MTH 235 SUPPLEMENTARY MATERIAL
JOHN MACHACEK
Abstract. This is a supplement to MTH 235: Differential Equations. In this
supplement we first learn how to solve certain higher order ODEs. We later
learn how to solve some recurrence relations using a characteristic polynomial.
1. Higer Order ODEs
In this first section we consider higher order ODEs. We will only consider the
homogeneous linear ODEs with constant coefficients. Like 2nd order homogeneous
linear ODEs, we will show higher order ODEs can also be solved by finding the
roots of a polynomial. Consider a linear nth-order homogeneous linear ODE with
constant coefficients
(1)
y (n) (t) + an−1 y (n−1) (t) + · · · + a1 y 0 (t) + a0 y = 0
then we have the characteristic polynomial
(2)
p(r) = rn + an−1 rn−1 + · · · + a1 r + a0 .
For any root r0 of the polynomial in (2), i.e. p(r0 ) = 0, we have that y(t) = er0 t
as a solution to ODE in (1). We can combine these solutions to find the general
solution. Like the 2nd order case some care must be taken with repeated roots, but
we will not discuss this here.
In class we specialized to the case of n = 2 and looked at the equation
y 00 (t) + a1 y 0 (t) + a0 y(t) = 0
because the characteristic polynomial p(r) = r2 + a1 r + a0 is quadratic and hence
always easy to solve. Higher degree polynomial equations are not always as easy
to solve as quadratic equations, but the following exercise has you find the general
solution a higher order ODE involving a cubic equation we can solve.
Exercise 1. Find the (real valued) general solution to the 3rd order ODE
y 000 (t) − y(t) = 0.
Hint: The characteristic polynomial is p(r) = r3 − 1 and factors as p(r) = (r −
1)(r2 + r + 1).
√
√
Solution: y(t) = C1 et + C2 e−t/2 cos( 3t/2) + C2 e−t/2 sin( 3t/2).
In the next exercise you solve an IVP using the general solution you just found.
Note that we had a 3rd order ODE. So, we have 3 constants and hence 3 initial
conditions are needed.
Exercise 2. Find the solution to the IVP y 000 (t) − y(t) = 0 with y(0) = 1, y 0 (0) =
0, y 00 (0) = 0.
√
Solution: y(t) = (1/3)et + (2/3)e−t/2 cos( 3t/2).
1
2
JOHN MACHACEK
2. Recurrence Relations
We have considered a function y(t) where t is a real variable and y is a realvalued function on t. We now consider discrete functions. Recall sequences (think
back to MTH 133: Calc II) are lists of numbers. We can write a sequence as
x0 , x1 , x2 , . . .
with nth term xn . We can also write this sequence as {xn }n≥0 . These sequence can
be though of as functions with the domain of the natural numbers N = {0, 1, 2, . . . }.
For example, the sequence xn = n is
0, 1, 2 . . .
which can be thought of as the identity function. Another example is xn = 3n − 1
which looks like
−1, 2, 8, 26 . . .
where again we have a explicit formula for the nth term.
When studying differential equations we take information about how a function
y relates to its derivatives and try to deduce what that function y must be. This is
the continuous/differentiable world. Now, in the discrete world, we consider some
sequence {xn }n≥0 . We may not have an explicit formula the the term xn , but
instead know how xn relates to previous terms. This is called a recurrence relation.
We can try to solve a recurrence relation to obtain an explicit formula for a sequence.
Solving a recurrence relation can be similar to solving a differential equation. In
fact, later we will see similar techniques work for solving both recurrence relations
and differential equations. Try to solve the following simple recurrence relations in
the following exercises. These simple examples can be solved just by writing out
the first terms terms and finding a pattern.
Exercise 3 (Arithmetic Progression). Solve the recurrence relation xn = xn−1 + 10
if x0 = 2.
Solution: xn = 10n + 2.
In general if xn = xn−1 + a for some constant a we have that xn = na + x0 and
the sequence is called an arithmetic progression.
Exercise 4 (Geometric Progression). Solve the recurrence relation xn = 2xn−1 if
x0 = 3.
Solution: xn = 3 · 2n .
In general if xn = axn−1 for some constant a we have that xn = x0 an and the
sequence is called an geometric progression.
Note the proceeding two exercises only depended on 1 previous term (and only
required 1 initial condition). These are analogous to 1st order ODEs. Now we
discuss an analog to 2nd order ODEs. The following theorem should look very
familiar when compared to homogeneous linear ODEs with constant coefficients.
Theorem 5. If a1 and a0 are constants, then the general solution to the recurrence
relation
xn + a1 xn−1 + a0 xn−2 = 0
is given by
• C1 r1n + C2 r2n (r1 6= r2 )
• C1 rn + C2 nrn (r1 = r2 = r)
MTH 235 SUPPLEMENTARY MATERIAL
3
where r1 and r2 are roots of p(r) = r2 + a1 r + a0
If 2 initial values are specified one can use the above theorem to find the explicit
solution to a recurrence relation of this kind. Now solve the recurrence for the
famous Fibonacci numbers in the exercise below.
Exercise 6 (Fibonacci Numbers). Solve the recurrence relation xn = xn−1 + xn−2
if x0 = 0, x1 = 1. √
√
√
√
Solution: xn = (1/ 5)((1 + 5)/2)n − (1/ 5)((1 − 5)/2)n .