r−1−
2∞∑
n=0cnx
n+r = 0
∞∑n=1
(n + r − 1)(n + r − 2)cn−1xn+r−1 −
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 3
∞∑n=0
(n +
r)cnxn+r−1 − 2
∞∑n=1
cn−1xn+r−1 = 0
44
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 32)Find solutions at the point x = 0 for the ODE:
x(x− 1)y′′ + 3y′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r− 1)cnxn+r−
∞∑n=0
(n + r)(n + r− 1)cnxn+r−1 + 3
∞∑n=0
(n + r)cnxn+r−1−
2∞∑
n=0cnx
n+r = 0
∞∑n=1
(n + r − 1)(n + r − 2)cn−1xn+r−1 −
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 3
∞∑n=0
(n +
r)cnxn+r−1 − 2
∞∑n=1
cn−1xn+r−1 = 0
n = 0 : −r(r − 1)c0 + 3rc0 = 0, −r2 + 4r = 0, r1 = 0, r2 = 4
45
The Series Solutions of ODE (5.2 Solutions about Singular Points )
Problem (5.2: 32)Find solutions at the point x = 0 for the ODE:
x(x− 1)y′′ + 3y′ − 2y = 0.
Solution:x = 0 is a regular singular point. Frobenius method:
y =
∞∑n=0
cnxn+r y′ =
∞∑n=0
(n + r)cnxn+r−1 y′′ =
∞∑n=0
(n + r)(n + r− 1)cnxn+r−2
∞∑n=0
(n + r)(n + r− 1)cnxn+r−
∞∑n=0
(n + r)(n + r− 1)cnxn+r−1 + 3
∞∑n=0
(n + r)cnxn+r−1−
2∞∑
n=0cnx
n+r = 0
∞∑n=1
(n + r − 1)(n + r − 2)cn−1xn+r−1 −
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 3
∞∑n=0
(n +
r)cnxn+r−1 − 2
∞∑n=1
cn−1xn+r−1 = 0
n = 0 : −r(r − 1)c0 + 3rc0 = 0, −r2 + 4r = 0, r1 = 0, r2 = 4
r2−r1 = 2 it’s an integer number (it might have 2 power series solutions)46
47
The Series Solutions of ODE (5.2 Solutions about Singular Points )
∞∑n=1
(n + r − 1)(n + r − 2)cn−1xn+r−1 −
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 3
∞∑n=0
(n +
r)cnxn+r−1 − 2
∞∑n=1
cn−1xn+r−1 = 0
Solve for r1 = 0:
48
The Series Solutions of ODE (5.2 Solutions about Singular Points )
∞∑n=1
(n + r − 1)(n + r − 2)cn−1xn+r−1 −
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 3
∞∑n=0
(n +
r)cnxn+r−1 − 2
∞∑n=1
cn−1xn+r−1 = 0
Solve for r1 = 0:∞∑
n=1(n− 1)(n− 2)cn−1x
n−1 −∞∑
n=0n(n− 1)cnx
n−1 + 3∞∑
n=0ncnx
n−1 − 2∞∑
n=1cn−1x
n−1 = 0
49
The Series Solutions of ODE (5.2 Solutions about Singular Points )
∞∑n=1
(n + r − 1)(n + r − 2)cn−1xn+r−1 −
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 3
∞∑n=0
(n +
r)cnxn+r−1 − 2
∞∑n=1
cn−1xn+r−1 = 0
Solve for r1 = 0:∞∑
n=1(n− 1)(n− 2)cn−1x
n−1 −∞∑
n=0n(n− 1)cnx
n−1 + 3∞∑
n=0ncnx
n−1 − 2∞∑
n=1cn−1x
n−1 = 0
∞∑n=1
(n− 1)(n− 2)cn−1xn−1 −
∞∑n=1
n(n− 1)cnxn−1 + 3
∞∑n=1
ncnxn−1 − 2
∞∑n=1
cn−1xn−1 = 0
50
The Series Solutions of ODE (5.2 Solutions about Singular Points )
∞∑n=1
(n + r − 1)(n + r − 2)cn−1xn+r−1 −
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 3
∞∑n=0
(n +
r)cnxn+r−1 − 2
∞∑n=1
cn−1xn+r−1 = 0
Solve for r1 = 0:∞∑
n=1(n− 1)(n− 2)cn−1x
n−1 −∞∑
n=0n(n− 1)cnx
n−1 + 3∞∑
n=0ncnx
n−1 − 2∞∑
n=1cn−1x
n−1 = 0
∞∑n=1
(n− 1)(n− 2)cn−1xn−1 −
∞∑n=1
n(n− 1)cnxn−1 + 3
∞∑n=1
ncnxn−1 − 2
∞∑n=1
cn−1xn−1 = 0
(n− 4)cn = (n− 3)cn−1, c1 =2
3, c2 =
1
2c1 =
1
3c0, c3 = 0
51
The Series Solutions of ODE (5.2 Solutions about Singular Points )
∞∑n=1
(n + r − 1)(n + r − 2)cn−1xn+r−1 −
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 3
∞∑n=0
(n +
r)cnxn+r−1 − 2
∞∑n=1
cn−1xn+r−1 = 0
Solve for r1 = 0:∞∑
n=1(n− 1)(n− 2)cn−1x
n−1 −∞∑
n=0n(n− 1)cnx
n−1 + 3∞∑
n=0ncnx
n−1 − 2∞∑
n=1cn−1x
n−1 = 0
∞∑n=1
(n− 1)(n− 2)cn−1xn−1 −
∞∑n=1
n(n− 1)cnxn−1 + 3
∞∑n=1
ncnxn−1 − 2
∞∑n=1
cn−1xn−1 = 0
(n− 4)cn = (n− 3)cn−1, c1 =2
3, c2 =
1
2c1 =
1
3c0, c3 = 0
Can we find the value of c4 - ?
52
The Series Solutions of ODE (5.2 Solutions about Singular Points )
∞∑n=1
(n + r − 1)(n + r − 2)cn−1xn+r−1 −
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 3
∞∑n=0
(n +
r)cnxn+r−1 − 2
∞∑n=1
cn−1xn+r−1 = 0
Solve for r1 = 0:∞∑
n=1(n− 1)(n− 2)cn−1x
n−1 −∞∑
n=0n(n− 1)cnx
n−1 + 3∞∑
n=0ncnx
n−1 − 2∞∑
n=1cn−1x
n−1 = 0
∞∑n=1
(n− 1)(n− 2)cn−1xn−1 −
∞∑n=1
n(n− 1)cnxn−1 + 3
∞∑n=1
ncnxn−1 − 2
∞∑n=1
cn−1xn−1 = 0
(n− 4)cn = (n− 3)cn−1, c1 =2
3, c2 =
1
2c1 =
1
3c0, c3 = 0
Can we find the value of c4 - ? (It’s an arbitrary number)
c5 = 2c4, c6 = 3c4...
53
The Series Solutions of ODE (5.2 Solutions about Singular Points )
∞∑n=1
(n + r − 1)(n + r − 2)cn−1xn+r−1 −
∞∑n=0
(n + r)(n + r − 1)cnxn+r−1 + 3
∞∑n=0
(n +
r)cnxn+r−1 − 2
∞∑n=1
cn−1xn+r−1 = 0
Solve for r1 = 0:∞∑
n=1(n− 1)(n− 2)cn−1x
n−1 −∞∑
n=0n(n− 1)cnx
n−1 + 3∞∑
n=0ncnx
n−1 − 2∞∑
n=1cn−1x
n−1 = 0
∞∑n=1
(n− 1)(n− 2)cn−1xn−1 −
∞∑n=1
n(n− 1)cnxn−1 + 3
∞∑n=1
ncnxn−1 − 2
∞∑n=1
cn−1xn−1 = 0
(n− 4)cn = (n− 3)cn−1, c1 =2
3, c2 =
1
2c1 =
1
3c0, c3 = 0
Can we find the value of c4 - ? (It’s an arbitrary number)
c5 = 2c4, c6 = 3c4...
General solution is:
y(x) = c0(1 +1
2x +
1
3x2) + c4x
4(1 + 2x + 3x2...)
54
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 2)Find the general solution for the ODE:
x2y′′ + xy′ + (x2 − 1)y = 0.
Solution:
55
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 2)Find the general solution for the ODE:
x2y′′ + xy′ + (x2 − 1)y = 0.
Solution:
56
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 2)Find the general solution for the ODE:
x2y′′ + xy′ + (x2 − 1)y = 0.
Solution:ν = 1 is an integer the solution is given as (page 262 (11))
y(x) = c1J1(x) + c2Y1(x)
57
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 12)Solved in the solution manual.
58
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 44)Find the Legendre polynomials P6 and P7 and the differential equationsfor them.
Solution:
59
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 44)Find the Legendre polynomials P6 and P7 and the differential equationsfor them.
Solution:page 269, substitute n = 6 into (26):
P6(x) =1
16[231x6 − 315x4 + 105x2 − 5]
n = 6, (1− x2)y′′ − 2xy′ + 42y = 0
60
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 46)Change variables x = cos(θ) in
sin(θ)d2y
dθ2+ cos(θ)
dy
dθ+ n(n + 1)(sin(θ))y = 0
Solution:
61
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 46)Change variables x = cos(θ) in
sin(θ)d2y
dθ2+ cos(θ)
dy
dθ+ n(n + 1)(sin(θ))y = 0
Solution:
dy
dθ=
dy
dx
dx
dθ=
dy
dx(− sin(θ))
62
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 46)Change variables x = cos(θ) in
sin(θ)d2y
dθ2+ cos(θ)
dy
dθ+ n(n + 1)(sin(θ))y = 0
Solution:
dy
dθ=
dy
dx
dx
dθ=
dy
dx(− sin(θ))
d2y
dθ2= (− cos(θ))
dy
dx+ (sin(θ))2
d2y
dx2
63
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 46)Change variables x = cos(θ) in
sin(θ)d2y
dθ2+ cos(θ)
dy
dθ+ n(n + 1)(sin(θ))y = 0
Solution:
dy
dθ=
dy
dx
dx
dθ=
dy
dx(− sin(θ))
d2y
dθ2= (− cos(θ))
dy
dx+ (sin(θ))2
d2y
dx2
sin(θ)(y′′ sin2(θ)− y′ cos(θ)) + cos(θ)(y′ sin(θ)) + n(n + 1) sin(θ)y = 0
64
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 46)Change variables x = cos(θ) in
sin(θ)d2y
dθ2+ cos(θ)
dy
dθ+ n(n + 1)(sin(θ))y = 0
Solution:
dy
dθ=
dy
dx
dx
dθ=
dy
dx(− sin(θ))
d2y
dθ2= (− cos(θ))
dy
dx+ (sin(θ))2
d2y
dx2
sin(θ)(y′′ sin2(θ)− y′ cos(θ)) + cos(θ)(y′ sin(θ)) + n(n + 1) sin(θ)y = 0
y′′ sin2(θ)− 2y′ cos(θ) + n(n + 1)y = 0
65
The Series Solutions of ODE (5.3 Special Functions )
Problem (5.3: 46)Change variables x = cos(θ) in
sin(θ)d2y
dθ2+ cos(θ)
dy
dθ+ n(n + 1)(sin(θ))y = 0
Solution:
dy
dθ=
dy
dx
dx
dθ=
dy
dx(− sin(θ))
d2y
dθ2= (− cos(θ))
dy
dx+ (sin(θ))2
d2y
dx2
sin(θ)(y′′ sin2(θ)− y′ cos(θ)) + cos(θ)(y′ sin(θ)) + n(n + 1) sin(θ)y = 0
y′′ sin2(θ)− 2y′ cos(θ) + n(n + 1)y = 0
y′′(1− x2)− 2y′x + n(n + 1)y = 0
Legendre equation.
66
See you next week :-) !
67
