Math 257/316 Assignment 2 Solutions
P
n
Problem 1. Find the first six non-zero terms in the power series y = ∞
n=0 an (x − x0 ) of
the general solution of the following second-order, linear, homogeneous ODEs, centred at
the indicated point x0 :
a)
y 00 + x2 y 0 + y = 0,
x0 = 0
b)
y 00 − xy 0 + y = 0,
x0 = 3,
writing the solutions in terms of a0 and a1 .
In each case, we have
y=
∞
X
n
0
an (x − x0 ) , y =
n=0
∞
X
nan (x − x0 )
n−1
00
, y =
n=1
=
∞
X
n(n − 1)an (x − x0 )n−2
n=2
∞
X
(n + 2)(n + 1)an+2 (x − x0 )n
n=0
(where we shifted in the index in the y 00 sum because we may need to do so below).
a)
y 00 + x2 y 0 + y = 0,
x0 = 0.
Using the shifted-index formula for y 00 as above, and multiplying through the y 0 series
by x2 :
∞
∞
∞
X
X
X
n
n+1
0=
(n + 2)(n + 1)an+2 x +
nan x
+
an xn .
n=0
n=1
n=0
Shifting the index in the middle series n 7→ n − 1 (or, if you prefer, changing the
index by k = n + 1 (n = k − 1) and then changing ”k” back to ”n”):
0=
∞
X
n
(n + 2)(n + 1)an+2 x +
n=0
= [2a2 + a0 ] + [6a3 + a1 ]x +
∞
X
(n − 1)an−1 x +
n=2
∞
X
n
∞
X
an xn
n=0
[(n + 2)(n + 1)an+2 + (n − 1)an−1 + an ]xn .
n=2
1
This requires
1
a2 = − a0 ,
2
1
a3 = − a1 ,
6
and the recurrence relation
an+2 = −
(n − 1)an−1 + an
,
(n + 1)(n + 2)
n = 2, 3, 4, 5, . . . .
So a0 and a1 are free, and
a4 = −
1
1
a1 + a2
= a0 − a1 ,
12
24
12
a5 = −
2a2 + a3
1
1
= a0 +
a1 ,
20
20
120
so
1 2
1 4
1 5
y(x) = a0 1 − x + x + x + · · ·
2
24
20
1 4
1 5
1 3
x + ··· ,
+ a1 x − x − x +
6
12
120
or, if you prefer,
y(x) = a0 + a1 x −
b)
y 00 − xy 0 + y = 0,
a0 2 a1 3 a0
a1 4 a0
a1 5
x − x +
−
x +
+
x + ··· .
2
6
24 12
20 120
x0 = 3,
Plugging into the ODE yields
0=
∞
X
n
(n + 2)(n + 1)an+2 (x − 3) − x
n=0
∞
X
n−1
nan (x − 3)
+
n=1
∞
X
an (x − 3)n .
n=0
To multiply the second sum through by x, we write x = (x − 3) + 3, so:
0=
=
=
∞
X
n
(n + 2)(n + 1)an+2 (x − 3) − (x − 3)
n=0
∞
X
∞
X
n=0
∞
X
n=1
∞
X
n−1
nan (x − 3)
n=1
(n + 2)(n + 1)an+2 (x − 3)n −
n
(n + 2)(n + 1)an+2 (x − 3) −
n=0
= [2a2 − 3a1 + a0 ] +
∞
X
n=1
∞
X
−3
∞
X
n−1
nan (x − 3)
n=1
nan (x − 3)n −
n
nan (x − 3) −
∞
X
n=1
∞
X
3nan (x − 3)n−1 +
2
an (x − 3)n
n=0
∞
X
an (x − 3)n
n=0
n
3(n + 1)an+1 (x − 3) +
n=0
[(n + 2)(n + 1)an+2 − 3(n + 1)an+1 − (n − 1)an ](x − 3)n
n=1
+
∞
X
∞
X
n=0
an (x − 3)n
(or we could have observed that the lower limit in the second sum could be changed
to 0 without changing anything, and not written the constant term separately, saving
a little bit of effort). This requires
1
a2 = (3a1 − a0 ),
2
and the recurrence relation
an+2 =
3(n + 1)an+1 + (n − 1)an
,
(n + 1)(n + 2)
n = 1, 2, 3, 4, . . . .
So a0 and a1 are free, and
3
1
1
1 9
1
5
a3 = a2 = (3a1 − a0 ), a4 =
(9a3 + a2 ) =
( (3a1 − a0 ) + (3a1 − a0 )) =
(3a1 − a0 ),
3
2
4·3
4·3 2
2
4·3
1
6
1
(12a4 + 2a3 ) =
(5(3a1 − a0 ) + (3a1 − a0 )) =
(3a1 − a0 ), . . .
a5 =
5·4
5·4
5·4
so
1
1
y(x) = a0 + a1 (x − 3) + (3a1 − a0 )(x − 3)2 + (3a1 − a0 )(x − 3)3
2
2
5
6
+
(3a1 − a0 )(x − 3)4 +
(3a1 − a0 )(x − 3)5 + · · ·
4·3
5·4
or, if you prefer,
1
1
5
6
2
3
4
5
y(x) = a0 1 − (x − 3) − (x − 3) −
(x − 3) −
(x − 3) + · · ·
2
2
4·3
5·4
3
3
3·5
3·6
2
3
4
5
+ a1 x − 3 + (x − 3) + (x − 3) +
(x − 3) +
(x − 3) + · · · .
2
2
4·3
5·4
Notice something funny here? If we take, say a0 = 3 and a1 = 1 (so 3a1 − a0 = 0),
all terms quadratic and higher cancel out, leaving just y(x) = x, which is indeed a
solution! (And if we wanted, we could take y(x) = x as one of our fundamental set of
solutions, and, say, the function in the square brackets multiplying a0 as the other.)
Problem 2. Determine a power series solution centred about x0 = 0 for the following initial
value problem
(x2 − 4x + 4)y 00 + (4x − 8)y 0 + 2y = 0,
Can you express your answer as a familiar function?
3
y(0) = 1/2,
y 0 (0) = 1/4.
P
n
0
00
As ever, substituting y(x) = ∞
n=0 an x , and the expressions for y and y , into the
ODE, and then multiplying through by the powers of x where appropriate, yields
0 = x2
∞
X
n(n − 1)an xn−2 − 4x
n=2
∞
X
+ 4x
=
nan xn−1 − 8
+
∞
X
∞
X
nan xn−1 + 2
∞
X
4nan xn −
n=1
n(n − 1)an xn−2
an xn
n=0
4n(n − 1)an xn−1 +
n=2
∞
X
∞
X
n=2
n=1
n(n − 1)an xn −
n=2
∞
X
n(n − 1)an xn−2 + 4
n=2
n=1
∞
X
∞
X
∞
X
4n(n − 1)an xn−2
n=2
8nan xn−1 +
n=1
∞
X
2an xn .
n=0
In the terms with powers xn−1 , shift the index n 7→ n + 1 (or make a change of index
k = n − 1), and in the term with power xn−2 , shift the index n 7→ n + 2 (or change index
k = n − 2), remembering to also shift the starting point of the sum appropriately, to find
0=
∞
X
n
n(n − 1)an x −
n=2
∞
X
+
n=1
∞
X
n
4(n + 1)(n)an+1 x +
n=1
4nan xn −
∞
X
∞
X
4(n + 2)(n + 1)an+2 xn
n=0
8(n + 1)an+1 xn +
n=0
∞
X
2an xn
n=0
= [8a2 − 8a1 + 2a0 ] + [−8a2 + 24a3 + 4a1 − 16a2 + 2a1 ]x
∞
X
+
[n(n − 1)an − 4(n + 1)nan+1 + 4(n + 2)(n + 1)an+2 + 4nan − 8(n + 1)an+1 + 2an ]xn
n=2
= [2(4a2 − 4a1 + a0 )] + [2(12a3 + 3a1 − 12a2 )]x
∞
X
+
[4(n + 2)(n + 1)an+2 − (4n + 8)(n + 1)an+1 + (n + 1)(n + 2)an ]xn .
n=2
So
a2 = a1 − a0 /4,
a3 = a2 − a1 /4,
and the recurrence relation is
(4n + 8)an+1 − (n + 2)an
4(n + 2)
= an+1 − an /4,
n = 2, 3, 4, 5, . . . .
an+2 =
Now use the initial data:
1/4 = y 0 (0) = a1 .
1/2 = y(0) = a0 ,
4
Then
a2 = 1/8, a3 = 1/16, a4 = 1/32, a5 = 1/64, . . .
and so
∞
1X
1
1
1
y(x) = 1/2+x/4+x /8+x /16+x /32+· · · =
(x/2)n =
=
,
2
2 1 − x/2
2−x
2
3
4
(−2 < x < 2)
n=0
(recognizing the series as a geometric one).
Remark: it is no accident that the answer is so simple in the end, since this ODE
is an Euler equation (centred at x = 2), (x − 2)2 y 00 + 4(x − 2)y 0 + 2y = 0, with indicial
equation 0 = r(r − 1) + 4r + 2 = r2 + 3r + 2 = (r + 1)(r + 2) and hence general solution
y(x) = c1 |x − 2|−1 + c2 |x − 2|−2 .
Problem 3. For each of the following equations find all singular points, and determine
whether each one is regular or irregular:
a)
(x2 − 1)y 00 + y = 0
b)
x(x + 1)2 y 00 + y 0 = 0
c)
x2 y 00 + y 0 +
1
sin(x) y
=0
In each case we want to write the ODE, after dividing through by the coefficient of y 00 ,
in the form
y 00 + p(x)y 0 + q(x)y = 0,
and examine the coefficient functions p(x) and q(x) to find the singular points and determine whether or not they are ’regular’.
a) (x2 − 1)y 00 + y = 0
Here p(x) = 0, q(x) = 1/(x2 − 1), so the singular points are x = ±1. Obviously
(x − 1)p(x) = 0 is ’fine’ (analytic) at x = 1, as is (x + 1)p(x) = 0 at x = −1, so we are
only worried about q(x). Well,
(x − 1)2 q(x) =
(x − 1)2
(x − 1)2
x−1
=
=
2
x −1
(x − 1)(x + 1)
x+1
is analytic at x = 1 (it is a rational function whose denominator does not vanish there),
so x = 1 is a regular singular point. Similarly,
(x + 1)2 q(x) =
(x + 1)2
(x + 1)2
x+1
=
=
2
x −1
(x − 1)(x + 1)
x−1
is analytic at x = −1 (it is a rational function whose denominator does not vanish there),
so x = −1 is a regular singular point.
5
b) x(x + 1)2 y 00 + y 0 = 0
Here again q(x) = 0 (so we ignore it) and p(x) = 1/(x(x+1)2 ), so the singular points
are 0 and −1. Since
1
xp(x) =
(x + 1)2
is analytic at x = 0, x = 0 is a regular singular point. Since
(x + 1)p(x) =
1
x(x + 1)
is not analytic at x = −1 (it blows up), x = −1 is an irregular singular point.
1
c) x2 y 00 + y 0 + sin(x)
y=0
2
Here p(x) = 1/x , q(x) = cot(x)/x2 = 1/(x2 sin(x)). Both p and q are singular at
x = 0, and q is also singular at all points where sin(x) vanishes – namely kπ for every
integer k (which includes 0). Thus kπ, k an integer, are the singular points. For
x = 0,
1
xp(x) =
x
is not analytic at x = 0, and so x = 0 is an irregular singular point. At the other
singular points, p(x) is fine, so we can ignore it, and we consider q(x) at x = kπ, k 6= 0:
(x − kπ)2
1
(x − kπ)2
=
lim
x→kπ x2 sin(x)
(kπ)2 x→kπ sin(x)
1
2(x − kπ)
=
lim
= 0,
(kπ)2 x→kπ cos(x)
lim (x − kπ)2 q(x) = lim
x→kπ
is finite. So x = kπ, k 6= 0, are regular singular points.
Problem 4. Verify that the following equation has a regular singular point at x = 0, and
find the first three non-zero terms in each of two linearly independent series solutions (valid
for x > 0): 2x2 y 00 + 5xy 0 + (x + 1)y = 0
Dividing through, p(x) = 5/(2x) and q(x) = (x + 1)/(2x2 ), both of which are singular
at x = 0 (hence 0 is a singular point), but xp(x) = 5/2 and x2 q(x) = (x + 1)/2 are both
analytic at 0, so 0 is indeed a regular singular point.
Plugging
y(x) =
∞
X
n=0
an xn+r ,
y0 =
∞
X
(n + r)an xn+r−1 ,
n=0
y 00 =
∞
X
(n + r)(n + r − 1)an xn+r−2
n=0
6
into the ODE results in:
∞
∞
∞
X
X
X
0 = 2x2
(n + r)(n + r − 1)an xn+r−2 + 5x
(n + r)an xn+r−1 + (x + 1)
an xn+r
n=0
=
=
∞
X
n=0
∞
X
n=0
2(n + r)(n + r − 1)an xn+r +
2(n + r)(n + r − 1)an xn+r +
n=0
∞
X
n=0
∞
X
5(n + r)an xn+r +
5(n + r)an xn+r +
n=0
= [2r(r − 1) + 5r + 1]a0 xr +
∞
X
∞
X
n=0
∞
X
an xn+r+1 +
an−1 xn+r +
n=1
n=0
∞
X
n=0
∞
X
an xn+r
an xn+r
n=0
{[(n + r)(2(n + r − 1) + 5) + 1]an + an−1 }xn+r .
n=1
The first
(xr )
term gives the indicial equation:
0 = 2r2 + 3r + 1 = (2r + 1)(r + 1)
whose roots are r1 = −1/2 and r2 = −1. Then the higher coefficients give us the recurrance
relation:
−an−1
,
n = 1, 2, 3, . . . .
an =
[(n + r)(2(n + r) + 3) + 1]
Taking first the root r1 = −1/2, we have: a0 free, and for n = 1, 2, 3, . . .,
an =
−an−1
−an−1
−an−1
= 2
=
.
[(n − 1/2)(2n + 2) + 1]
2n + n
n(2n + 1)
So
1
1
1
1
1
a1 =
a0 , a3 = −
a2 = −
a0 , · · ·
a1 = − a0 , a2 = −
3
(2)(5)
(2)(3 · 5)
(3)(7)
(3!)(3 · 5 · 7)
and one solution (taking a0 = 1), for x > 0, is
1
1
1
−1/2
2
3
1− x+
y1 (x) = x
x −
x + ··· .
3
(2!)(3 · 5)
(3!)(3 · 5 · 7)
Now choosing the second root, r2 = −1, we find a0 free (as usual), and for n = 1, 2, 3, . . .,
an =
−an−1
−an−1
−an−1
= 2
=
.
[(n − 1)(2n + 1) + 1]
2n − n
n(2n − 1)
So
1
1
1
1
1
a1 = − a0 , a2 = −
a1 =
a0 , a3 = −
a2 = −
a0 , · · ·
1
(2)(3)
(2)(3)
(3)(5)
(3!)(3 · 5)
and a second (independent) solution, for x > 0, is
1
1
−1
2
3
y2 (x) = x
1−x+
x −
x + ··· .
(2!)(3)
(3!)(3 · 5)
7