MATB44 Assignment 3 2015–16 Solutions
Question 1.
Find a particular solution yp of each of the following equations.
(a) y 00 + 16y = e3x
(c) y 00 + 2y 0 − 3y = 1 + xex
Solution:
Question 1(a):
y 00 + 16y = e3x
Solution to homogeneous equation is
y1 = cos(4x), y2 = sin(4x)
Undetermined coefficients: Try
yp = Ae3x
Substitute in:
(9 + 16)Ae3x = e3x
1
yp = e3x
25
General solution
y = C1 cos(4x) + C2 sin(4x) +
1 3x
e
25
Question 1(b):
y 00 − y 0 − 6y = 2 sin(3x)
Solution to homogeneous equation:
r2 − r − 6 = 0
(r − 3)(r + 2) = 0
y1 = e3x , y2 = e−2x
Undetermined coefficients: use
yp = A sin(3x) + B cos(3x)
(sincce sin(3x) is not a solution of the homogeneous equation)
Substitute in:
1
2
−9A sin(3x)−9B cos(3x)−(3A cos(3x)−3B sin(3x))−6(A sin(3x)+B cos(3x)) = 2 sin(3x)
−9A + 3B − 6A = 2
−9B − 3A − 6B = 0
A = −5B
B = 1/40
A = −5/40
General soln
y = C1 e3x + C2 e−2x − (5/40) sin(3x) + (1/40) cos(3x)
Question 1(c):
y 00 + 2y 0 − 3y = xex
To use method of undetermined coefficients, check whether 1 is a root
of the characteristic equation:
r2 + 2r − 3 = 0
which is equivalent to
(r − 1)(r + 3) = 0
so 1 is a root of the equation. Thus our test solution must be
y = x(Ax + B)ex
in other words
y = (Ax2 + Bx)ex
y 0 = (2Ax + B)ex + (Ax2 + Bx)ex = Ax2 + (B + 2A)x + B ex
y 00 = 2Aex + 2(2Ax + B)ex + (Ax2 + Bx)ex
= Ax2 + (4A + B)x + 2(A + B) ex .
Thus
y 00 +2y 0 −3y = (A + 2A − 3A)x2 + (4A + B + 2B + 4A − 3B)x + (4B + 2A) ex
= 0 × x2 + 8Ax + (2A + 4B) ex .
To solve
y 00 + 2y 0 − 3y = xex
we equate coefficients of ex and coefficients of xex and of x2 ex : 4B +
2A = 0 (coefficient of ex ) or
2
B=− A
3
3
1
+8A = 1 (coefficient of xex ) or A = 18 , B = − 16
Solution is
1
1
y = ( x2 − x)ex
8
16
Question 1(d)
(d) y 00 + y = sin(x) + x cos(x)
Soln: The solution is the sum of the solutions to the problems
y 00 + y = sin(x)
(0.1)
and
y 00 + y = x cos(x).
(0.2)
Note that sin x and cos x are solutions of the homogeneous equation.
So the test solution for (0.1) is
y = x(A sin(x) + B cos(x))
This leads to
y 0 = A sin(x) + B cos(x) + x(A cos(x) − B sin(x))
y 00 = 2A cos(x) − 2B sin(x) + x(−A sin(x) − B cos(x))
Hence
y 00 + y = 2A cos(x) − 2B sin(x) = sin(x)
so B = − 12 and A = 0. Solution to (0.1) is y = − 12 x cos(x).
For (0.2), the test solution is
y = x(Ax + B) cos(x) + x(Cx + D) sin(x)
y 0 = (Ax + B) cos(x) + x(A) cos(x) − x(Ax + B) sin(x)
+(Cx + D) sin(x) + x(C sin x) + x(Cx + D) cos x
00
y = A cos x−(Ax+B) sin x+A cos x−Ax sin x−(Ax+B) sin x−Ax sin x−x(Ax+B) cos x
+(Cx+D) cos x+C sin x+C sin x+Cx cos x+(Cx+D) cos x+Cx cos x−x(Cx+D) sin x
y 00 + y = 2A cos x − 2(Ax + B) sin x − 2Ax sin x
+2C sin x + 2(Cx + D) cos x + 2Cx cos x
= 2(A + D) cos x + 2(C − B) sin x − 4Ax sin x + 4Cx cos x.
As this must be equal to x cos x we need
1
1
C = , A = D = 0, , B = C =
4
4
4
so the solution is
x
x2
cos x +
sin x
4
4
Question 2. Use the method of variation of parameters to find a
particular solution of the following differential equations:
Question 2(a)
y 00 + 9y = 2 sec(3x)
y=
Two linearly independent solutions to the homogeneous equation are
y1 = cos 3x and y2 = sin 3x. The Wronskian is W = 3. According to
variation of parameters, this means a particular solution is
Z
Z
y2 (t)g(t)
y1 (t)g(t)
Y (x) = −y1 (x)
dt + y2 (x)
dt
W (t)
W (t)
Z
Z
2 cos 3t
2 sin 3t
dt + sin 3x
dt
= − cos 3x
3 cos 3t
3 cos 3t
2
2
= cos 3x ln cos 3x + x sin 3x
9
3
Question 2(b)
y 00 − 2y 0 + y = x−2 ex
Two linearly independent solutions to the homogeneous equation are
y1 = ex , y2 = xex . So W = y1 y20 − y2 y10 = e2x (x + 1) − xe2x = e2x
Thus a particular solution is (by variation of parameters)
Z
Z t −2 t
tet t−2 et
et e
x
x
Y (x) = −e
dt + xe
2t
e
e2t
1
= −ex ln(x) − xex = −ex ln(x) − ex
x
(b) y 00 − 2y 0 + y = x−2 ex
(c) x2 y 00 − 3xy 0 + 4y = x4
Question 2(c).
x2 y 00 − 3xy 0 + 4x = x4
We need to find solution of the homogeneous equation
x2 y 00 − 3xy 0 + 4x = 0
5
This is an Euler equation: it has a solution y = xr where
r(r − 1) − 3r + 4 = 0
or
r2 − 4r + 4 = 0
or
(r − 2)2 = 0
in other words
y1 (x) = x2 .
Since this is a repeated root, there is an additional solution
y2 (x) = x2 ln(x).
To solve the inhomogeneous equation:
Wronskian
W (y1 , y2 ) = x2 (2x ln(x) + x2 /x − x2 ln(x)(2x))
= x3
The equation (normalized so coefficient of y 00 is 1) is
4
3
y 00 − y 0 + = x2
x
x
This gives the solution of the inhomogeneous equation (variation of
parameters) as
Z
Z
y2 (x)x2
y1 (x)x2
Y (x) = −y1 (x)
dx
+
y
(x)
dx
2
x3
x3
Z 2
Z 2 2
x ln(x)
xx
2
2
= −x
dx + x ln(x)
dx
x
x3
Z
Z
2
2
= −x
x ln(x)dx + x ln(x) xdx
Z
2
= −x
udv + x2 ln(x)(x2 /2)
where u = ln(x) and v = x2 /2 so du = dx/x and dv = xdx. Hence the
integral becomes
Z
2
−x (uv − vdu) + (x4 /2) ln(x)
Z
2
2
= −x x /2 ln(x) − xdx/2 + x4 /2 ln(x)
6
= x4 /4.
So the solution is
y = x4 /4.
Question 2(d)
x2 y 00 + xy 0 + y = ln(x)
Two linearly independent solutions to the homogeneous equation are
y1 = cos ln(x), y2 = sin ln(x) The Wronskian is
1
W =
x
By variation of parameters, the solution of the inhomogeneous equation
is
Z
Z
sin ln(t) ln(t)
cos ln(t) ln(t) −1
Y (x) = − cos ln(x)
dt + sin ln(x)
t dt
2
−1
tt
t2
Z
Z
= − cos ln(x) sin ln(t) ln(t)d ln(t) + sin ln(x) cos ln(t) ln(t)d ln(t)
Z
Z
= − cos ln(x) u sin(u)du + sin ln(x) u cos(u)du
= cos ln(x)(u cos u − sin u)|u=ln x + sin ln x(u sin u + cos u)|u=ln x
This simplifies to Y (x) = ln(x).
Question 3. Use the method of undetermined coefficients to find
particular solutions of the following equations:
Question 3(a)
y 00 + 9y = 4 cos(3x)
Soln: cos 3x is a solution of the homogeneous equation so we must use
the test solution
y = Ax sin 3x + Bx cos 3x
This gives
y 0 = A sin 3x + B cos 3x + x(3A cos 3x − 3B sin 3x)
y 00 = 2(3A cos 3x − 3B sin 3x) + x(−9A sin 3x − 9B cos 3x)
y 00 + 9y = 6A cos 3x − 6B sin 3x = 4 cos 3x
so A = 2/3 and B = 0.
Question 3 (b).
y 00 + 4y 0 + 4y = 3e−2x + e−x
7
Solution:
First solve
(0.3)
y 00 + 4y 0 + 4y = e−x
(0.4)
y 00 + 4y 0 + 4y = 3e−2x
For (0.1), try
y = Ce−x
y 0 = −Ce−x , y 00 = Ce−x
so
(C − 4C + 4C)e−x = e−x
which implies C = 1.
For (0.3), the solution is yA = e−x .
For (0.4), note e−2x and xe−2x are solutions of the homogeneous
equation (since −2 is a repeated root of the characteristic equation)
So substitute
y = Dx2 e−2x
y 0 = D(2xe−2x − 2x2 e−2x ) = D(2x − 2x2 )e−2x
y 00 = D(2 − 4x)e−2x − 2D(2x − 2x2 )e−2x
= D(2 − 8x + 4x2 )e−2x
So
y 00 + 4y 0 + 4y = D(2 − 8x + 4x2 ) + 4D(2x − 2x2 ) + 4Dx2 e−2x
= 2D + (−8D + 8D)x + (4D − 8D + 4D)x2 e−2x
= 2De−2x = 3e−2x
Hence D = 3/2. So the solution to (0.2) is
yB = (3/2)x2 e−2x
So the solution to this question is
y = yA + yB = e−x + (3/2)x2 e−2x .
Question 4. For x > 0, find the general solution of
2x2 y 00 + xy 0 − y = 3x − 5x2
8
The homogeneous equation is an Euler equation. To convert it to a
constant-coefficient equation, we substitute x = eu or u = ln(x) which
lead to
1
du
=
dx
x
Then we have
dy
dy du
1 dy
=
=
dx
du dx
x du
2
dy
1 dy 1 d dy
+
=− 2
dx2
x du x dx du
1 dy 1 1 d2 y
+
=− 2
x du x x du2
Hence the homogeneous equation becomes
1 d2 y
1 dy
1 dy
2
2x − 2
+
−y =0
+x
x du x2 du2
x du
in other words
2
or
dy dy
d2 y
−
2
+
−y =0
du2
du du
d2 y
dy
−
−y =0
2
du
du
The right hand side of the equation transforms into
2
3x − 5x2 = 3eu − 5e2u
So we need to solve the equation
dy
d2 y
−
− y = 3eu − 5e2u
2
du
du
We can do this using the method of undetermined coefficients. The
characteristic equation is
2
2r2 − r − 1 = 0
so its solutions are r = 1 and r = − 21
To solve
d2 y 1 dy 1
3
−
− y = eu
2
du
2 du 2
2
we need to substitute
y = Aueu
9
(the extra factor u is because eu is a solution of the homogeneous
equation) So
3
d2 y 1 dy 1
− y = A(3/2)eu = eu
−
2
du
2 du 2
2
so A = 1.
To solve
d2 y 1 dy 1
5
− y = − e2u
−
2
du
2 du 2
2
we substitute
y = Be2u
(here no extra factor u is required because 2 is not a root of the homogeneous equation) This gives
1
5
B(4 − 1 − ) = −
2
2
or
B = −1
So the solution is
y = ueu − e2u
or
y = x ln(x) − x2
Question 5
y 00 + xy = 0
This is very similar to the Airy equation (Example 2, Chap. 5.2 of
textbook). The Airy Equation is solved in the textbook and was also
solved in class. The Airy Equation is
y 00 − xy = 0
The recurrence is (compare text, Section 5.2, (18))
∞
∞
X
X
n
(n + 2)(n + 1)an+2 x = −
an xn+1 .
n=0
n=0
This leads to (cf. equation following (18) on p. 259 in text)
10
2 · 1a2 +
∞
X
n
(n + 2)(n + 1)an+2 x = −
∞
X
n=1
an−1 xn .
n=1
This leads to (cf. (16) Section 5.2 text)
(n + 2)(n + 1)an+2 = −an−1 n ≥ 1.
Hence the solution is (compare second last equation p. 243)
a0
a3n = (−1)n
2 · 3 · 5 · 6 . . . (3n − 4)(3n − 3)(3n − 1 )(3n)
and (compare first equation, p. 344)
a3n+1 = (−1)n
a1
3 · 4 · 6 · 7 . . . (3n − 3)(3n − 2)(3n)(3n + 1)
Hence the solution to Question 1 is the same as the solution to the
Airy equation, except that the signs of a3n and a3n+1 are (−1)n (whereas
for the Airy equation the signs of all nonzero an are positive).
Question 6
(2x − x2 )y 00 − 6(x − 1)y 0 − 4y = 0
Substitute t = x − 1 since initial conditions are given at x = 1
dy
dy
=
dx
dt
and
d2 y
d2 y
= 2
dx2
dt
This substitution gives
(−t2 + 1)y 00 − 6ty 0 − 4y = 0
(where now y 0 means
Put
dy
dt
and y 00 means
y=
∞
X
d2 y
).
dt2
an tn ,
n=0
so
y0 =
∞
X
n=0
nan tn−1
11
and
00
y =
∞
X
n(n − 1)an tn−2
n=0
Thus we have
∞
∞
∞
∞
X
X
X
X
n−2
n
n
n(n − 1)an t
−
n(n − 1)an t − 6
an (n)t − 4
an tn = 0.
n=2
n=0
n=0
n=0
Reindex first sum: n − 2 = m, n = m + 2.
∞
∞
∞
∞
X
X
X
X
m
n
n
(m+2)(m+1)am+2 t −
n(n−1)an t −6
an (n)t −4
an tn = 0.
m=0
n=0
n=0
n=0
This gives the recurrence
(n + 2)(n + 1)an+2 = (n2 + 5n + 4)an
or
an+2 =
(n + 4)an
.
n+2
The initial condition was
a0 = 0, a1 = 1
so all
a2n = 0
and
5
a3 = a1
3
7
7
a5 = a3 = a1
5
3
For odd n,
(0.5)
an =
n+2
a1
3
(Proof by induction: the previous equation is true for n = 3. Assume
this equation is true for n − 2 and prove it for n. We assume
n
an−2 = a1 .
3
Then
n+2
n+2
an =
an =
a1
n
3
as was to be shown.)
12
In other words, our formula is
a2m+1 =
−2(t + 1)t
∞
X
2m + 3
a1 .
3
an (n + r)(n + r − 1)tn+r−2
n=0
−2
∞
X
am+1 (m+1+r)(m+r)tm+r −2
m=−1
∞
X
an (n+r)(n+r−1)tn+r −6
n=0
∞
X
an (n+r)tn+r −4
n=0
m = −1 term from first sum:
a0 r(−1 + r) = 0
so r = 0 or 1
n ≥ 0:
−2an+1 (n + 1 + r)(n + r) − 2an (n + r)(n + r − 1) − 6an (n + 4) − 4an = 0
(n + r)(n + r − 1) − 3(n + 4) − 2
(n + 1 + r)(n + r)
If r = 1: a0 = 1, and y(t = 0) = 0 is automatically satisfied. So choose
r=1
Then
(n + 1)n − 3(n + 1) − 2
an+1 = −an
(n + 2)(n + 1)
(n + 1)(n − 3) − 2
= −an
(n + 2)(n + 1)
−3 − 2
a1 = −a0
2×1
2×1−3×2−2
a2 = −a1
3×2
(1 × (−3) − 2) × (2 × (−2) − 2) × . . . (n × (n − 4) − 2)
an = (−1)n
(n + 1)!n!
(when n ≥ 1).
P
n
So y = ∞
n=0 an (x − 1) with an as above.
Question 7
y 00 + xy 0 + y = 0
an+1 = −an
∞
X
n=0
an tn+r = 0.
13
Substitute y =
P∞
n=0
an x n ,
0
y =
∞
X
nan xn−1 ,
n=0
y 00 =
∞
X
n(n − 1)an xn−2
n=0
So
∞
X
n(n − 1)an x
n−2
+
n=0
∞
X
n
nan x +
n=0
∞
X
an xn = 0.
n=0
Reindex first sum:
∞
∞
X
X
n(n − 1)an xn−2 =
(m + 2)(m + 1)am+2 xm
n=0
m=−2
=
∞
X
(m + 2)(m + 1)am+2 xm
m=0
∞
X
(n + 2)(n + 1)an+2 xn
=
n=0
So
∞
X
(n + 2)(n + 1)an+2 + nan + an = 0.
n=0
So
(n + 1)an
an
=−
.
(n + 2)(n + 1)
n+2
If y1 (0) = 1, then a0 = 1 and we may take a1 = 0. So
a0
a2 = −
2
a2
a0
a0
a4 = − =
= 2
4
2×4
2 (1 × 2)
Claim
(−1)m a0
a2m =
2m m!
Proof by induction: True for m = 0. If true for m − 1, the induction
assumption is that
an+2 = −
a2(m−1) =
(−1)m−1 a0
.
2m−1 (m − 1)!
14
We have from the recurrence relation that
a2(m−1)
a2m = −
2m
m
(−1) a0
=
2m m!
which completes the proof.
Hence the series is
∞
X
(−1)m x2m
y1 =
2m m!
m=0
since a0 = 1.
If y2 (0) = 0, we are forced to have y20 (0) 6= 0, because we chose
y10 (0) = 0; take y20 (0) = 1, or a1 = 1.
Then
a2m−1
a2m+1 = −
2m + 1
(by recurrence relation). Claim
(−1)m a1
.
1 × 3 × 5 × · · · × (2m + 1)
Proof by an induction similar to the proof for the formula for y1 . So
∞
X
(−1)m x2m+1
.
y2 =
1
×
3
×
5
×
·
·
·
×
(2m
+
1)
m=0
a2m+1 =
(b) Ratio test to verify convergence of series:
For y1 , the m-th term in the series is
bm = (−1)m
x2m
2m m!
The (m-1)-th term is
bm−1 =
So
|
Similarly for y2 .
bm
|=
(−1)m−1 x2(m−1)
2m−1 (m − 1)!
x2m 2m−1 (m − 1)!
2m m!x2(m−1)
bm−1
x2
=
→ 0 as m → ∞
2m
15
(c)
m
∞ X
−x2
1
2
y1 =
= e−x /2 .
2
m!
m=0
Reduction of order: A second solution is
y2 (x) = v(x)y1 (x)
where
y1 v 00 + (2y10 + py1 )v 0 = 0
for an equation
y 00 + p(x)y 0 + q(x)y = 0.
So here p(x) = x.
v 00 + (2
Putting u = v 0 ,
y10
+ p)v 0 = 0
y1
y10
+ p)u = 0
y1
y0
du
= −(2 1 + p)dx
u
y1
Z
ln |u| = −2 ln |y1 | − p
u0 + (2
R
u = y1−2 e− p ,
Z
Z
x2
pdx = xdx = .
2
So
2
2
u = ex eZ−x /2 = ex
2
v = ex /2 dx
or
−x2 /2
Z
y2 (x) = e
x
2 /2
2 /2
et
dt.
0
Question 8 Determine whether x = 0 is an ordinary point, a regular
singular point, or an irregular singular point. If it is a regular singular
point, find the exponents of the differential equation at x = 0.
(a) xy 00 + (x − x3 )y 0 + (sin x)y = 0 This is equivalent to
sin x
y=0
y 00 + (1 − x2 )y 0 +
x
16
Note that the function
everywhere else) since
sin x =
sin x
x
is infinitely differentiable (at x = 0 and
∞
X
(−1)n x2n+1
n=0
(2n + 1)!
= x − x3 /3! + . . .
So x = 0 is an ordinary point.
(b)
x2 y 00 + cos xy 0 + xy = 0
This is in the form
P (x)y 00 + Q(x)y 0 + R(x)y = 0
so P (x) = x2 , Q(x) = cos(x), R(x) = x Then
x cos x
cos x
xQ(x)/P (x) =
=
2
x
x
which tends to ∞ as x → 0. Hence x = 0 is an irregular singular point.
(c)
x(1 + x)y 00 + 2y 0 + 3xy = 0
The equation is in the form
P (x)y 00 + Q(x)y 0 + R(x)y = 0
where P (x) = x(1 + x), Q(x) = 2, R(x) = 3x. So xQ(x)/P (x) =
2x
= 2/(1 + x) which remains finite as x → 0. Also x2 R(x)/P (x) =
x(1+x)
3xx2
1+x
→ 0 as x → 0. So x = 0 is a regular singular point. (It is a
singular point because P (0) = 0.)
Exponents for 3(c): Multiplying by x2 /x(1 + x), the equation becomes
2x2
3xx2
2 00
0
xy +
y +
y=0
x(1 + x)
x(1 + x)
so
2x
3x2
x2 y 00 +
y0 +
y=0
(1 + x)
(1 + x)
The Euler equation which approximates this is
x2 y 00 + 2xy 0 = 0
17
1
since 1+x
= 1 − x + x2 . . . .. so we obtain this Euler equation by taking
2x
3x2
the leading order terms in x in (1+x)
and (1+x)
(note that to order 0 in
x the coefficient of y is 0). Exponents:
r(r − 1) + 2r = 0
r
(substituting y = x ) So r(r + 1) = 0 or r = 0 or −1.
Question 9
2x2 y 00 + xy 0 − (1 + 2x2 )y = 0
P
n+r
Take y = ∞
So
n=0 an x
2
∞
X
(n+r)(n+r−1)an x
n+r
n=0
∞
∞
∞
X
X
X
n+r
n+r
+ (n+r)an x −
an x −2
an xn+r+2 = 0
n=0
n=0
Reindex last sum: it becomes (using m = n + 2)
∞
X
am−2 xm+r .
−2
m=2
So collecting terms multiplying xn+r for n ≥ 2, we have
an [2(n + r)(n + r − 1) + (n + r) − 1] − 2an−2 = 0
So for n ≥ 2, we get
an =
2an−2
(n + r) [2(n + r) − 1] − 1
For n = 1:
a1 [2(1 + r)r + r] = 0
a1 [r(3 + 2r)] = 0
or r = 0 or r = −3/2 assuming a1 6= 0.
For n = 0:
a0 [2r(r − 1) + r − 1] = 0
a0 (r − 1)(2r + 1) = 0
r = 1 or r = −1/2 assuming a0 6= 0.
If a0 6= 0, take r = 1:
2an−2
an =
(n + 1)(2n + 1) − 1
n=0
18
2an−2
2an−2
=
2
2n + 3n
n(2n + 3)
=
So
a2m =
So
y 1 = a0
a0
m!(4m + 3)(4[m − 1] + 3) . . . (4 + 3)
∞
X
x2m
m!(4m + 3)(4[m − 1] + 3) . . . (4 + 3)
m=0
If r = − 12 ,
an =
2an−2
2an−2
=
(n − 1/2)(2n − 2) − 1
(2n − 1)(n − 1) − 1
2an−2
2an−2
=
2
2n − 3n
n(2n − 3)
=
So
2a0
2(4 − 3)
a0
=
m!(4m − 3) . . . (4 − 3)
a2 =
a2m
So
y 2 = a0
∞
X
1
x2m− 2
x−1/2 +
m!(4m − 3) . . . (4 − 3)
m=1
!
.
Question 10
2xy 00 − y 0 − y = 0
P∞
P
n+r
n
. So
Put y = xr ∞
n=0 an x
n=0 an x =
0
y =
∞
X
an (n + r)xn+r−1 ,
n=0
00
y =
∞
X
an (n + r)(n + r − 1)xn+r−2 .
n=0
So
∞
∞
∞
X
X
X
2
an (n+r)(n+r−1)an xn+r−1 −
an (n+r)xn+r−1 −
an xn+r = 0.
n=0
n=0
n=0
19
Reindex first 2 sums: m = n − 1
∞
∞
∞
X
X
X
m+r
m+r
2
am+1 (m+1+r)(m+r)x
−
am+1 (m+1+r)x
−
am xm+r = 0
m=−1
m=−1
m=0
So for m = −1 this gives
a0 (2r(−1 + r) − r) = 0
Assuming a0 6= 0, we get
−2r + 2r2 − r = 0
r = 0 or r = 3/2.
For m ≥ 0 we get the recurrence relation
am+1 [2(m + 1 + r)(m + r) − (m + 1 + r)] = am
am
am+1 =
(m + r + 1)[2(m + r) − 1]
For r = 0 this becomes
am
am+1 =
(m + 1)(2m − 1)
For r = 3/2 it becomes
am
am+1 =
(m + 5/2)(2m + 2)
In the case r = 0, we have
a0
am = −
m!)(2m − 3)(2m − 5) . . . 1
or
∞
X
xm
y1 = −a0
m!(2m − 3)(2m − 5) . . . 1
m=0
In the case r = 3/2, we get
am =
or
a0
5/2(1 + 5/2) . . . (m + 3/2)2m m!
∞
X
xm
y2 = a0 (1 +
5/2(1 + 5/2) + · · · + (m + 3/2)2m m!
m=1