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Transcript using power series
17
SECOND-ORDER
DIFFERENTIAL EQUATIONS
SECOND-ORDER DIFFERENTIAL EQUATIONS
17.4
Series Solutions
In this section, we will learn how to solve:
Certain differential equations
using the power series.
SERIES SOLUTIONS
Equation 1
Many differential equations can’t be solved
explicitly in terms of finite combinations of
simple familiar functions.
This is true even for a simple-looking
equation like:
y’’ – 2xy’ + y = 0
SERIES SOLUTIONS
However, it is important to be able to solve
equations such as Equation 1 because
they arise from physical problems.
In particular, they occur in connection with
the Schrödinger equation in quantum mechanics.
USING POWER SERIES
In such a case, we use the method of
power series.
That is, we look for a solution
of the form
y f x cn x
n
n 0
c0 c1 x c2 x c3 x
2
3
USING POWER SERIES
The method is to substitute this expression
into the differential equation and determine
the values of the coefficients c0, c1, c2, …
This technique resembles the method
of undetermined coefficients discussed
in Section 17.2
USING POWER SERIES
Before using power series to solve Equation 1,
we illustrate the method on the simpler
equation y’’ + y = 0 in Example 1.
It’s true that we already know how to solve
this equation by the techniques of Section 17.1
Still, it’s easier to understand the power series
method when it is applied to this simpler equation.
E. g. 1—Equation 2
USING POWER SERIES
Use power series to solve
y’’ + y = 0
We assume there is a solution
of the form
y c0 c1 x c2 x c3 x ... cn x
2
3
n 0
n
E. g. 1—Equation 3
USING POWER SERIES
We can differentiate power series term by
term.
So,
y ' c1 2c2 x 3c3 x ... ncn x
2
n 1
n 1
y '' 2c2 2 3c3 x ... n n 1 cn x
n 1
n2
USING POWER SERIES
E. g. 1—Equation 4
To compare the expressions for y and y’’
more easily, we rewrite y’’ as:
y '' n 2 n 1 cn 2 x
n 0
n
E. g. 1—Equation 5
USING POWER SERIES
Substituting the expressions in Equations 2
and 4 into the differential equation,
we obtain:
n 2 n 1 c
n 0
or
n2
x cn x 0
n
n
n 0
n 2 n 1 c
n 0
n2
cn x 0
n
E. g. 1—Equation 6
USING POWER SERIES
If two power series are equal, then the
corresponding coefficients must be equal.
So, the coefficients of xn in Equation 5
must be 0:
n 2 n 1 cn2 cn 0
cn 2
cn
n 1 n 2
n 0,1, 2,3...
RECURSION RELATION
Example 1
Equation 6 is called a recursion
relation.
If c0 and c1 are known, it allows us to determine
the remaining coefficients recursively by putting
n = 0, 1, 2, 3, … in succession, as follows.
RECURSION RELATION
c0
Put n = 0: c2
1 2
c1
Put n = 1: c3
23
Example 1
c0
c0
c2
Put n = 2: c4
3 4 1 2 3 4 4!
RECURSION RELATION
Example 1
c
c
c
3
1
Put n = 3: c5
1
4 5 2 3 4 5 5!
c0
c0
c4
Put n = 4: c6
56
4!5 6
6!
c5
c1
c1
Put n = 5: c7
6 7 5!6 7
7!
USING POWER SERIES
Example 1
By now, we see the pattern:
c0
For the even coefficients, c2 n 1
2n !
n
c1
For the odd coefficients, c2 n 1 1
2n 1!
n
Putting these values back into Equation 2,
we write the solution as follows.
Example 1
USING POWER SERIES
y c0 c1 x c2 x c3 x c4 x c5 x
2
3
4
5
x
x
x
n x
c0 1 1
2n !
2! 4! 6!
3
5
7
2 n 1
x
x
x
x
n
c1 x 1
3! 5! 7!
2n 1!
2
4
2n
6
2n
2 n 1
x
x
n
c0 1
c1 1
2n ! n 0
2n 1!
n 0
n
Notice that there are two arbitrary constants, c0 and c1.
NOTE 1
We recognize the series obtained in
Example 1 as being the Maclaurin series
for cos x and sin x.
See Equations 15 and 16 in Section 11.10
NOTE 1
Therefore, we could write the solution
as:
y(x) = c0 cos x + c1 sin x
However, we are not usually able to express
power series solutions of differential equations
in terms of known functions.
Example 2
USING POWER SERIES
Solve y’’ – 2xy’ + y = 0
We assume there is a solution
of the form
y cn x
n 0
n
Example 2
USING POWER SERIES
Then, as in Example 1,
y ' ncn x
n 1
n 0
and
y '' n n 1 cn x
n2
n2
n 2 n 1 cn 2 x
n 0
n
Example 2
USING POWER SERIES
Substituting in the differential equation,
we get:
n 0
n 1
n 0
n
n 1
n
n
2
n
1
c
x
2
x
nc
x
c
x
n2
n
n 0
n 0
n 1
n 0
n
n
n
n
2
n
1
c
x
2
nc
x
c
x
n2 n n 0
n 2 n 1 c
n 0
n2
2n 1 cn x 0
n
USING POWER SERIES
E. g. 2—Equation 7
The equation is true if the coefficient of xn
is 0:
(n + 2)(n + 1)cn+2 – (2n – 1)cn = 0
cn 2
2n 1
cn
n 1 n 2
n 0, 1, 2, 3, ...
USING POWER SERIES
Example 2
We solve this recursion relation by putting
n = 0, 1, 2, 3, … successively in Equation 7:
1
Put n = 0: c2
c0
1 2
1
Put n = 1: c3
c1
23
USING POWER SERIES
Example 2
3
3
3
Put n = 2: c4
c2
c0 c0
3 4
1 2 3 4
4!
5
1 5
1 5
Put n = 3: c5
c3
c1
c1
45
2 3 4 5
5!
7
3 7
37
Put n = 4: c6
c4
c0
c0
56
4!5 6
6!
USING POWER SERIES
Example 2
9
1
5
9
1
5
9
Put n = 5: c7
c5
c1
c1
67
5!6 7
7!
11
3 7 11
Put n = 6: c8
c6
c0
7 8
8!
13
1 5 9 13
Put n = 7: c9
c7
c1
89
9!
USING POWER SERIES
Example 2
In general,
The even coefficients are given by:
3 7 11 (4n 5)
c2 n
c0
(2n)!
The odd coefficients are given by:
1 5 9 (4n 3)
c2 n 1
c1
(2n 1)!
USING POWER SERIES
Example 2
The solution is:
y c0 c1 x c2 x 2 c3 x 3 c4 x 4
1 2 3 4 3 7 6 3 7 11 8
c0 1 x x
x
x
4!
6!
8!
2!
1 3 1 5 5 1 5 9 7 1 5 9 13 9
c1 x x
x
x
x
3!
5!
7!
9!
USING POWER SERIES
E. g. 2—Equation 8
Simplifying,
1 2 3 7 4n 5 2 n
y c0 1 x
x
2n !
n2
2!
1 5 9 4n 3 2 n 1
c1 x
x
2n 1!
n 1
NOTE 2
In Example 2, we had to assume that the
differential equation had a series solution.
Now, however, we could verify directly that
the function given by Equation 8 is indeed
a solution.
NOTE 3
Unlike the situation of Example 1,
the power series that arise in the solution
of Example 2 do not define elementary
functions.
NOTE 3
The functions
1 2 3 7 4n 5 2 n
y1 x 1 x
x
2!
2n !
n2
and
1 5 9 4n 3 2 n 1
y2 x x
x
2n 1!
n 1
are perfectly good functions.
However, they can’t be expressed
in terms of familiar functions.
NOTE 3
We can use these power series
expressions for y1 and y2 to compute
approximate values of the functions
and even to graph them.
NOTE 3
The figure shows the first few partial sums
T0, T2, T4, … (Taylor polynomials) for y1(x).
We see how they
converge to y1.
NOTE 3
Thus, we can graph both y1 and y2
as shown.
NOTE 4
Suppose we were asked to solve
the initial-value problem
y’’ – 2xy’ + y = 0
y(0) = 0
y’(0) = 1
NOTE 4
We would observe from Theorem 5
in Section 11.10 that:
c0 = y(0) = 0
c1 = y’(0) = 1
This would simplify the calculations in Example 2,
since all the even coefficients would be 0.
NOTE 4
The solution to the initial-value problem
is:
1 5 9 4n 3 2 n 1
y ( x) x
x
2n 1!
n 1