Assignment 1
Due date: July 14, 2015
1. Solve y 0 = 2xy 2 and y(1) = 1.
Answer.
Since
dy
= y 0 = xy, then
dx
1
dy = 2xdx.
y2
So we get
1
− =
y
Z
1
dy =
y2
Z
2x dx = x2 + C.
Since y(1) = 1, then −1 = 1 + C, that is,
C = −2.
So we get −
1
= x2 − 2, which implies that the solution to y 0 = 2xy 2 and y(1) = 1 is given by:
y
y=−
x2
1
.
−2
2
2. Find the solution to y 0 − 2xy = ex and y(0) = 0.
Answer.
First, let’s find the solution to the homogeneous equation: y 0 − xy = 0, that is,
dy
= 2xy.
dx
So either y(x) = 0 or
1
dy = x dx. Then
y
Z
ln |y| =
Then y = ±ex
2 +C
1
dy =
y
Z
x dx = x2 + C.
2
= ±eC · ex . So the general solution to y 0 − xy = 0 is given by:
2
y = Cex .
2
2
Let yp (x) = C(x)ex be a particular solution to y 0 − 2xy = ex , then
2
yp0 = C 0 (x)ex + 2xC(x)ex
ex
2
2
= yp0 − 2xyp
2
2
= C 0 (x)ex + 2xC(x)ex − 2xC(x)ex
2
= C 0 (x)ex .
2
2
2
2
Then ex = C 0 (x)ex , that is, C 0 (x) = 1. So C(x) = x + C. Hence the general solution to y 0 − 2xy = ex is:
2
2
2
2
y(x) = C(x)ex = (x + C)ex = Cex + xex .
2
Since y(0) = 1, then 1 = C. So the solution to y 0 − 2xy = ex and y(0) = 1 is:
2
2
y = ex + xex .
3. Let α be a real constant, find a fundamental set of solutions to y 00 − 4y 0 + α(4 − α)y = 0. (Hint: You need to
consider two cases which depend on α.)
Answer.
Let y = erx be a solution to y 00 − 4y 0 + α(4 − α)y = 0, then
y 0 = rerx
y 00 = r2 erx
0 = y 00 − 4y 0 + α(4 − α)y
= r2 erx − 4rerx + α(4 − α)erx
= erx [r2 − 4r + α(4 − α)].
Then we get r2 − 4r + α(4 − α) = 0, which implies that
r1 = α,
and r2 = 4 − α.
We have the following two cases:
I. If r1 = r2 , that is, α = 4 − α, then α = 2. Then y1 (x) = e2x and y2 (x) = xe2x form a fundamental set of
solutions to y 00 − 4y 0 + 4y = 0.
II. If r1 6= r2 , that is, α 6= 4 − α, then α 6= 2. Then y1 (x) = eαx and y2 (x) = e(4−α)x form a fundamental set
of solutions to y 00 − 4y 0 + α(4 − α) = 0.
4. Find the general solution to y 00 + 10y 0 + 25y = 0.
Answer.
Let y = erx be a solution to y 00 + 10y 0 + 25y = 0, then
y 0 = rerx
y 00 = r2 erx
0 = y 00 + 10y 0 + 25y
= r2 erx + 10rerx + 25erx
= erx (r2 + 10r + 25).
Then r2 + 10r + 25 = 0, that is,
r1 = r2 = −5.
Therefore, the general solution to y 00 + 10y 0 + 25y = 0 is given by:
y = C1 e−5x + C2 xe−5x .
Page 2
5. Find a fundamental set of solutions to 2y 00 + 4y 0 + 9y = 0.
Answer.
Let y = erx be a solution to 2y 00 + 4y 0 + 9y = 0, then
y 0 = rerx
y 00 = r2 erx
0 = 2y 00 + 4y 0 + 9y
= 2r2 erx + 4rerx + 9erx
= erx (2r2 + 4r + 9).
Then 2r2 + 4r + 9 = 0, that is,
r1 =
−4 +
√
−56
4
√
14
= −1 +
i,
2
√
14
i.
2
and r2 = −1 −
Notice that
e
r1 x
= e
√
−1+
142
i
x
√
−x+i 214 x
√
−x
i 214 x
= e
·e
"
= e
√
14
x
2
= e−x cos
√
Then y1 (x) = e
−x
cos
14
x
2
√
!
+ i sin
√
!
−x
and y2 (x) = e
sin
6. Find a particular solution of y 00 − y = 2 sin(x2 ).
Let y = erx be a solution to y 00 − y = 0, then
y 0 = rerx
y 00 = r2 erx
0 = y 00 − y
= r2 erx − erx
= erx (r2 − 1).
Then r2 − 1 = 0, that is,
r1 = 1,
and r2 = −1.
Then the general solution to y 00 − y = 0 is:
y(x) = C1 ex + C2 e−x .
Page 3
!#
.
!
14
x form a fundamental set of solutions to 2y 00 +
2
4y 0 + 9y = 0.
Answer.
14
x
2
Let yp (x) = C1 (x)ex + C2 (x)e−x be a particular solution to y 00 − y = 2 sin(x2 ), then
yp0 (x) = C10 (x)ex + C1 (x)ex + C20 (x)e−x − C2 (x)e−x
Set
C10 (x)ex + C20 (x)e−x = 0.
Then we have
yp0 (x) = C1 (x)ex − C2 (x)e−x
yp00 (x) = C10 (x)ex + C1 (x)ex − C20 (x)e−x + C2 (x)e−x
So we have
yp00 − yp = C10 (x)ex + C1 (x)ex − C20 (x)e−x + C2 (x)e−x − C1 (x)ex + C2 (x)e−x
= C10 (x)ex − C20 (x)e−x
= sin(x2 )
In summary, we know that C10 (x) and C20 (x) satisfy
( 0
C1 (x)ex + C20 (x)e−x = 0
C10 (x)ex − C20 (x)e−x = 2 sin(x2 )
Solve C10 (x) and C20 (x), we get
C10 (x) = e−x sin(x2 ),
So
Z
C1 (x) =
e
−x
and C20 (x) = −ex sin(x2 ).
Z
2
and C2 (x) = −
sin(x ) dx,
ex sin(x2 ) dx.
Therefore, a particular solution to y 00 − y = 2 sin(x2 ) can be:
yp (x) = e
x
Z
e
−x
2
sin(x ) dx − e
−x
Z
ex sin(x2 ) dx .
7. Find the solution to x2 y 00 + 6xy 0 + 6y = 0, y(1) = 0 and y 0 (1) = 1.
Answer.
Let y = xr be a solution to x2 y 00 + 6xy 0 + 6y = 0, then
y 0 = rxr−1
y 00 = r(r − 1)xr−2
0 = x2 y 00 + 6xy 0 + 6y
= r(r − 1)xr + 6rxr + 6xr
= xr [r(r − 1) + 6r + 6].
Page 4
Then r(r − 1) + 6r + 6 = 0, then r2 + 5r + 6 = 0, that is,
r1 = −2,
and r2 = −3.
So the general solution to x2 y 00 + 6xy 0 + 6y = 0 is:
y = C1 x−2 + C2 x−3 .
Notice that
y 0 = −2C2 x−3 − 3C2 x−4 .
Since y(1) = 0 and y 0 (1) = 1, then
C1 + C2 = 0,
−2C2 − 3C2 = 1.
Then C1 = 1 and C2 = −2. So the solution to x2 y 00 + 6xy 0 + 6y = 0, y(1) = 0 and y 0 (1) = 1 is:
y = x−2 − x−3 .
8. Find the general solution to x2 y 00 + 5xy 0 + 4y = 0 for x > 0.
Answer.
Let y = xr be a solution to x2 y 00 + 5xy 0 + 4y = 0, then
y 0 = rxr−1
y 00 = r(r − 1)xr−2
0 = x2 y 00 + 5xy 0 + 4y
= r(r − 1)xr + 5rxr + 4xr
= xr [r(r − 1) + 5r + 4].
Then r(r − 1) + 5r + 4 = 0, then r2 + 4r + 4 = 0, that is,
r1 = r2 = −2.
So the general solution to x2 y 00 + 5xy 0 + 4y = 0 is:
y = C1 x−2 + C2 x−2 ln x .
9. Find the general solution to x2 y 00 + 3xy 0 + 2y = 0 for x > 0.
Answer.
Let y = xr be a solution to x2 y 00 + 3xy 0 + 2y = 0, then
y 0 = rxr−1
y 00 = r(r − 1)xr−2
0 = x2 y 00 + 3xy 0 + 2y
= r(r − 1)xr + 3rxr + 2xr
= xr [r(r − 1) + 3r + 2].
Page 5
Then r(r − 1) + 3r + 2 = 0, then r2 + 2r + 2 = 0, that is,
r1 = −1 + i,
and r2 = −1 − i.
Notice that
xr1
= x−1+i
= x−1 · xi
= x−1 · ei ln x
= x−1 [cos(ln x) + i sin(ln x)]
= x−1 cos(ln x) + ix−1 sin(ln x).
So the general solution to x2 y 00 + 3xy 0 + 2y = 0 is:
y = C1 x−1 cos(ln x) + C2 x−1 sin(ln x) .
10. Let α, β be constants, find the condition that α and β should satisfy for which all solutions of x2 y 00 + (α +
1)xy 0 + βy = 0 for x > 0 approach zero as x → 0. (Hint: lim xr ln x = 0 for all r > 0.)
x→0
Answer.
Let y = xr be a solution to x2 y 00 + (α + 1)xy 0 + βy = 0, then
y 0 = rxr−1
y 00 = r(r − 1)xr−2
0 = x2 y 00 + (α + 1)xy 0 + βy
= r(r − 1)xr + (α + 1)rxr + βxr
= xr [r(r − 1) + (α + 1)r + β]
= xr (r2 + αr + β).
Then r2 + αx + β = 0, that is,
r1 =
−α +
p
α2 − 4β
,
2
and r2 =
−α −
p
α2 − 4β
.
2
So we have the following three cases:
I. If α2 −4β > 0, that is, r1 and r2 are two real numbers. Then the general solution to x2 y 00 +(α+1)xy 0 +βy =
0 is given by:
y = C1 xr1 + C2 xr2 .
Since we need all solutions to approach
zero as x → 0, then r1 > 0 and r2 > 0. Since r1 > r2 , then we
p
−α − α2 − 4β
> 0. So we get
only need r2 > 0, that is,
2
p
α + α2 − 4β < 0.
Page 6
α
II. If α2 − 4β = 0, that is, r1 = − . Then the general solution to x2 y 00 + (α + 1)xy 0 + βy = 0 is given by:
2
α
α
y = C1 x− 2 + C2 x− 2 ln x.
Since we need all solutions to approach zero as x → 0, then −
α
> 0, that is,
2
α < 0.
III. If α2 − 4β < 0, that is, r1 and r2 are two complex numbers, and
p
α
4β − α2
r1 = − + i
.
2
2
Notice that
√
x
r1
= x
−α
+i
2
= x
−α
2
α
4β−α2
2
√
i
·x√
= x− 2 · e"i
α
= x− 2
α
= x− 2
4β−α2
2
4β−α2
2
ln x
!
!#
p
p
4β − α2
4β − α2
ln x + i sin
ln x
· cos
2
2
!
!
p
p
4β − α2
4β − α2
−α
cos
ln x + ix 2 sin
ln x .
2
2
Then the general solution to x2 y 00 + (α + 1)xy 0 + βy = 0 is given by:
!
!
p
p
α
4β − α2
4β − α2
−
−α
ln x + C2 x 2 sin
ln x .
y = C1 x 2 cos
2
2
Since we need all solutions to approach zero as x → 0, then −
α
> 0, that is,
2
α < 0.
In summary, in order to make all solutions to x2 y 00 + (α + 1)xy 0 + βy = 0 approach zero as x → 0, α and β
should satisfy the following two conditions:
(a). α < 0.
and
(b). α2 − 4β < α2
11. Find the power series solution about x = 1 to the problem y 00 + 4y 0 + 6xy = 0, y(1) = 0 and y 0 (1) = 1.
Page 7
Answer. Since 6x = 6(x − 1) + 6, it’s easy to see that x = 1 is an ordinary point of y 00 + 4y 0 + 6xy = 0. Let
∞
X
y=
ck (x − 1)k be a solution to y 00 + 4y 0 + 6xy = 0, then
k=0
y0 =
∞
X
kck (x − 1)k−1
k=0
=
∞
X
kck (x − 1)k−1
k=1
=
∞
X
(k + 1)ck+1 (x − 1)k
k=0
y 00 =
∞
X
k(k + 1)ck+1 (x − 1)k−1
k=0
=
∞
X
k(k + 1)ck+1 (x − 1)k−1
k=1
=
∞
X
(k + 1)(k + 2)ck+2 (x − 1)k
k=0
00
0 = y + 4y 0 + 6xy
= y 00 + 4y 0 + [6 + 6(x − 1)]y
∞
∞
∞
X
X
X
ck (x − 1)k
(k + 1)ck+1 (x − 1)k + [6 + 6(x − 1)]
=
(k + 1)(k + 2)ck+2 (x − 1)k + 4
=
k=0
k=0
∞
X
∞
X
(k + 1)(k + 2)ck+2 (x − 1)k + 4
(k + 1)ck+1 (x − 1)k + 6
∞
X
ck (x − 1)k
k=0
k=0
k=0
+6
k=0
∞
X
ck (x − 1)k+1
k=0
=
∞
X
k
(k + 1)(k + 2)ck+2 (x − 1) + 4
k=0
+6
∞
X
(k + 1)ck+1 (x − 1) + 6
k=0
∞
X
k
∞
X
ck (x − 1)k
k=0
ck−1 (x − 1)k
k=1
= 1 · 2 · c2 + 4 · 1 · c1 + 6c0 +
∞
X
[(k + 1)(k + 2)ck+2 + 4(k + 1)ck+1 + 6ck + 6ck−1 ] (x − 1)k
k=1
= 2c2 + 4c1 + 6c0 +
∞
X
[(k + 1)(k + 2)ck+2 + 4(k + 1)ck+1 + 6ck + 6ck−1 ] (x − 1)k .
k=1
Then
2c2 + 4c1 + 6c0 = 0,
and
(k + 1)(k + 2)ck+2 + 4(k + 1)ck+1 + 6ck + 6ck−1 = 0,
Page 8
∀k ≥ 1.
So we have
c2 = −3c0 − 2c1
4(k + 1)ck+1 + 6ck + 6ck−1
ck+2 = −
,
(k + 2)(k + 1)
∀k ≥ 1.
Since y(1) = 0 and y 0 (1) = 1, then c0 = 0 and c1 = 1, which implies that
c2 = −2
−4 · 2 · c2 + 6c1 + 6c0
c3 = −
3·2
−8 · (−2) + 6 · 1 + 6 · 0
= −
6
11
= − .
3
So the power series solution about x = 1 to the problem y 00 + 4y 0 + 6xy = 0, y(1) = 0 and y 0 (1) = 1 is given
by:
11
y(x) = (x − 1) − 2(x − 1)2 − (x − 1)3 + · · · .
3
12. Consider the differential equation 2(x − 1)y 00 + y 0 + y = 0 for x > 0.
(a) Find the general power series solutions at x = 0 (find the first four non-zero terms). What should be the
minimal radius of convergence of this series?
Answer. It’s easy to see that x = 0 is an ordinary point of 2(x − 1)y 00 + y 0 + y = 0. Since 2(x − 1)y 00 +
y 0 + y = 0, then
1
1
y0 +
y = 0.
y 00 +
2(x − 1)
2(x − 1)
1
, solve 2(z − 1) = 0, that is, z = 1, then f (z) is complex differentiable except at
2(z − 1)
1
about x = 0 is:
z = 1, which implies that the radius of convergence of the Taylor series for
2(x − 1)
Let f (z) =
R = |1 − 0| = 1.
Therefore, a series solution of the form
∞
X
n
cn x converges at least for |x| < 1. Let y =
n=0
∞
X
k=0
Page 9
ck xk be a
solution to 2(x − 1)y 00 + y 0 + y = 0, then
y0 =
∞
X
kck xk−1
k=0
=
=
∞
X
k=1
∞
X
kck xk−1
(k + 1)ck+1 xk
k=0
y 00 =
∞
X
k(k + 1)ck+1 xk−1
k=0
=
∞
X
k(k + 1)ck+1 xk−1
k=1
=
∞
X
(k + 1)(k + 2)ck+2 xk
k=0
0 = 2(x − 1)y 00 + y 0 + y
∞
∞
∞
X
X
X
k
k
ck xk
(k + 1)ck+1 x +
(k + 1)(k + 2)ck+2 x +
= 2(x − 1)
=
=
k=0
∞
X
k=0
k=0
k=0
∞
X
2(k + 1)(k + 2)ck+2 x
k+1
−
∞
X
∞
∞
X
X
k
ck xk
(k + 1)ck+1 x +
2(k + 1)(k + 2)ck+2 x +
k
2k(k + 1)ck+1 xk −
k=1
= −2 · 2c2 + c1 + c0 +
k=0
k=0
k=0
∞
X
2(k + 1)(k + 2)ck+2 xk +
∞
X
(k + 1)ck+1 xk +
k=0
k=0
∞
X
∞
X
ck xk
k=0
[2k(k + 1)ck+1 − 2(k + 1)(k + 2)ck+2 + (k + 1)ck+1 + ck ]xk
k=1
= c0 + c1 − 4c2 +
∞
X
[−2(k + 1)(k + 2)ck+2 + (k + 1)(2k + 1)ck+1 + ck ]xk .
k=1
Then we have
c0 + c1 − 4c2 = 0,
and
− 2(k + 1)(k + 1)ck+2 + (k + 2)(2k + 1)ck+1 + ck = 0,
So we get
c2 =
c0 + c1
,
4
and
ck+2 =
(k + 1)(2k + 1)ck+1 + ck
,
2(k + 1)(k + 2)
Page 10
∀k ≥ 1.
∀k ≥ 1.
So we get
c2 =
c3 =
=
=
=
=
1
1
c0 + c1
4
4
(1 + 1)(2 · 1 + 1)c2 + c1
2(1 + 1)(1 + 2)
6c2 + c1
12
1
1
c2 + c1
2 12
1 1
1
1
c0 + c1 + c1
2 4
4
12
1
5
c0 + c1 .
8
24
(b) Find the power series solutions at x = 0 (find the first four non-zero terms) such that y(0) = 0 and
y 00 (0) = 2.
Answer.
By the result of part (a), we know that the general power series solution at x = 0 has the
∞
X
ck xk such that
form
k=0
c2 =
c0 + c1
,
4
and
ck+2 =
(k + 1)(2k + 1)ck+1 + ck
,
2(k + 1)(k + 2)
Since y(0) = 0, then c0 = 0. Since y 00 (0) = 2, then c2 =
and c2 = 1, then
c1 = 4.
f 00 (0)
2
1
1
= = 1. Since c2 = c0 + c1 , c0 = 0
2!
2
4
4
By the result of part (a), then
1
5
5
c3 = c0 + c1 = .
8
24
6
For c4 , we have
c4 =
=
=
=
=
(2 + 1)(2 · 2 + 1)c3 + c2
2(2 + 1)(2 + 2)
15c3 + c2
24
5
1
c3 + c2
8
24
5 5
1
· +
·1
8 6 12
29
.
48
Page 11
∀k ≥ 1.
13. The Chebyshev differential equation is
(1 − x2 )y 00 − xy 0 + α2 y = 0,
where α is a real constant.
(a) Determined two linearly independent solutions in powers of x for |x| < 1.
Answer.
It’s easy to see that x = 0 is an ordinary point of
(1 − x2 )y 00 − xy 0 + α2 y
= 0. Let y =
k=0
be a solution to (1 − x2 )y 00 − xy 0 + α2 y = 0, then
y
0
=
∞
X
kck x
k−1
=
k=0
∞
X
∞
X
kck xk−1
k=1
∞
X
=
(k + 1)ck+1 xk
y 00 =
k=0
∞
X
k(k + 1)ck+1 xk−1 =
∞
X
k(k + 1)ck+1 xk−1
k=1
k=0
∞
X
(k + 1)(k + 2)ck+2 xk
=
k=0
0 = (1 − x2 )y 00 − xy 0 + α2 y
∞
∞
∞
X
X
X
2
k
k
2
= (1 − x )
(k + 1)(k + 2)ck+2 x − x
(k + 1)ck+1 x + α
ck xk
k=0
k=0
k=0
∞
∞
X
X
k
=
(k + 1)(k + 2)ck+2 x −
(k + 1)(k + 2)ck+2 xk+2
k=0
∞
X
k=0
∞
X
k=0
k=0
(k + 1)ck+1 xk+1 + α2
−
ck xk
∞
∞
∞
∞
X
X
X
X
=
(k + 1)(k + 2)ck+2 xk −
(k − 1)kck xk −
kck xk + α2
ck xk
k=0
k=2
= 1 · 2c2 + 2 · 3c3 x +
∞
X
(k + 1)(k + 2)ck+2 xk −
k=2
−1 · c1 x −
∞
X
k=1
∞
X
k=0
(k − 1)kck xk
k=2
kck xk + α2 c0 + α2 c1 x + α2
k=2
∞
X
ck xk
k=2
∞
X
= 2c2 + α2 c0 + (6c3 − c1 + α2 c1 )x +
[(k + 1)(k + 2)ck+2 − (k − 1)kck − kck + α2 ck ]xk
= 2c2 + α2 c0 + [6c3 + (α2 − 1)c1 ]x +
k=2
∞
X
[(k + 1)(k + 2)ck+2 + (α2 − k 2 )ck ]xk .
k=2
Page 12
ck xk
So we get
α2 c0 + 2c2 = 0,
6c3 + (α2 − 1)c1 = 0,
(k + 1)(k + 2)ck+2 + (α2 − k 2 )ck = 0,
and
That is, we have
c2 = −
α2
c0 ,
2
c3 = −
α2 − 1
c1 ,
6
and
ck+2 =
k 2 − α2
ck ,
(k + 2)(k + 1)
∀k ≥ 2.
So we get
(2n − 2)2 − α2
c2n−2
(2n)(2n − 1)
(2n − 2)2 − α2 (2n − 4)2 − α2
·
c2n−4
=
(2n)(2n − 1) (2n − 2)(2n − 3)
n
Y
(2k − 2)2 − α2
= c2
2k(2k − 1)
c2n =
k=2
α2
n
Y
(2k − 2)2 − α2
· c0 ·
= −
2
2k(2k − 1)
k=2
n
Y
(2k − 2)2 − α2
= c0
2k(2k − 1)
∀n ≥ 2
k=1
(2n − 1)2 − α2
c2n−1
(2n + 1)(2n)
(2n − 1)2 − α2 (2n − 3)2 − α2
=
·
c2n−3
(2n + 1)(2n) (2n − 1)(2n − 2)
n
Y
(2k − 1)2 − α2
= c3
(2k + 1)(2k)
c2n+1 =
k=2
n
Y (2k − 1)2 − α2
α2 − 1
= −
· c1 ·
6
(2k + 1)(2k)
k=2
n
Y
(2k − 1)2 − α2
,
= c1
(2k + 1)(2k)
∀n ≥ 2.
k=1
In summary, we have
c2n
n
Y
(2k − 2)2 − α2
= c0
2k(2k − 1)
and c2n+1
k=1
n
Y
(2k − 1)2 − α2
= c1
,
(2k + 1)(2k)
k=1
Page 13
∀n ≥ 1.
∀k ≥ 2.
So we get
y = c0 + c1 x +
∞
X
c2n x2n +
n=1
∞
X
∞
X
c2n+1 x2n+1
n=1
!
!
n
∞
n
2 − α2
Y
X
Y
(2k − 2)2 − α2
(2k
−
1)
= c0 + c1 x + c0
x2n + c1
x2n+1
2k(2k − 1)
(2k + 1)(2k)
n=1 k=1
n=1 k=1
"
"
!
#
!
#
∞
n
∞
n
2
2
2 − α2
X Y (2k − 2) − α
X
Y
(2k
−
1)
= c0 1 +
x2n + c1 x +
x2n+1 .
2k(2k − 1)
(2k + 1)(2k)
n=1
n=1
k=1
k=1
Let
∞
X
y1 (x) = 1 +
y2 (x) = x +
n
Y
(2k − 2)2 − α2
2k(2k − 1)
!
(2k − 1)2 − α2
(2k + 1)(2k)
!
n=1
∞
X
k=1
n
Y
n=1
k=1
x2n
x2n+1 .
It’s clear that y1 (x) and y2 (x) are linearly independent, then y1 (x) and y2 (x) form a fundamental set of
solutions to y 00 − xy = 0.
(b) Show that if α is a non-negative integer n, then there is a polynomial solution of degree n.
Answer. If α = 0, it’s easy to see that y(x) = 1 is a solution to (1−x2 )y 00 −xy 0 +α2 y = (1−x2 )y 00 −xy 0 =
0. If α = 1, it’s easy to see that y(x) = x is a solution to (1 − x2 )y 00 − xy 0 + α2 y = (1 − x2 )y 00 − xy 0 + y = 0.
Let α be an integer n ≥ 2, if n is even, that is, n = 2N for some integer N , by the result of part (a), then
!
∞
m
X
Y
(2k − 2)2 − α2
y1 (x) = 1 +
x2m .
2k(2k − 1)
m=1
Then
m
Y
k=1
2)2
α2
(2k −
−
2k(2k − 1)
=
m
Y
k=1
2)2
k=1
(2N )2
(2k −
−
2k(2k − 1)
y1 (x) = 1 +
N
X
m=1
= 0 for all m ≥ N + 1, which implies that
m
Y
(2k − 2)2 − α2
2k(2k − 1)
!
x2m ,
k=1
which is a polynomial solution of degree 2N = α = n.
If n is odd, that is, n = 2N + 1 for some integer N , by the result of part (a), then
!
∞
m
X
Y
(2k − 1)2 − α2
x2m+1 .
y2 (x) = x +
(2k + 1)(2k)
m=1
Then
k=1
m
m
Y
Y
(2k − 1)2 − α2
(2k − 1)2 − (2N + 1)2
=
= 0 for all m ≥ N + 1, which implies that
(2k + 1)(2k)
(2k + 1)(2k)
k=1
k=1
!
N
m
Y
X
(2k − 1)2 − α2
y2 (x) = x +
x2m+1 ,
(2k + 1)(2k)
m=1
k=1
is a polynomial solution of degree 2N + 1 = α = n.
Page 14
(c) Find a polynomial solution for each of the case α = n = 0, 1, 2, 3.
Answer.
By the result of part (b), then
• When α = 0, then y(x) = 1 is a polynomial solution.
• When α = 1, then y(x) = x is a polynomial solution.
02 − 22 2
• When α = 2, then y(x) = 1 +
x = 1 − 2x2 is a polynomial solution.
2·1
12 − 32 3
4
• When α = 3, then y(x) = x +
x = x − x3 is a polynomial solution.
3·2
3
Page 15