Linear n-th Order DifferentialEquations

General Form; Properties A linear n-th order differentialequation is an equation of the form

dny

dxn+ p1(x)

dn−1y

dxn−1+ ... + pn−1(x)

dy

dx+ pn(x) y = g(x),

where the pk(x), k = 1, 2, ..., n and g(x) are known continuous,or at least piecewise continuous, functions defined on some in-

terval a < x < b (which may be (−∞,∞). We make the usual

distinction between homogeneous and non-honogeneous equa-

tions, according as g(x) is, or is not, identically equal to zeroon (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 prop-erties 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 andz(x) is a solution of the corresponding homogeneous equation

dnz

dxn+ p1(x)

dn−1z

dxn−1+ ... + pn−1(x)

dz

dx+ pn(x) z = 0,

then y2(x) = y1(x) + z(x) is also a solution of the inhomoge-

neous 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 equa-

tion 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 so-lutions 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

dny

dxn+ p1(x)

dn−1y

dxn−1+ ... + pn−1(x)

dy

dx+ pn(x) y = g(x)

= αg1(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 funda-

mental 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 in-

homogenous 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.

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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 sat-

isfy those conditions. We will see that, just as in the second

order case, this property depends only on the choice of the so-lutions 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

(L y)(x) =dny

dxn+ p1(x)

dn−1y

dxn−1+ ... + pn−1(x)

dy

dx+ pn(x) y.

It will also be convenient, for later purposes, to denote the dif-

ferential operator d kydx k by Dk y and write

L y = Dn y + p1(x)Dn−1 y + ... + pn−1(x)D y + pn(x) y,

or, symbolically,

L = Dn + p1(x)Dn−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 lin-

ear homogeneous n-th order equation with constant coefficients

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to be an equation of the form

dny

dxn+ p1

dn−1y

dxn−1+ ... + pn−1

dy

dx+ pn y = 0,

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 formy(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 Funda-

mental Theorem of Algebra we know that this equation has a

total of n roots, real or complex, r1, r2, ..., rn. the list may con-

tain 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,which has s = −2 as a double root. Then solving r2 = −2 we

obtain each of the roots r = ±i√

2 as a double complex root of

the original equation.

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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 thesecases 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 becausep(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 equa-

tion to yield an equation of lower degree. Thus, since

p(r)

r − 1= r2 + 1 ≡ p̂(r),

we can immediately see that the remaining roots of the corre-

sponding 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 thesecond order case, but we will point out some new features.

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Example 1 We consider the differential equation

d 4y

dx 4+ 4

d 2y

dx 2+ 4 y = 0.

We have seen that the characteristic equation r4 + 4 r2 + 4 = 0

has the roots ±i√

2, each of them a root of double multiplicity.This means that the functions ei

√2x and e−i

√2x are both solu-

tions and, because each root is double, the functions x ei√

2x and

x e−i√

2x are also solutions. When we go to the real parts of thesesolutions we see that the equation has the general solution

y(x, c1, c2, c3, c4) =

c1 sin√

2x + c2 cos√

2x + c3 x sin√

2x + c4 x cos√

2x.

A new feature appearing here is the existence of solutions ofthe 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 doit 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

dx 6− 64 y = 0.

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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 = k π3 . We obtain the desired 6 roots

by successively letting k = 0, 1, 2, 3, 4, 5. Thus the roots are

2 eikπ

3 , k = 0, 1, 2, 3, 4, 5, i.e.,

2, 2 (1

2+ i

√3

2), 2 (−1

2+ i

√3

2),

−2, 2 (−1

2− i

√3

2), 2 (

1

2− i

√3

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 involvedconstitute a fundamental set.

Example 3 Consider the differential equation

d 3y

dx 3− 2

d 2y

dx 2+ 3

dy

dx− 2 y = 0.

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 thatp(r) = (r − 1)(r2 − r + 2).

The quadratic formula then gives the roots of r2 − r + 2 = 0

as

r2 =1

2+ i

√7

2, r3 =

1

2− i

√7

2.

Thus the general solution takes the form

y(x, c1, c2, c3) = c1 ex + ex/2(c2 sin

√7x

2+ c3 cos

√7x

2).

Example 4 Find the general solution of

d 6y

dx 6− 2

d 4y

dx 4+ 2

d 2y

dx 2− y = 0.

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, whichin 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

s2 =1

2+ i

√3

2= eiπ/3

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and

s3 =1

2− i

√3

2= e−iπ/3.

These are easily seen to have the square roots

r3 = eiπ/6 =

√3

2+

i

2, r4 = − r3 = −

√3

2− i

2,

r5 = e−iπ/6 =

√3

2− i

2, r6 = − r5 = −

√3

2+

i

2.

Accordingly, the general solution takes the form

y(x, c1, c2, c3, c4, c5, c6) = c1 ex + c2 e−x + e√

3x

2 (c3 sinx

2+ c4 cos

x

2)

+ e−√

3x

2 (c5 sinx

2+ c6 cos

x

2).

Example 5 Find the general solution of

2d 5y

dx 5− 5

d 4y

dx 4+ 2

dy

dx− 5 y = 0.

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

4,

3π

4,

5π

4,

7π

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corresponding to roots

r2 =1√2(1 + i), r3 =

1√2(− 1 + i),

r4 =1√2(− 1 − i), r5 =

1√2(1 − i).

We can then read off the general solution in the form

y(x, c1, c2, c3, c4, c5) = c1 e5x

2 + ex√

2 (c2 sinx√2

+ c3 cosx√2)

+ e− x

√

2(c4 sinx√2

+ c5 cosx√2).

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