Math 308
Worksheet 2: Sections 5.1 - 5.3
09 July 2012
Question. Suppose that
P5 (x) = 1 + 7x2 − 13x3 + 13 x4 − 2x5 ,
Q5 (x) = x − 6x2 + x3 + 2x4 − 21 x5
are the degree 5 Taylor polynomials (a.k.a the order 5 approximations) at xo = 0
of two solutions, call them f (x) and g(x), to a second–order, linear, homogeneous
differential equation
y 00 + p(x)y 0 + q(x)y = 0 .
(1)
If y(x) is a third solution to the differential equation, and
y(0) = 3 and y 0 (0) = −2 ,
(2)
then what is y (3) (0)?
Solution. Step 1. The fact that P5 (x) is the fifth–order approximation of f (x) at
x = 0 tells us that
f (0) = P5 (0) ,
(3)
f (3) (0) = P5 (0) ,
f 00 (0) = P500 (0) ,
f 0 (0) = P50 (0) ,
(4)
f (4) (0) = P5 (0) ,
(5)
f (5) (0) = P5 (0) .
So, computing P5k (0) tells us
(3)
f (0) = 1 ,
f (3) (0) = −78 ,
f 0 (0) = 0 ,
f (4) (0) = 8 ,
f 00 (0) = 14 ,
f (5) (0) = −240 .
Likewise, Q5 (x) tells us
(4)
g(0) = 0 ,
g (3) (0) = 6 ,
g 0 (0) = 1 ,
g (4) (0) = 48 ,
g 00 (0) = −12 ,
g (5) (0) = −60 .
Step 2. Because the differential equation (1) is linear and homogeneous, the Principle
of Superposition tells us that y(x) = af (x) + bg(x) is also a solution. Equations (3)
and (4) tell us that
y(0) = a and y 0 (0) = b .
Equation (2) tells us to choose a = 3 and b = −2, so that
y(x) = 3 f (x) − 2 g(x) .
Step 3. Now equations (3) and (4) tell us
y (3) (0) = 3 f (3) (0) − 2 g (3) (0) = -246 .
1
Definition. Take a differentiable function f (x). The Taylor series of f (x) at x = xo
is the infinite sum
∞
X
1 (k)
f (xo )(x − xo )k .
P∞ (x) =
k!
k=0
For convenience, set
ak =
1 (k)
f (xo ) .
k!
If the limit
ak+1 = L,
lim
k→∞ ak P∞
k
exists, then the Ratio Test says that the Taylor series P∞ (x) =
k=0 ak (x − xo )
converges (the sum of infinitely many terms adds up to a finite number) for any value
x such that |x − xo | < 1/L. We call 1/L the radius of convergence, and say that
the function f (x) is analytic at x = xo .
Example. Familiar analytic functions that you worked with in Calculus include the
following.
◦ Trig functions sin(x) and cos(x), the exponential ex , and polynomials; all of these
are analytic at every point xo and have infinite radius of convergence.
◦ Logarithmic functions, such as ln(x), are analytic at points xo > 0, with radius of
convergence 1/xo .
◦ If p(x) and q(x) are polynomials, then the rational function f (x) = p(x)/q(x) is
analytic at xo as long as q(xo ) 6= 0.
Definition. Start with a generic second–order, linear differential equation R(x)y 00 +
P (x)y 0 + Q(x)y = 0 and divide through by R(x) to put the equation in normal form
(5)
y 00 + p(x) y 0 + q(x) y = 0 .
Pick a initial point xo . If both functions p(x) and q(x) are analytic at xo , then xo is
an ordinary point of the differential equation; otherwise, xo is a singular point.
Key Idea. At
we can find analytic solutions. The will be of the
Pan ordinary point,
n
a
(x
−
x
)
.
To
find y(x) we proceed in two steps: (i) use the
form y(x) = ∞
o
n=0 n
initial values to determine a0 and a1 ; and (ii) use the differential equation (5) to find
a recursive relationship for the other an .
P
n
Exercise 1 (§5.2, 7). Find the solution y(x) = ∞
n=0 an x to the initial value problem
y 00 + xy 0 + 2y = 0 ,
with y(0) = 1 and y 0 (0) = 0 .
Exercise 2 (§5.2, 7). Find the solution y(x) =
y 00 + xy 0 + 2y = 0 ,
P∞
n=0
an xn to the initial value problem
with y(0) = 0 and y 0 (0) = 1 .
Homework. As part of your preparation for Quiz 1 (Wednesday, July 11) and the
Final Exam (Tuesday, August 07), I recommend the following:
(A) Section 5.2: For each of Exercises
10, 12 and 13 do:
P
n
0
◦ Find the solution y(x) = ∞
a
n x with initial values y(0) = 1 and y (0) = 0.
Pn=0
n
0
◦ Find the solution y(x) = ∞
n=0 an x with initial values y(0) = 0 and y (0) =
−1.
(B) Section 5.2: Exercise 21(b). (The Hermite Equation.)
(C) Section 5.3: Exercises 5 and 6.
(D) Section 5.3: Exercises 22 and 23.
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Colleen Robles
robles@math.tamu.edu