Linear equations with constant coefficients

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
If the original equation (15.1) has f(x) = 0 (i.e. it is homogeneous) then of
course the complementary function yc (x) in (15.3) is already the general solution.
If, however, the equation has f(x) = 0 (i.e. it is inhomogeneous) then yc (x) is only
one part of the solution. The general solution of (15.1) is then given by
y(x) = yc (x) + yp (x),
(15.7)
where yp (x) is the particular integral, which can be any function that satisfies (15.1)
directly, provided it is linearly independent of yc (x). It should be emphasised for
practical purposes that any such function, no matter how simple (or complicated),
is equally valid in forming the general solution (15.7).
It is important to realise that the above method for finding the general solution
to an ODE by superposing particular solutions assumes crucially that the ODE
is linear. For non-linear equations, discussed in section 15.3, this method cannot
be used, and indeed it is often impossible to find closed-form solutions to such
equations.
15.1 Linear equations with constant coefficients
If the am in (15.1) are constants rather than functions of x then we have
an
dn y
dn−1 y
dy
+ a0 y = f(x).
+ an−1 n−1 + · · · + a1
n
dx
dx
dx
(15.8)
Equations of this sort are very common throughout the physical sciences and
engineering, and the method for their solution falls into two parts as discussed
in the previous section, i.e. finding the complementary function yc (x) and finding
the particular integral yp (x). If f(x) = 0 in (15.8) then we do not have to find
a particular integral, and the complementary function is by itself the general
solution.
15.1.1 Finding the complementary function yc (x)
The complementary function must satisfy
an
dn y
dn−1 y
dy
+ a0 y = 0
+ an−1 n−1 + · · · + a1
dxn
dx
dx
(15.9)
and contain n arbitrary constants (see equation (15.3)). The standard method
for finding yc (x) is to try a solution of the form y = Aeλx , substituting this into
(15.9). After dividing the resulting equation through by Aeλx , we are left with a
polynomial equation in λ of order n; this is the auxiliary equation and reads
an λn + an−1 λn−1 + · · · + a1 λ + a0 = 0.
492
(15.10)
15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
In general the auxiliary equation has n roots, say λ1 , λ2 , . . . , λn . In certain cases,
some of these roots may be repeated and some may be complex. The three main
cases are as follows.
(i) All roots real and distinct. In this case the n solutions to (15.9) are exp λm x
for m = 1 to n. It is easily shown by calculating the Wronskian (15.6)
of these functions that if all the λm are distinct then these solutions are
linearly independent. We can therefore linearly superpose them, as in
(15.3), to form the complementary function
yc (x) = c1 eλ1 x + c2 eλ2 x + · · · + cn eλn x .
(15.11)
(ii) Some roots complex. For the special (but usual) case that all the coefficients
am in (15.9) are real, if one of the roots of the auxiliary equation (15.10)
is complex, say α + iβ, then its complex conjugate α − iβ is also a root. In
this case we can write
c1 e(α+iβ)x + c2 e(α−iβ)x = eαx (d1 cos βx + d2 sin βx)
sin
= Aeαx
(βx + φ),
cos
(15.12)
where A and φ are arbitrary constants.
(iii) Some roots repeated. If, for example, λ1 occurs k times (k > 1) as a root
of the auxiliary equation, then we have not found n linearly independent
solutions of (15.9); formally the Wronskian (15.6) of these solutions, having
two or more identical columns, is equal to zero. We must therefore find
k − 1 further solutions that are linearly independent of those already found
and also of each other. By direct substitution into (15.9) we find that
xeλ1 x ,
x2 eλ1 x ,
...,
xk−1 eλ1 x
are also solutions, and by calculating the Wronskian it is easily shown that
they, together with the solutions already found, form a linearly independent
set of n functions. Therefore the complementary function is given by
yc (x) = (c1 + c2 x + · · · + ck xk−1 )eλ1 x + ck+1 eλk+1 x + ck+2 eλk+2 x + · · · + cn eλn x .
(15.13)
If more than one root is repeated the above argument is easily extended.
For example, suppose as before that λ1 is a k-fold root of the auxiliary
equation and, further, that λ2 is an l-fold root (of course, k > 1 and l > 1).
Then, from the above argument, the complementary function reads
yc (x) = (c1 + c2 x + · · · + ck xk−1 )eλ1 x
+ (ck+1 + ck+2 x + · · · + ck+l xl−1 )eλ2 x
+ ck+l+1 eλk+l+1 x + ck+l+2 eλk+l+2 x + · · · + cn eλn x .
493
(15.14)
HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
Find the complementary function of the equation
dy
d2 y
−2
+ y = ex .
dx2
dx
(15.15)
Setting the RHS to zero, substituting y = Aeλx and dividing through by Aeλx we obtain
the auxiliary equation
λ2 − 2λ + 1 = 0.
The root λ = 1 occurs twice and so, although ex is a solution to (15.15), we must find
a further solution to the equation that is linearly independent of ex . From the above
discussion, we deduce that xex is such a solution, so that the full complementary function
is given by the linear superposition
yc (x) = (c1 + c2 x)ex . Solution method. Set the RHS of the ODE to zero (if it is not already so), and
substitute y = Aeλx . After dividing through the resulting equation by Aeλx , obtain
an nth-order polynomial equation in λ (the auxiliary equation, see (15.10)). Solve
the auxiliary equation to find the n roots, λ1 , λ2 , . . . , λn , say. If all these roots are
real and distinct then yc (x) is given by (15.11). If, however, some of the roots are
complex or repeated then yc (x) is given by (15.12) or (15.13), or the extension
(15.14) of the latter, respectively.
15.1.2 Finding the particular integral yp (x)
There is no generally applicable method for finding the particular integral yp (x)
but, for linear ODEs with constant coefficients and a simple RHS, yp (x) can often
be found by inspection or by assuming a parameterised form similar to f(x). The
latter method is sometimes called the method of undetermined coefficients. If f(x)
contains only polynomial, exponential, or sine and cosine terms then, by assuming
a trial function for yp (x) of similar form but one which contains a number of
undetermined parameters and substituting this trial function into (15.9), the
parameters can be found and yp (x) deduced. Standard trial functions are as
follows.
(i) If f(x) = aerx then try
yp (x) = berx .
(ii) If f(x) = a1 sin rx + a2 cos rx (a1 or a2 may be zero) then try
yp (x) = b1 sin rx + b2 cos rx.
(iii) If f(x) = a0 + a1 x + · · · + aN xN (some am may be zero) then try
yp (x) = b0 + b1 x + · · · + bN xN .
494
15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
(iv) If f(x) is the sum or product of any of the above then try yp (x) as the
sum or product of the corresponding individual trial functions.
It should be noted that this method fails if any term in the assumed trial
function is also contained within the complementary function yc (x). In such a
case the trial function should be multiplied by the smallest integer power of x
such that it will then contain no term that already appears in the complementary
function. The undetermined coefficients in the trial function can now be found
by substitution into (15.8).
Three further methods that are useful in finding the particular integral yp (x) are
those based on Green’s functions, the variation of parameters, and a change in the
dependent variable using knowledge of the complementary function. However,
since these methods are also applicable to equations with variable coefficients, a
discussion of them is postponed until section 15.2.
Find a particular integral of the equation
d2 y
dy
−2
+ y = ex .
dx2
dx
From the above discussion our first guess at a trial particular integral would be yp (x) = bex .
However, since the complementary function of this equation is yc (x) = (c1 + c2 x)ex (as
in the previous subsection), we see that ex is already contained in it, as indeed is xex .
Multiplying our first guess by the lowest integer power of x such that the result does not
appear in yc (x), we therefore try yp (x) = bx2 ex . Substituting this into the ODE, we find
that b = 1/2, so the particular integral is given by yp (x) = x2 ex /2. Solution method. If the RHS of an ODE contains only functions mentioned at the
start of this subsection then the appropriate trial function should be substituted
into it, thereby fixing the undetermined parameters. If, however, the RHS of the
equation is not of this form then one of the more general methods outlined in subsections 15.2.3–15.2.5 should be used; perhaps the most straightforward of these is
the variation-of-parameters method.
15.1.3 Constructing the general solution yc (x) + yp (x)
As stated earlier, the full solution to the ODE (15.8) is found by adding together
the complementary function and any particular integral. In order to illustrate
further the material discussed in the last two subsections, let us find the general
solution to a new example, starting from the beginning.
495
HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
Solve
d2 y
+ 4y = x2 sin 2x.
dx2
(15.16)
First we set the RHS to zero and assume the trial solution y = Aeλx . Substituting this into
(15.16) leads to the auxiliary equation
λ2 + 4 = 0
⇒
λ = ±2i.
(15.17)
Therefore the complementary function is given by
yc (x) = c1 e2ix + c2 e−2ix = d1 cos 2x + d2 sin 2x.
(15.18)
We must now turn our attention to the particular integral yp (x). Consulting the list of
standard trial functions in the previous subsection, we find that a first guess at a suitable
trial function for this case should be
(ax2 + bx + c) sin 2x + (dx2 + ex + f) cos 2x.
(15.19)
However, we see that this trial function contains terms in sin 2x and cos 2x, both of which
already appear in the complementary function (15.18). We must therefore multiply (15.19)
by the smallest integer power of x which ensures that none of the resulting terms appears
in yc (x). Since multiplying by x will suffice, we finally assume the trial function
(ax3 + bx2 + cx) sin 2x + (dx3 + ex2 + fx) cos 2x.
(15.20)
Substituting this into (15.16) to fix the constants appearing in (15.20), we find the particular
integral to be
x3
x2
x
cos 2x +
sin 2x +
cos 2x.
12
16
32
The general solution to (15.16) then reads
yp (x) = −
(15.21)
y(x) = yc (x) + yp (x)
= d1 cos 2x + d2 sin 2x −
x3
x2
x
cos 2x +
sin 2x +
cos 2x. 12
16
32
15.1.4 Linear recurrence relations
Before continuing our discussion of higher-order ODEs, we take this opportunity
to introduce the discrete analogues of differential equations, which are called
recurrence relations (or sometimes difference equations). Whereas a differential
equation gives a prescription, in terms of current values, for the new value of a
dependent variable at a point only infinitesimally far away, a recurrence relation
describes how the next in a sequence of values un , defined only at (non-negative)
integer values of the ‘independent variable’ n, is to be calculated.
In its most general form a recurrence relation expresses the way in which un+1
is to be calculated from all the preceding values u0 , u1 , . . . , un . Just as the most
general differential equations are intractable, so are the most general recurrence
relations, and we will limit ourselves to analogues of the types of differential
equations studied earlier in this chapter, namely those that are linear, have
496
15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
constant coefficients and possess simple functions on the RHS. Such equations
occur over a broad range of engineering and statistical physics as well as in the
realms of finance, business planning and gambling! They form the basis of many
numerical methods, particularly those concerned with the numerical solution of
ordinary and partial differential equations.
A general recurrence relation is exemplified by the formula
un+1 =
N−1
ar un−r + k,
(15.22)
r=0
where N and the ar are fixed and k is a constant or a simple function of n.
Such an equation, involving terms of the series whose indices differ by up to N
(ranging from n−N +1 to n), is called an Nth-order recurrence relation. It is clear
that, given values for u0 , u1 , . . . , uN−1 , this is a definitive scheme for generating the
series and therefore has a unique solution.
Parallelling the nomenclature of differential equations, if the term not involving
any un is absent, i.e. k = 0, then the recurrence relation is called homogeneous.
The parallel continues with the form of the general solution of (15.22). If vn is
the general solution of the homogeneous relation, and wn is any solution of the
full relation, then
un = vn + wn
is the most general solution of the complete recurrence relation. This is straightforwardly verified as follows:
un+1 = vn+1 + wn+1
N−1
N−1
=
ar vn−r +
ar wn−r + k
r=0
=
N−1
r=0
ar (vn−r + wn−r ) + k
r=0
=
N−1
ar un−r + k.
r=0
Of course, if k = 0 then wn = 0 for all n is a trivial particular solution and the
complementary solution, vn , is itself the most general solution.
First-order recurrence relations
First-order relations, for which N = 1, are exemplified by
un+1 = aun + k,
with u0 specified. The solution to the homogeneous relation is immediate,
un = Can ,
497
(15.23)
HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
and, if k is a constant, the particular solution is equally straightforward: wn = K
for all n, provided K is chosen to satisfy
K = aK + k,
i.e. K = k(1 − a)−1 . This will be sufficient unless a = 1, in which case un = u0 + nk
is obvious by inspection.
Thus the general solution of (15.23) is
#
Can + k/(1 − a) a = 1,
(15.24)
un =
u0 + nk
a = 1.
If u0 is specified for the case of a = 1 then C must be chosen as C = u0 −k/(1−a),
resulting in the equivalent form
un = u0 an + k
1 − an
.
1−a
(15.25)
We now illustrate this method with a worked example.
A house-buyer borrows capital B from a bank that charges a fixed annual rate of interest
R%. If the loan is to be repaid over Y years, at what value should the fixed annual payments
P , made at the end of each year, be set? For a loan over 25 years at 6%, what percentage
of the first year’s payment goes towards paying off the capital?
Let un denote the outstanding debt at the end of year n, and write R/100 = r. Then the
relevant recurrence relation is
un+1 = un (1 + r) − P
with u0 = B. From (15.25) we have
un = B(1 + r)n − P
1 − (1 + r)n
.
1 − (1 + r)
As the loan is to be repaid over Y years, uY = 0 and thus
P =
Br(1 + r)Y
.
(1 + r)Y − 1
The first year’s interest is rB and so the fraction of the first year’s payment going
towards capital repayment is (P − rB)/P , which, using the above expression for P , is equal
to (1 + r)−Y . With the given figures, this is (only) 23%. With only small modifications, the method just described can be adapted to
handle recurrence relations in which the constant k in (15.23) is replaced by kαn ,
i.e. the relation is
un+1 = aun + kαn .
(15.26)
As for an inhomogeneous linear differential equation (see subsection 15.1.2), we
may try as a potential particular solution a form which resembles the term that
makes the equation inhomogeneous. Here, the presence of the term kαn indicates
498
15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
that a particular solution of the form un = Aαn should be tried. Substituting this
into (15.26) gives
Aαn+1 = aAαn + kαn ,
from which it follows that A = k/(α − a) and that there is a particular solution
having the form un = kαn /(α − a), provided α = a. For the special case α = a, the
reader can readily verify that a particular solution of the form un = Anαn is appropriate. This mirrors the corresponding situation for linear differential equations
when the RHS of the differential equation is contained in the complementary
function of its LHS.
In summary, the general solution to (15.26) is
#
C1 an + kαn /(α − a) α = a,
(15.27)
un =
C2 an + knαn−1
α = a,
with C1 = u0 − k/(α − a) and C2 = u0 .
Second-order recurrence relations
We consider next recurrence relations that involve un−1 in the prescription for
un+1 and treat the general case in which the intervening term, un , is also present.
A typical equation is thus
un+1 = aun + bun−1 + k.
(15.28)
As previously, the general solution of this is un = vn + wn , where vn satisfies
vn+1 = avn + bvn−1
(15.29)
and wn is any particular solution of (15.28); the proof follows the same lines as
that given earlier.
We have already seen for a first-order recurrence relation that the solution to
the homogeneous equation is given by terms forming a geometric series, and we
consider a corresponding series of powers in the present case. Setting vn = Aλn in
(15.29) for some λ, as yet undetermined, gives the requirement that λ should satisfy
Aλn+1 = aAλn + bAλn−1 .
Dividing through by Aλn−1 (assumed non-zero) shows that λ could be either of
the roots, λ1 and λ2 , of
λ2 − aλ − b = 0,
(15.30)
which is known as the characteristic equation of the recurrence relation.
That there are two possible series of terms of the form Aλn is consistent with the
fact that two initial values (boundary conditions) have to be provided before the
series can be calculated by repeated use of (15.28). These two values are sufficient
to determine the appropriate coefficient A for each of the series. Since (15.29) is
499
HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
both linear and homogeneous, and is satisfied by both vn = Aλn1 and vn = Bλn2 , its
general solution is
vn = Aλn1 + Bλn2 .
If the coefficients a and b are such that (15.30) has two equal roots, i.e. a2 = −4b,
then, as in the analogous case of repeated roots for differential equations (see
subsection 15.1.1(iii)), the second term of the general solution is replaced by Bnλn1
to give
vn = (A + Bn)λn1 .
Finding a particular solution is straightforward if k is a constant: a trivial but
adequate solution is wn = k(1 − a − b)−1 for all n. As with first-order equations,
particular solutions can be found for other simple forms of k by trying functions
similar to k itself. Thus particular solutions for the cases k = Cn and k = Dαn
can be found by trying wn = E + Fn and wn = Gαn respectively.
Find the value of u16 if the series un satisfies
un+1 + 4un + 3un−1 = n
for n ≥ 1, with u0 = 1 and u1 = −1.
We first solve the characteristic equation,
λ2 + 4λ + 3 = 0,
to obtain the roots λ = −1 and λ = −3. Thus the complementary function is
vn = A(−1)n + B(−3)n .
In view of the form of the RHS of the original relation, we try
wn = E + Fn
as a particular solution and obtain
E + F(n + 1) + 4(E + Fn) + 3[E + F(n − 1)] = n,
yielding F = 1/8 and E = 1/32.
Thus the complete general solution is
un = A(−1)n + B(−3)n +
n
1
+ ,
8 32
and now using the given values for u0 and u1 determines A as 7/8 and B as 3/32. Thus
un =
1
[28(−1)n + 3(−3)n + 4n + 1] .
32
Finally, substituting n = 16 gives u16 = 4 035 633, a value the reader may (or may not)
wish to verify by repeated application of the initial recurrence relation. 500
15.1 LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS
Higher-order recurrence relations
It will be apparent that linear recurrence relations of order N > 2 do not present
any additional difficulty in principle, though two obvious practical difficulties are
(i) that the characteristic equation is of order N and in general will not have roots
that can be written in closed form and (ii) that a correspondingly large number
of given values is required to determine the N otherwise arbitrary constants in
the solution. The algebraic labour needed to solve the set of simultaneous linear
equations that determines them increases rapidly with N. We do not give specific
examples here, but some are included in the exercises at the end of the chapter.
15.1.5 Laplace transform method
Having briefly discussed recurrence relations, we now return to the main topic
of this chapter, i.e. methods for obtaining solutions to higher-order ODEs. One
such method is that of Laplace transforms, which is very useful for solving
linear ODEs with constant coefficients. Taking the Laplace transform of such an
equation transforms it into a purely algebraic equation in terms of the Laplace
transform of the required solution. Once the algebraic equation has been solved
for this Laplace transform, the general solution to the original ODE can be
obtained by performing an inverse Laplace transform. One advantage of this
method is that, for given boundary conditions, it provides the solution in just
one step, instead of having to find the complementary function and particular
integral separately.
In order to apply the method we need only two results from Laplace transform
theory (see section 13.2). First, the Laplace transform of a function f(x) is defined
by
f̄(s) ≡
∞
e−sx f(x) dx,
(15.31)
0
from which we can derive the second useful relation. This concerns the Laplace
transform of the nth derivative of f(x):
f (n) (s) = sn f̄(s) − sn−1 f(0) − sn−2 f (0) − · · · − sf (n−2) (0) − f (n−1) (0),
(15.32)
where the primes and superscripts in parentheses denote differentiation with
respect to x. Using these relations, along with table 13.1, on p. 455, which gives
Laplace transforms of standard functions, we are in a position to solve a linear
ODE with constant coefficients by this method.
501
HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
Solve
d2 y
dy
−3
+ 2y = 2e−x ,
dx2
dx
subject to the boundary conditions y(0) = 2, y (0) = 1.
(15.33)
Taking the Laplace transform of (15.33) and using the table of standard results we obtain
s2 ȳ(s) − sy(0) − y (0) − 3 [sȳ(s) − y(0)] + 2ȳ(s) =
2
,
s+1
which reduces to
(s2 − 3s + 2)ȳ(s) − 2s + 5 =
2
.
s+1
(15.34)
Solving this algebraic equation for ȳ(s), the Laplace transform of the required solution to
(15.33), we obtain
ȳ(s) =
2s2 − 3s − 3
1
2
1
=
+
−
,
(s + 1)(s − 1)(s − 2)
3(s + 1) s − 1 3(s − 2)
(15.35)
where in the final step we have used partial fractions. Taking the inverse Laplace transform
of (15.35), again using table 13.1, we find the specific solution to (15.33) to be
y(x) = 13 e−x + 2ex − 13 e2x . Note that if the boundary conditions in a problem are given as symbols, rather
than just numbers, then the step involving partial fractions can often involve
a considerable amount of algebra. The Laplace transform method is also very
convenient for solving sets of simultaneous linear ODEs with constant coefficients.
Two electrical circuits, both of negligible resistance, each consist of a coil having selfinductance L and a capacitor having capacitance C. The mutual inductance of the two
circuits is M. There is no source of e.m.f. in either circuit. Initially the second capacitor
is given a charge CV0 , the first capacitor being uncharged, and at time t = 0 a switch in
the second circuit is closed to complete the circuit. Find the subsequent current in the first
circuit.
Subject to the initial conditions q1 (0) = q̇1 (0) = q̇2 (0) = 0 and q2 (0) = CV0 = V0 /G, say,
we have to solve
Lq̈1 + Mq̈2 + Gq1 = 0,
Mq̈1 + Lq̈2 + Gq2 = 0.
On taking the Laplace transform of the above equations, we obtain
(Ls2 + G)q̄1 + Ms2 q̄2 = sMV0 C,
Ms2 q̄1 + (Ls2 + G)q̄2 = sLV0 C.
Eliminating q̄2 and rewriting as an equation for q̄1 , we find
MV0 s
[(L + M)s2 + G ][(L − M)s2 + G ]
V0
(L + M)s
(L − M)s
.
=
−
2
2
2G (L + M)s + G (L − M)s + G
q̄1 (s) =
502