(x 2 + 1)y' = xy; y(0) = 1
Solving
By
The Power Series Method
∞
Assume that y can be expressed as
∑a x
n
n
. Since
y(0) = 1 , a0 = 1. Re-write
n=0
∞
the differential equation as
x y'+ y' = xy . Since y = ∑ an x n , x 2 y'+ y' = xy
2
n=0
∞
implies that
x
2
∑ na x
∞
n−1
n
n=1
factors
+ ∑ nan x
∞
n−1
n=1
= x ∑ an x n and, distributing the
∞
∞
n=0
∞
n=1
n=1
n=0
x 2 and x: ∑ nan x n+1 + ∑ nan x n−1 = ∑ an x n+1 . Write each sum term-
by-term, with constant term in red, linear terms in green, quadratic terms in blue, and
third-degree terms in purple.
[a1 x + 2a2 x + ... ] + [a1 + 2a2 x + 3a 3 x + 4a 4 x + ...]=[a 0 x + a1 x + a2 x + ...] .
2
3
2
3
2
3
a1 , so for the equation to be true for all x, a1 = 0 .
a0 1
Looking at the (green ) linear terms in the equation, 2a =a , so a =
= . Now
2 2
look at the (blue) quadratic terms to get a + 3a =a , which implies that a3 = 0 .
The only constant (red ) term is
2
1
3
0
2
1
Continuing to the third-degree (purple ) terms, 2a2 + 4a 4 =a2 , which implies that
a4
=
− a2
4
=
−1
8
Now, look at each series more generally:
a1 x + 2a2 x + ... +na n x
2
3
n+1
+ ...
a1 + 2a2 x + 3a 3 x + 4a 4 x + ...+(n+2)a n+2 x
2
3
a 0 x + a1 x + a2 x + ...+a n x
2
3
n+1
+ ...
n+1
+ ... and
Notice that the middle series is written so that the power of its general term is n+1, to
make the comparisons easier to make and the corresponding equation easier to write.
These three general terms all apply for n > 0.
Looking at the power n+1 (black) terms above, we get the equation:
+ (n+2)a n+2 = a n
na n
and solving for a n+2 , a n+2
=
the odd-numbered coefficients must equal zero.
(1− n)a n
n+2
. Since
a3 = 0 , all
We have already determined the even-numbered terms
1
−1
, a3 = 0 , a =
, and now using the recursion formula
2
8
a0 = 1, a1 = 0 , a2 =
4
above,
a 4+2
=
(1− 4)a 4
4+2
or a 6
=
(−3)
-1
8 . Still using the recursion formula – without
6
simplifying (in order to see any pattern emerging,
(−3)
a8
=
(−5)
-1
8
6
and
8
(−3)
(−5)
a10
=
a10
=
a 2n
=
(−7)
-1
8
6
8
10
(-1)(−3)(−5)(−7)
. The pattern is clear!
, and we conjecture that, more generally,
(2)(4)(6)(8)(10)
(-1)(−3)(−5)(−7) ... (−[2n − 3])
(2)(4)(6)(8)(10) ... (2n)
= (−1)n+1
(1)(3)(5)(7) ... (2n − 3)
(2)(4)(6)(8)(10) ... (2n)
, which
can be established by using the Principle of Mathematical Induction along with the
recursion formula. The series generated to solve the equation is
1 2 ∞
(1)(3)(5)(7) ... (2n − 3) 2n
1+ x + ∑ (−1)n+1
x . Using the Ratio Test, the
2
(2)(4)(6)(8)(10)
...
(2n)
n=2
series can be shown to converge for |x| < 1 and to diverge for |x| > 1.
Endnotes.
1. The above is an awful lot of hard work to get a solution that we can obtain much
more easily (and in a handier form, too) by just separating the variables in the
original DE. Moreover, the solution function is defined for all real numbers.
2. So, in summary, this is not a good way to solve the DE; it’s just a good way to
illustrate the process of using a power series to solve a DE.
3. If an initial condition had not been given, the solution process would have been
only slightly more difficult. Instead of having y(0) = 1 , we would have
y(0) = c , for some constant c. Then we would have a0 = c , a2 =
a4
=
−c
8
c
,
2
, and, continuing, the solution would be just multiplied by this constant
factor c. To see this independently, suppose that y is any solution function for the
DE in this problem. Then
x 2 (cy)'+ (cy)' = x 2 (cy') + cy' = c(x 2 y'+ y') = cxy = x(cy) ; i.e.,
cy is also a solution to the DE.
4. When this sort of thing happens (y a solution implies cy is a solution) the
equation is said to be homogeneous, so it’s important to be able to identify
homogeneous equations.