Linear n-th Order Differential
Equations
General Form; Properties A linear n-th order differential
equation is an equation of the form
d n−1 y
dy
d ny
+ p1 (x) n−1 + ... + pn−1 (x)
+ pn (x) y = g(x),
n
dx
dx
dx
where the pk (x), k = 1, 2, ..., n and g(x) are known continuous,
or at least piecewise continuous, functions defined on some interval a < x < b (which may be (−∞, ∞). We make the usual
distinction between homogeneous and non-honogeneous equations, according as g(x) is, or is not, identically equal to zero
on (a, b), respectively. A solution is an n times differentiable
function, y(x), defined on (a, b), which, on substitution into the
equation, reduces the equation to an identity. The general properties are essentially the same as we have already listed for the
inhomogeneous equation in the second order case:
•1) If y1 (x) is a solution of the inhomogeneous equation and
z(x) is a solution of the corresponding homogeneous equation
d nz
d n−1 z
dz
+
p
(x)
+
...
+
p
(x)
+ pn (x) z = 0,
1
n−1
dx n
dx n−1
dx
then y2(x) = y1 (x) + z(x) is also a solution of the inhomogeneous equation.
•2) If y1(x) and y2(x) are solutions of the inhomogeneous
equation just indicated, then the difference, z(x) = y2 (x) −
y1(x), is a solution of the corresponding homogeneous equation.
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Equivalently, there is a solution z(x) of the homogeneous equation such that y2(x) = y1(x) + z(x).
•3) If g(x) = α g1(x) + β g2(x) and y1(x) and y2(x) are solutions of the inhomogeneous equation with g(x) replaced by
g1 (x) and g2 (x), respectively, then y(x) = α y1(x) + β y2(x) is
a solution of
d ny
d n−1 y
dy
+ pn (x) y = g(x)
+
p
(x)
+
...
+
p
(x)
1
n−1
dx n
dx n−1
dx
= α g1 (x) + β g2 (x).
•4) The general solution of the homogeneous equation takes
the form
z(x, c1 , c2 , ..., cn ) = c1 z1 (x) + c2 z2 (x) + ... + cn zn (x),
where c1 , c2 , ..., cn are arbitrary constants and z1 (x), z2 (x), ..., zn (x)
are n solutions of the homogeneous equation forming a fundamental set of solutions - a concept which will explain a little
later. Letting yp(x) be a particular solution (i.e., any solution)
of the inhomogeneous equation, the general solution of the inhomogenous equation is given by
y(x, c1 , c2 , ..., cn ) = c1 z1 (x) + c2 z2 (x) + ... + cn zn (x) + yp (x).
An initial value problem for an n-th order equation such as
we are studying here consists of that differential equation and a
specification of the values of y(x) and its first n − 1 derivatives
at some point x0 in the interval (a, b):
y(x0 ) = y0 , y (k) (x0 ) = yk , k = 1, 2, ..., n − 1.
2
The basic requirement for n solutions z1 (x), z2 (x), ..., zn (x) to
constitute a fundamental set of solutions for the homogeneous
equation is that for each such specification of initial conditions it
should be possible to choose a unique set of coefficients c1 , c2 , ..., cn
such that the linear combination c1 z1 (x) + c2 z2 (x) + ... +
cn zn (x), again a solution of the homogeneous equation, will satisfy those conditions. We will see that, just as in the second
order case, this property depends only on the choice of the solutions z1 (x), z2 (x), ..., zn (x) and not on the particular point x0
in question.
We will abbreviate the general n-th order linear equation by
L y = g, where
d ny
d n−1 y
dy
(L y)(x) =
+
p
(x)
+
...
+
p
(x)
+ pn (x) y.
1
n−1
dx n
dx n−1
dx
It will also be convenient, for later purposes, to denote the difd ky
k
ferential operator dx
k by D y and write
L y = D n y + p1 (x) D n−1 y + ... + pn−1 (x) D y + pn (x) y,
or, symbolically,
L = D n + p1 (x) D n−1 + ... + pn−1 (x) D + pn (x) I,
where I denotes the identity operator with the property I y = y.
This will enable a rather algebraic treatment of the equations
as we progress.
Constant Coefficient Homogeneous Equations As an
immediate extension of the second order case, we define the linear homogeneous n-th order equation with constant coefficients
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to be an equation of the form
d ny
d n−1 y
dy
+
p
+
...
+
p
+ pn y = 0,
1
n−1
dx n
dx n−1
dx
where p1 , p2 , ..., pn are constants; we will always take them to be
real in our work here but one can equally well treat equations
with complex coefficients. Substituting a solution of the form
y(x) = erx one obtains
erx (rn + p1 rn−1 + ... + pn−1 r + pn ) = 0,
necessitating, as in the previous second order case, that r should
satisfy the characteristic equation
p(r) ≡ rn + p1 rn−1 + ... + pn−1 r + pn = 0.
This is an n-th degree polynomial equation; from the Fundamental Theorem of Algebra we know that this equation has a
total of n roots, real or complex, r1 , r2 , ..., rn . the list may contain repetitions; if rk is repeated m times we say it is a root
of multiplicity m. If the coefficients p1 , p2 , ..., pn are all real,
any complex roots must appear in conjugate pairs; i.e., if the
list includes rk = ρ + iσ then somewhere it must also include
rj = ρ − iσ. Complex roots can also be repeated. For example,
the equation
r4 + 4 r2 + 4 = 0
becomes, on setting s = r2 , the equation s2 + 4 s + 4 = 0,
2
which has s = −2 as a double root.
√ Then solving r = −2 we
obtain each of the roots r = ±i 2 as a double complex root of
the original equation.
4
It is not generally possible to find closed algebraic forms for
the roots of an n-th degree polynomial equation. This is possible
for n ≤ 4; the procedures for n = 1, 2 are familiar while the
procedures for n = 3, 4 are very complicated. All of these
cases are included in Mathematica and Maple. If the equations
are worked by hand the most important tool is the Remainder
Theorem: the binomial r − a is a factor of p(r) if and only if
p(a) = 0. Thus, e.g., in the case of the equation
p(r) = r3 − r2 + r − 1 = 0
we can see immediately that r − 1 must be a factor because
p(1) = 13 − 12 + r − 1 = 0. In the same way one can see that
r + 1 must be a factor of
q(r) = r3 + r2 + r + 1 = 0.
Once a factor has been found, it can be divided out of the equation to yield an equation of lower degree. Thus, since
p(r)
= r2 + 1 ≡ p̂(r),
r −1
we can immediately see that the remaining roots of the corresponding equation p(r) = 0 are those of the equation p̂(r) =
r2 + 1 = 0, and thus are ±i.
Once the roots of the characteristic equation of a constant
coefficient differential equation Ly = 0 have been found, the
rules for obtaining solutions are essentially the same as in the
second order case, but we will point out some new features.
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Example 1
We consider the differential equation
d 4y
d 2y
+ 4 2 + 4 y = 0.
dx 4
dx
4
2
We have seen that
√the characteristic equation r + 4 r + 4 = 0
has the roots ±i 2, each of them√a root of double
multiplicity.
√
i 2x
−i 2x
and e
are both
This means that the functions e
√ solui 2x
tions√and, because each root is double, the functions x e
and
−i 2x
xe
are also solutions. When we go to the real parts of these
solutions we see that the equation has the general solution
c1 sin
√
y(x, c1 , c2 , c3 , c4 ) =
√
√
√
2x + c2 cos 2x + c3 x sin 2x + c4 x cos 2x.
A new feature appearing here is the existence of solutions of
the form x sin x, x cos x of the homogeneous equation; that did
not happen in the second order constant coefficient homogeneous
case because second degree (i.e., quadratic polynomial equations
with real coefficients) cannot have repeated complex roots.
To check that this really is the general solution we have to
verify that the functions involved form a fundamental set. We
will postpone this until later because there are easier ways to do
it than to form the Wronskian determinant by brute force as we
did in the case of second order equations.
Example 2
Consider the differential equation
d 6y
− 64 y = 0.
dx 6
6
Here the characteristic equation is r6 − 64 = 0. The solutions
are then the sixth roots of 64. We let
r = ρ eiθ , 64 = 26 ei2kπ .
Then we have
ρ6 ei6θ = 26 ei2kπ
which gives ρ = 2 and θ = k3π . We obtain the desired 6 roots
by successively letting k = 0, 1, 2, 3, 4, 5. Thus the roots are
kπ
2 ei 3 , k = 0, 1, 2, 3, 4, 5, i.e.,
√
√
1
3
1
3
2, 2 ( + i ), 2 (− + i ),
2
2
2
2
√
√
1
3
1
3
−2, 2 (− − i ), 2 ( − i ).
2
2
2
2
Taking account of the way these occur in complex conjugate
pairs, the general solution takes the form
√
√
y(x, c1 , c2 , c3 , c4 , c5 , c6 ) = c1 e2x + ex (c2 sin 3x + c3 cos 3x)
√
√
+ c4 e−2x + e−x (c5 sin 3x + c6 cos 3x).
We will again postpone verification that the functions involved
constitute a fundamental set.
Example 3
Consider the differential equation
d 2y
dy
d 3y
−2 2 +3
− 2 y = 0.
3
dx
dx
dx
Here the characteristic polynomial is
p(r) = r3 − 2 r2 + 3 r − 2.
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It is easy to see that p(1) = 0, so the remainder theorem says
that r − 1 is a factor of p(r), i.e., r1 = 1 is a root of the
characteristic equation p(r) = 0. Carrying out the division, we
see that
p(r) = (r − 1)(r2 − r + 2).
The quadratic formula then gives the roots of r2 − r + 2 = 0
as
√
√
1
1
7
7
, r3 = − i
.
r2 = + i
2
2
2
2
Thus the general solution takes the form
√
√
7x
7x
y(x, c1 , c2 , c3 ) = c1 ex + ex/2(c2 sin
+ c3 cos
).
2
2
Example 4
Find the general solution of
d 4y
d 2y
d 6y
− 2 4 + 2 2 − y = 0.
dx 6
dx
dx
Since the characteristic equation p(r) = r6 − 2 r4 + 2 r2 − 1 = 0
contains only even powers of r, we let s = r2 and consider
instead the cubic equation q(s) = s3 − 2 s2 + 2 s − 1 = 0. We
readily verify that q(1) = 0 so we have a root s1 = 1, which
in turn yields the roots r1 = 1 and r2 = −1 of the original
equation p(r) = 0. Factoring s − 1 out of q(s) we find that
q(s) = (s − 1)(s2 − s + 1).
Applying the quadratic formula we obtain two additional roots
√
1
3
= eiπ/3
s2 = + i
2
2
8
and
√
1
3
s3 = − i
= e−iπ/3 .
2
2
These are easily seen to have the square roots
√
√
3
i
3
i
+ , r4 = − r3 = −
− ,
r3 = eiπ/6 =
2
2
2
2
√
√
3
i
3
i
r5 = e−iπ/6 =
− , r6 = − r5 = −
+ .
2
2
2
2
Accordingly, the general solution takes the form
x
−x
y(x, c1 , c2 , c3 , c4 , c5 , c6 ) = c1 e + c2 e
−
+e
Example 5
√
3x
2
(c5 sin
+e
√
3x
2
(c3 sin
x
x
+ c4 cos )
2
2
x
x
+ c6 cos ).
2
2
Find the general solution of
d 5y
d 4y
dy
2 5 −5 4 +2
− 5 y = 0.
dx
dx
dx
Here the characteristic equation is
p(r) = 2 r5 − 5 r4 + 2 r − 5 = 0.
Here there is no pattern of exclusively even, or exclusively odd,
powers of r, but the repetition of the 2, -5 pattern offers a clue.
Inspection shows that
p(r) = (2 r − 5)(r4 + 1).
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Accordingly, we have the real root r1 = 52 . The remaining roots
are obtained by solving r4 = −1. Writing r = ρ eiθ , −1 =
ei(π + 2kπ) , we have
ρ4 ei 4θ = (1) ei(π + 2kπ) .
Since ρ, the absolute value of r, must be non-negative, we have
ρ = 1. Taking k = 0, 1, 2, 3, we obtain the values
θ =
π 3π 5π 7π
,
,
,
4 4
4
4
corresponding to roots
1
1
r2 = √ (1 + i), r3 = √ (− 1 + i),
2
2
1
1
r4 = √ (− 1 − i), r5 = √ (1 − i).
2
2
We can then read off the general solution in the form
5x
x
x
√x
y(x, c1 , c2 , c3 , c4 , c5 ) = c1 e 2 + e 2 (c2 sin √ + c3 cos √ )
2
2
x
x
√x
+ e− 2 (c4 sin √ + c5 cos √ ).
2
2
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