MATH 2280-002
Lecture Notes: 1/29/2013-1/30/2013
Math 2280 Section 002 [SPRING 2013]1
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Homogeneous Equations with Constant Coefficients
The more general version of an nth order homogeneous linear differential equation with constant
coefficients is:
an y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 y = 0.
We guess based on our experience with linear DE’s (back in HW 1) that y = erx may work for some
values of r. So we’ll plug y = erx in and see what works:
an rn erx + an−1 rn−1 erx + · · · a1 rerx + a0 erx = 0.
Now, if we divide both sides by erx , which we know we can do for any value of x as erx is never 0,
we get the equation
an rn + an−1 rn−1 + · · · + a1 r + a0 = 0.
which is true if and only if r is a root of the characteristic polynomial for our differential equation
an xn + an−1 xn−1 + · · · + a1 x + a0 .
So, to repeat, if r is a root of the characteristic polynomial for our differential equation, then erx
is a solution of our differential equation. You should’ve heard the term “characteristic polynomial”
before when you studied eigenvalues for matrices back in linear algebra. This is not a coincidence.
We talk about this more when we get to Chapter 5.
Now, the fundamental theorem of algebra says that for any polynomial
an xn + an−1 xn−1 + · · · a1 x + a0
can be factored it as
(x − r1 )n1 (x − r2 )n2 · · · (x − rm )nm
where n1 + n2 + · · · + nm = n, and the constants ri could be complex numbers. This presents us
with a number of possible situations.
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Notes adapted from Dylan Zwick’s Spring 2009 lecture notes.
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MATH 2280-002
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Lecture Notes: 1/29/2013-1/30/2013
Possible Situations
We can break down our possible situations into four distinct cases of varying complexity.
Case 1. All roots of the characteristic polynomial are real and distinct (m = n).
If this is the case then we have n distinct solutions of the form {er1 x , er2 x , . . . , ern x }, and all
our solutions can be written in the form
y = c1 er1 x + c2 er2 x + · · · + cn ern x
The Wronskian for these solutions, evaluated at 0, is
1
1
r1
r2
2
r2
r
W (0) = 1
2
..
..
.
.
n−1 n−1
r
r
2
1
···
···
···
..
.
1
rn
rn2
..
.
···
rnn−1
.
This is something called the “Vandemonde determinant”, and it’s never 0 when the ri are
distinct. This is proven in one of the suggested homework problems.
Case 2. All roots are equal (m = 1) and real. We’ll set r = r1 = ... = rm to be the root of
the characteristic polynomial.
Let’s first just look at the case when our polynomial factors as
(x − r)n .
In this case our differential equation can be written as
(D − r)k y = 0.
where D is the “differential operator”, an operator that when applied to a function takes the
derivative of that function. It’s a linear operator, and if you raise it to any power, that just
means you take that number of derivatives. For example,
(D − 1)2 y = (D2 − 2D + 1)y = y 00 − 2y 0 + y.
Now, any solution y to our differential equation can be written in the form y(x) = u(x)erx . If
we apply the operator (D − r) to this solution we get:
(D − r)u(x)erx = u0 (x)erx + ru(x)erx − ru(x)erx = u0 (x)erx .
If we repeat this process n times we get:
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MATH 2280-002
Lecture Notes: 1/29/2013-1/30/2013
(D − r)n u(x)erx = u(n) (x)erx .
If u(x)erx is a solution to our differential equation then we have u(n) (x)erx = 0, which implies
that u(n) = 0, which means
u(x) = c0 + c1 x + c2 x2 + · · · + cn−1 xn−1
This give us a spanning set of independent functions {1, x, x2 , . . . , xn−1 }, and a corresponding
set of solutions to our ODE {erx , xerx , x2 erx , . . . , xn−1 erx }.
We’ll discuss the mixed case later.
Case 3. Complex Roots
Suppose that the characteristic polynomial for our polynomial is second order and can be
factored as
(x − r)(x − r).
where r is complex and r is its complex conjugate. Note that for any polynomial
an xn + an−1 xn−1 + · · · a1 x + a0 .
if the ai are real and if r is a complex root, then r must be a complex root as well. So, for any
characteristic equation if we get a complex root, we’ll get another root by taking its complex
conjugate.
OK, so if r is a complex root, then we can write r as r = a + bi, and we get a solution to our
differential equation
erx = e(a+ib)x = eax eibx = eax (cos (bx) + i sin (bx)),
where we’ve used Euler’s formula:
eiθ = cos θ + i sin θ.
Now, the fact that r = a − ib is also a solution to our characteristic polynomial give us another
solution
er = eax (cos (bx) − i sin (bx)).
A linear combination of these two can be written as
k1 erx + k2 erx = eax ((k1 + k2 ) cos (bx) + (k1 − k2 )i sin (bx)).
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MATH 2280-002
Lecture Notes: 1/29/2013-1/30/2013
If we allow k1 , k2 to be complex, then we can choose them so that the above equation is equal
to
eax (c1 cos (bx) + c2 sin (bx)).
where the coefficients c1 and c2 are arbitrary real constants. So, we get two linearly independent
solutions to our ODE eax cos (bx) and eax sin (bx), and any other solution can be written as a
linear combination of these two.
We’ll incorporate the case of when you have repeated complex roots or multiple complex roots
in our general case.
Case 4. The General Case
Suppose we have a linear homogeneous ordinary differential equation with constant coefficients
an y (n) + an−1 y (n−1) + · · · + a1 y 0 + a0 = 0.
To solve this problem, in general, what we do is we first calculate the roots of the characteristic
polynomial
an xn + an−1 xn−1 + · · · + a1 x + a0 = (x − r1 )k1 (x − r2 )k2 · · · (x − rm )km .
Now, there will be some set of these roots, say r1 through rl , which are real. For each of these
we will get a number of solutions equal to the degree of the root. So, for example, for our first
root r1 we will get solutions
c1 er1 x , c2 xer1 x , . . . , ck1 xk1 −1 er1 x .
Now, as for the other roots, rl+1 through rm , these will come in complex conjugate pairs, and
each element of these pairs will have the same degree. So, for example, for the root rl+1 if we
arrange out terms appropriately rl+2 will be its complex conjugate, and we’ll have kl+1 = kl+2 .
For this pair of roots we’ll get the set of solutions
c1 eal+1 x cos (bl+1 x), c2 xeal+1 x cos (bl+1 x), . . . , ckl+1 xkl+1 −1 eal+1 x cos (bl+1 x)
and
a
x
a
x
l+1
l+1
d1 e
sin (bl+1 x), d2 xe
sin (bl+1 x), . . . , dkl+1 xkl+1 −1 eal+1 x sin (bl+1 x),
where the ci and di are unknown constants, and the constants al+1 and bl+1 are the real part
and complex part, respectively, of the root rl+1 = al+1 + ibl+1 .
Clear as mud? Yeah, in its full generality it’s kind of confusing to state, but just know that this
more general statement just builds up from the particular cases we’ve seen so far, and in this
class and in most situations we’ll only be dealing with relatively small order ODEs, so the full
machinery of the more general statement won’t be necessary in practice, but it’s good to know.
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MATH 2280-002
Lecture Notes: 1/29/2013-1/30/2013
To summarize, each root corresponds to a “piece” of the general solution. A root with multiplicity
k (the root appears k times in the characteristic polynomial) corresponds to k linearly independent
solutions of a total n linearly independent solutions to an nth order linear DE. Here’s what these
solutions look like based on the multiplicity of the root and whether the root has imaginary parts or
not:
I. r is real with multiplicity 1.
y(x) = c1 erx +
.
II. r is real with multiplicity k.
y(x) = (c1 + c2 x + c3 x2 + ... + ck xk−1 )erx +
.
III. r1 = a + bi and r2 = a − bi are complex (b 6= 0) with multiplicity 1.
y(x) = eax (c1 cos(bx) + c2 sin(bx)) +
(C1 may not equal c1 and C2 may not equal c2 .)
IV. r1 = a + bi and r2 = a − bi are complex (b 6= 0) with multiplicity k.
y(x) =eax [(c1 cos(bx) + d1 sin(bx)) + x(c2 cos(bx) + d2 sin(bx))+
... + xk−1 (ck cos(bx) + dk sin(bx))] +
The “+
” just means that this may be only part of the general solution. Let’s work through a
couple of examples.
Example. Write down the general solution to the differential equation
y (6) + 3y (5) − 6y 000 − 3y 00 + 3y 0 + 2y = 0
using the hint that
r6 + 3r5 − 6r3 − 3r2 + 3r + 2 = (r + 1)3 (r − 1)2 (r + 2).
Because the factorization of the characteristic polynomial
r6 + 3r5 − 6r3 − 3r2 + 3r + 2
is given, it is easy to write down the roots. We can classify the roots of the characteristic polynomial
as follows:
• r = 2 is a distinct, real root (multiplicity 1);
• r = 1 is a real root with multiplicity 2; and
• r = −1 is a real root with multiplicity 3.
That means our general solution is
y(x) = c1 e2x + (c2 + c3 x)ex + (c4 + c5 x + c6 x2 )e−x .
Notice that our 6th order linear DE has 6 linearly independent solutions: e2x , ex , xex , e−x , xe−x , x2 e−x .
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MATH 2280-002
Lecture Notes: 1/29/2013-1/30/2013
Example. Write down the general solution to the differential equation
y (5) + 3y (4) + 5y 000 + 5y 00 + 3y 0 + y = 0
using the hint that
r5 + 3r4 + 53 + 5r2 + 3r + 1 = (r2 + r + 1)2 (r + 1).
The quadratic formula tells us that the roots of r2 + r + 1 are the complex conjugates
√
3
1
±
i,
2
2
so we can classify the roots of the characteristic polynomial as follows:
• r = 1 is a distinct, real root (multiplicity 1); and
• r=
1
2
√
±
3
2 i
are complex conjugate roots with multiplicity 2.
That means our general solution is
"
√ !
1
3
x
x
c2 cos
x + c3 sin
y(x) = c1 e +e 2
2
√ !!
3
x
+ x c4 cos
2
What are the 5 linearly independent solutions to our DE?
6
√
3
x
2
√
!
+ c5 sin
3
x
2
!!#
.