Power Series Solutions to Differential Equations
Let’s consider the simple differential equation y 0 = y. (Of course you know already a
solution to this equation, right?) Although this can be solved by separation of variables, let’s
try a different approach. We will suppose that y can be written as a power series:
y = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + · · ·
with a positive radius of convergence R. If we differentiate, the new series will also have radius
of convergence R:
y 0 = a1 + 2a2 x + 3a3 x2 + 4a4 x3 + 5a5 x4 + · · ·
Since y must satisfy the differential equation y = y 0 , we have:
a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + · · · = a1 + 2a2 x + 3a3 x2 + 4a4 x3 + 5a5 x4 + · · ·
We now invoke the following fact: Theorem: If two convergent power series are equal, the
coefficients of equal powers of x must be equal. Thus, we must have:
a1
2a2
3a3
4a4
5a5
and in general
an
=
=
=
=
=
:
=
a0
a1
a2
a3
a4
= a0 so a2 = 12 a0
1
= 12 a0 so a3 = 3·2
a0
1
1
= 3·2 a0 so a4 = 4·3·2
a0
1
1
= 4·3·2 a0 so a5 = 5·4·3·2 a0
1
a.
n! 0
Thus, we can write all the coefficients in terms of a0 . If we call a0 the arbitrary constant C,
then we have:
C 2 C 3 C 4
C
x + x + x + · · · + xn + · · ·
2!
3!
4!
n!
1 2 1 3 1 4
1
= C(1 + x + x + x + x + · · · + xn + · · · )
2!
3!
4!
n!
x
= Ce .
y = C + Cx +
In this case, not only did we find a power series solution, but we were able to identify it as a
well-known function. Note that the answer is not unique, but contains an arbitrary constant
C corresponding to the constant term a0 , which equals y(0). If we were given the initial value
y(0) = 3.5, for example, our solution would be y = 3.5ex .
Exercise 1 In a similar way, find the general solution to the (non-separable) differential equation y 0 = y + x. Try to write your solution in terms of ex . Also, find the particular solution
corresponding to the initial value y(0) = 3.
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Exercise 2 Sometimes it is too difficult to identify the function given by the power series you
get when you use this technique. For the initial value problem y 0 = xy + x + 1, y(0) = 1 just
find the first 5 terms of a power series solution.
You can use the same technique to solve second (or higher) order differential equations.
For example, consider y 00 = y: a function which equals its second derivative. As before, assume
y = a0 + a1 x + a2 x2 + a3 x3 + a4 x4 + · · ·
so
00
y = 2a2 + 3 · 2a3 x + 4 · 3a4 x2 + 5 · 4a5 x3 + 6 · 5a6 x4 + · · ·
This gives the equations:
a0
a1
a2
a3
a4
a5
=
=
=
=
=
=
anything, say C
anything, say D
1
a = 12 C
2 0
1
1
a = 3·2
D
3·2 1
1
1
a = 4·3·2 C
4·3 2
1
1
a = 5·4·3·2
D
5·4 3
etc.
This gives us the series:
1
1
1
1
y = C + Dx + 12 Cx2 + 3·2
Dx3 + 4·3·2
Cx4 + 5·4·3·2
Dx5 + 6·5·4·3·2
Cx6 + · · ·
= C(1 + 2!1 x2 + 4!1 x4 + 6!1 x6 + · · · ) + D(x + 3!1 x3 + 5!1 x5 + · · · )
where C and D are arbitrary constants – they are determined once we know initial values
y(0) and y 0 (0).
Exercise 3 The hyperbolic sine and cosine, denoted sinh and cosh, are defined as follows:
ex − e−x
2
ex + e−x
cosh(x) =
2
sinh(x) =
Use the computation just made to show that the general solution to y 00 = y is y = A sinh(x) +
B cosh(x). You’ll need to find power series for sinh and cosh, which you get easily from the
series for ex and e−x (put −x for x in the series for ex ). Note that A = 2C and B = 2D.
Suppose y(0) = K and y 0 (0) = L; what will A and B equal?
Exercise 4 Repeat what was just done, except for the differential equation y 00 = −y: show
that the general solution is y = A sin(x) + B cos(x). If y(0) = K and y 0 (0) = L, what will A
and B equal?
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