- - - - - - --·
3.3. Homogeneous Equation: General Solution
Homogeneous Equation: General Solution
· General solution. We studied the first-order linear homogeneous equation
y'
+ p(x)y =
0
(1)
Chapter 2, where p( x) is continuous on the x interval of interest, I, and found
- solution to be
y(x) = ce-fp(x)dx,
(2)
C is an arbitrary constant. If we append to ( 1) an initial condition y( a) = b,
a is a point in I, then we obtain, from (2),
(3)
was shown in Section 2.2.
The solution (2) is really a family of solutions because of the arbitrary constant
We showed that (2) contains all solutions of (1), so we called it a general
of (1). In contrast, (3) was only one member of that family, so we called it
. particular solution.
· Now we tum to the nth-order linear equation
any
dxn
an-ly .
dy
+ PI(x) dxn-l + .. · + Pn-l(x) dx + Pn(x)y =
0,
(4)
once again we are interested in general and particular solutions. By a general
sohltio1D of (4), on an interval I, we mean a family of solutions that contains every
of (4) on that interval, and by a particular solution of (4), we mean any
member of that family of solutions.
We begin with a fundamental existence and uniqueness theorem.*
••~"'"'•'""•"'" 3.3.1 Existence and Uniqueness for Initial-Value Problem
. Pl(x), ... ,pn(x) are continuous on a closed interval I, then the initial-value
:&: PJ~om.em consisting of the differential equation
dny
dX
n
dn-ly
dy
+ Pl(x)-d
n
+ · · · + Pn-l(x)-dX + Pn(x)y = 0,
X- 1
(5a)
together with initial conditions
(5b)
*For a more complete sequence of theorems, and their proofs, we refer the interested reader to the
little book by J. C. Burkill, The Theory of Ordinary Differential Equations (Edinburgh: Oliver and
Boyd, 1956) or to William E. Boyce and Richard C. DiPrima, Elementary Differential Equations and
Boundary Value Problems, 5th ed. (New York: Wiley, 1992).
----·---.
83
84
Chapter 3. Linear Differential Equations of Second Order and Higher
where the initial point a is in/, has a solution on/, and that solution is unique.
Notice how the initial conditions listed in (Sb) are perfect-not too few and not.
too many -in narrowing the general solution of (Sa) down to a unique particular •
solution, for (Sa) gives y<n>(x) as a linear combination of y(n-l)(x), ... , y(x), the·
derivative of (Sa) gives y(n+l)(x) as a linear combination of y(n)(x), ... , y(x),
and so on. Thus, knowing y( a), ... , y< n- 1) (a) we can use the differential equation
(Sa) and its derivatives to compute y<n>(a), y(n+l)(a), and so on, and therefore to
develop the Taylor series of y(x) about the point a; that is, to determine y(x).
Let us leave the initial-value problem (S) now, and turn our attention to determining the nature of the general solution of the nth-order linear homogeneous
equation (4). We begin by re-expressing (4) in the compact form
L[y] =0,
where
dn
~-1
· L = -d
+
P1(x)-d
xn
xn-· 1
d
+ .. · + Pn-1(x)-dX + Pn(x)
(7)
is called an nth-order differential operator and
L[y] =
~
(
dxn
~-1
d
+ P1(x) dxn- 1 + .. · + Pn-1(x) dx + Pn(x)
~
)
[y]
~-1
=-dxn y(x) + Pl(x)-d
xn- y(x) + · · · + Pn(x)y(x)
1
(8)
defines the action of L on any n-times differentiable function y. L[y] is itself
a function of x, with values L[y](x). For instance, if n = 2, PI(x) = sinx,
p 2(x) = 5x, and y(x) = x 2, then L[y](x) = (x 2 )" + (sinx)(x 2 )' + 5x(x2 ) =
2 + 2xsinx + 5x3 .
The key property of the operator L defined by (8) is that
IL [au+ ,Bv] = aL [u] + ,BL [v] I
(9)
for any (n-times differentiable) functions u, v and any constants a, ,8. Let us verify
(9) for the representative case where L is of second order:
L [au + ,Bv] = (
~2 + Pl ddx + P2) (au + ,Bv)
=(au+ ,Bv)" + P1 (au+ ,Bv)' + P2 (au+ ,Bv)
= au" + ,Bv" + Pl au' + P1.Bv' + P2au + P2.Bv
1
1
=a (u" + P1U + P2u) + ,8 (v" + PIV + P2v)
= aL [u] + ,BL [v] .
Similarly for n ;::: 3.
eq
(10)
E~
90
Chapter 3. Linear Differential Equations of Second Order and Higher
(f) y""
+ 2y" + y = 0;
{cos x, sinx, x cos x, x sin x}
3. Are the following general solutions of x y" + xy' - 4y
on 0 < x < oo? On -oo < x < oo? Explain.
2
(~) Crx 2
(c) Ct(x 2 + x- 2 )
(b.) Crx 2
+ C2(x 2 - x- 2)
=0
+ C2x- 2
(~ y" + 2y' + 3y = 0;
y(O) = 5, y'(O) = -1
(b) y" + 2y' + 3y = 0; y(3) = 2, y'(3) = 37
(c) y" + xy' - y = 0; y(3) = y'(3) = 0
@ xy"' + xy' - y = 0; y( -1) = y' ( -1) = 0, y" ( -1)
(e) x 2 y"- y'- y = 0; y(6) = 0, y'(6) = 1
(f) (sin x )y"" + xy"' = 0; y(2) = y' (2) = y" (2) = 0,
y"'(2) = -9
=4
4. Are the following bases for the equation x 2 y"- xy' + y = 0 10. Verify that (22) is indeed a general solution of (21 ).
on 0 < x < oo? On -oo < x < 0? On -oo < x < oo? On
11. Consider the boundary-value problem consisting of the
differential equation y" + y = 0 plus the boundary conditions
given. Does the problem have any solutions? If so, find them.
Is the solution unique? HINT: A general solution of the differential equation is y = C 1 cosx + C 2 sinx.
< x < 10? Explain.
U!) {x, x 2 }
6
(b) {ex, e-x}
(f.) {x, x In lxl}
(d) { x + x In lxl, x - x In Ixi}
(S!) y(O) = 0, y(2) = 0
5. Show whether or not the following is a general solution of (b.) y(O) = 0, y(27r) = -3
y(vii) - 4y(vi) - 14y(v) + 56y(iv) + 49y 111 - 196y11 - 36y' + (c) y(1) = 1, y(2) = 2
144y = 0.
(d) y'(O) = 0, y(5) = 1
(e)
y'(O) = 0, y'(1r) = 0
(S!) Ctex + C2e-x + C3e2x + C4e-2x + Cse3x + C6e-3x
(f) y'(O) = 0, y'(61r) = 0
(b) Crex + C2e-x + C3e 2x + C4e- 2x + Cse 3x + C6e-Jx +
(g) y'(O) = 0, y'(27r) = 38
c7 sinh X + Cs cosh 2x
12. Consider the boundary-value problem consisting of the·.
Show that y 1 = 1 and Y2
2 are solutions of
0 plus the boundary
differential
equation y"" + 2y" + y
(y 3 - 6y 2 + lly- 6) y" = 0. Is y = Yt + Y2 = 1 + 2 = 3
conditions given. Does the problem have any solutions? If.
a solution as well? Does your result contradict the sentence
so, find them. Is the solution unique? HINT: A general sopreceding Example 2? Explain.
lution of the differential equation is y = ( C 1 + C 2 x) cos x +
7_. Show that each of the functions Y1
3x 2 - x and (C3 + C4 x) sinx.
2
2
y 2 = x - xis a solution of the equation x y"- 2y = 2x. Is U!) y (0 ) = y '( 0 ) = 0, y ( 7r ) = 0, y'( 7r ) = 2
the ~inear combination C1Y1 + C2Y2 a solution as well, for all (b) y(O) = y'(O) = y"(O) = o, y(1r) = 1
chotces of the constants Cr and C2?
(c) y(O) = y"(O) = o, y(1r) = o = y"(1r) = o
6.
=
=
=
~
0•3
7-ee {j.)
8. _(Taylor series method) Use the Taylor series method ~e-~0) = y"(O) = 0, y(1r)
scnbed below Theorem 3.3.1 to solve each of the followmg
initial-value problems for y(x), up to and including terms of
fifth order. NOTE: The term JCn>(a)(x- a)n jn! in the Taylor
~es of f(x) about x = a is said to be of nth-order.
y(O) = 4, y'(O) = 3
(b) y" _ 4y = 0; y(O) = -1, y'(O) = 0
(c) y" + 5y' + 6y = 0; y(O) = 2, y'(O) = -5
(Q) y" + xy = 0; y(O) = 1, y'(O) = 0
(e) y" + x 2 y = 0; y(O) = 2, y'(O) = -3
(f) y" - 3y = 0; y(5) = 4, y'(5) = 6
HINT: Expand
about x = 5.
(g) y" +3y' -y = 0; y(1) = 2, y'(1) = 0 HINT: Expand
about x = 1.
(h) y"'- y' + 2y = 0; y(O) = 0, y'(O) = 0, y"(O) = 1
(i) y"'- xy = 0; y(O) = 0, y'(O) = 3, y"(O) = -2
(.w}y" + y = 0;
9. Does the problem stated have a unique solution? No solution? A nonunique solution? Explain.
13.
= y"(1r) =
3
ove that the linearity property ( lO) is equivalent to the
proQerties
I~ )()
~ell !J!
L [u
+ v] = L [u] + L [v] ,
L [au] = aL [u].
(l3.lb)
That is, show that the truth of (I 0) implies the truth of (13.1},
and conversely.
14. We showed that ( 11) holds for the case k = 3, but did not
prove it in general. Here, we ask you to prove ( ll) for any integer k 2: 1. HINT: It is suggested that you use mathematical
induction, whereby a proposition P(k), fork 2: 1, is proved
by first showing that it holds fork = 1, and then showing that ·
if it holds for k then it must also hold for k + 1. In the present
example, the proposition P( k) is the equation ( 11 ).
15. (Example 4, Continued) (a) Verify that each of (18a) .
108
Chapter 3. Linear Differential Equations of Second Order and Higher
Computer software. To obtain a general solution of y'" - 9y'
use the command
dsolve( { diff(y(x), x, x, x)- 9 * diff(y(x), x)
=
0 using Maple,
= 0}, y(x));
and to solve the ODE subjectto the initial conditions y(O) = 5, y'(O)
-4, use the command
dsolve( {diff(y(x), x, x, x)- 9 * diff(y(x), x)
D(y)(O) = 2, D(D(y))(O) = -4}, y(x));
= 2, y"(O) =
= 0, y(O) = 5,
In place of diff(y( x), x, x, x) we could use diff(y( x), x$3), for brevity.
EXERCISES 3.4
1. Use whichever of equations (5)-(8) are needed to derive
these relations between the circular and hyperbolic functions:
(b) sin (ix) = i sinhx
(d) sinh (ix) i sinx
(a) cos (ix) = coshx
(c) cosh (ix) =cos x
=
2. Use equations (6) and/or (7) to derive or verify
(a) equation (9)
(c) equation (13a)
(e) equation (13c)
(b) equation (10)
(d) equation (l3b)
(f) equation (13d)
~-Theorem 3.4.2 states that eA 1 x, xeA 1 x, ... , xk-leA 1 x are LI.
Prove that claim.
4. (Nonrepeated roots) Find a general solution of each of the
following equations, and a particular solution satisfying the
given· conditions, if such conditions are given.
+ 5y' = 0
(b) y" - y' = 0
(c) y" + y' = 0; y(O)
(Q) y"- 3y' + 2y = 0;
y"- 4y'- 5y = 0;
1
II + y - 12y = 0;
g y"- 4y' + 5y = 0;
(ii) y"- 2y' + 3y = 0;
(i) y" - 2y' + 2y = 0;
(j) y" + 2y' + 3y = 0;
(k} y"' +3y' -4y = 0;
(1) y"'- y" + 2y' = 0;
(m.) y"' + y" - 2y = 0
(!!) y"
(n) y(iv) - y
=0
y'(O) = 0
y(1) = 1, y'(1) = 0
y(1) = 1, y'(1) = 0
y( -1) = 2, Y1 ( -1) = 5
y(O) = 2, y'(O) = 5
y(O) = 4, y'(O) = -1
y(O) = 0, y'(O) = -5
y(O) = 0, y'(O) = 3
y(O) = 0, y'(O) = 0, y"(O) = 6
y(O) = 1, y' (0) = 0, y" (0) = 0
= 3,
(o) y(iv) - 2y"- 3y = 0
(p) y(iv) + 6y 11 + 8y = 0
(q) y(iv) + 7y" + 12y = 0
(r) y(iv) - 2y"' - y" + 2y' = 0
rO(
5. (a)- (r) Solve the corresponding problem in Exercise 4
using computer software.
6. (Repeated roots) Find a general solution of each of the fol-
10.
usi
11.
the
lowing equations, and a particular solution satisfying the given
conditions, if such conditions are given.
=
(!!) y"
0; y( -3) = 5, y'( -3) = -1
(b) y" + 6y' + 9y 0; y(1)
e, y'(1) -2
(c) y"' = 0; y(O) = 3, y'(O)
-5, y"(O) = 1
+ 5y" 0; y(O) = 1, y'(O) 0, y"(O) = 0
(e) y"' + 3y" + 3y' + y = 0
ciDY"'
=
=
=
=
W..,y"'- 3y" + 3y'- y = 0
-y" -y' +y = 0
(h) y(iv) + 3y"' = 0
{i) y(iv) + y"' + y" = 0
(j) y(iv) + 8y" + 16y = 0
(k) y(vi} = 0; y(O) = y'(O)
y(iv)(O) = 0, y<v>(o) = 3
=
see
tim
=
(a)
~"'
whc
A2·
=
y"(O)
=
y'"(O)
=
7. (a)- (k) Solve the corresponding problem in Exercise 6
using computer software.
8. If the roots of the characteristic equation are as follows,
-then find the original differential equation and also a general
solution of it:
(!!)2,6
(c) 4- 2i, 4 + 2i
(b) 2i, -2i
(Q) -2, 3,5
iza1
as (
(D
Of]J
of{
(b)'
the
3.6. Solution of Homogeneous Equation: Nonconstant Coefficients
feel comfortable with erf( x) and regard it henceforth as a known tuncThough not included among the so-called "elementary functions," it is one
"special functions" that are now available in the engineering science and
tben11ati1CS literature.
We have seen, in tbis section, that nonconstant-coefficient equations can
in closed form only in exceptional cases. The most important of these is
the Cauchy-Euler equation
t .
n lff'y
X
dxn
+ C1X
n-llfl-ly
dxn-l
dy
+ · · · + Cn-lX dx + CnY = 0.
·····.Recall that a constant-coefficient equation necessarily admits at least one so·. in the form e~x, and that in the case of a repeated root of order k the
··
corresponding to that root can be found by reduction of order to be
Analogously, a Cauchy-Euler equation necesu,.., at least one solution in the form x~, and in the case of a repeated root
k the solutions corresponding to that root can be found by reduction of
to be [G1 + C2lnx + · · · + Ck(lnx)k-l] x~.
In fact, it turns out that the connection between constant-coefficient equations
Cauchy- Euler equations is even closer than that in as much as any given
,.,,.,..,.,_Euler equation can be reduced to a constant-coefficient equation by a
of independent variable .according to x = et. Discussion of that point is
for the exercises.
Beyond our striking success with the Cauchy-Euler equation, other successes
nonconstant-coefficient equations are few and far between. For instance, we
m.ight be able to obtain one solution by inspection and others, from it, by reducof order. Or, we might, in exceptional cases, be successful in factoring the
~diffe":nti:al operator but, again, such successes are exceptional. Thus, other lines of
~pproach will be needed for nonconstant-coefficient equations, and they are developed in Chapters 4 and 6.
+ C2x + ·· · + Gkxk-l) e~x.
llY ......
J. Derive a general solution for the given Cauchy-Eulerequa- (f) x 2y" + xy' + y = 0; y(1) = 1, y'(1) = 0
· .tion by seeking y(x) = x~. That is, derive the solution, rather (g) x 2y" + 3xy' + 2y = 0; y(1) = 0, y'(l) = 2
·thanmerelyuseanystatedresultsuchasTheorem3.6.1. In ad- (h)x2y"- 2y = O; y(-5) = 3, y'(-5) = 0
• dition, find the particular solution corresponding to the initial (i) 4x2y" + 5y = O; y(l) = 0, y'(l) = 1
conditions, if such conditions are given, and state the interval (j) x2y" + xy' + 4y = 0
ofvalidity of that solution.
(k) x2y" + 2xy' - 2y = O; y(3) = 2, y' (3) = 2
·· (!) xy' + y = 0
((i))x + 2) 2 y" - y = 0 lllNT: Let x + 2 = t.
(b)xy'- y = O; y(2) = 5
Ym)x2y"'- 2y' = 0; y(1) = 2, y'(1) = y"(1) = 0
·· ·~·xy" + y' = 0
(n) xy"'- y" = 0; y(1) = 1, y'(1) = y"(l) = 0
d y"- 4y' = O; y(1) = 0, y'(l) = 3
(o) x 2y" + xy' - tt2y = 0 (tt a constant)
·. e x 2y" + xy'- 9y = 0; y(2) = 1, y'(2) = 2
<19 x 3y"' + xy'- Y = 0.
131
148
Chapter 3. Linear Differential Equations of Second Order and Higher
then Yp is the sum of particular solutions Ypl, ... , Ypk corresponding to h, ... , fk,
respectively.
Having studied homogeneous equations earlier, we considered Yh as known
and focused our attention, in Sections 3.7.2 and 3.7.3, on methods for finding particular solutions yp: undetermined coefficients and variation of parameters. Of
these, the former is the e.asier to implement, but it is limited to cases where L is
of constant-coefficient type and where each /j(x) has only a finite number of LI
derivatives. The latter is harder to apply since it requires integrations, but is more
powerful since it works even if L has nonconstant coefficients, and for any functions /j(x).
1
EXERCISES 3.7
:r
1. Show whether or not the given forcing function satisfies 4. Obtain a general solution using the method of variation of
condition (ii), below equation (11). If so, give a finite family parameters.
of LI functions generated by it.
(~ x 2 cos x
(b) cos x sinh 2x
(c) In x
(Q) x 2 In x
x2
(f) e
(e) sinx I x
(!!)
y' - 3y = xe 2"' + 6
y' + y = x 4 + 2x
y' + 2y = 3e 2"' + 4sinx
y' - 3y = xe 3"' + 4
y' + y = 5 - e-x
(Qy' - y = x 2 e"'
(g) y" - y' = 5 sin 2x
(h) y" + y' = 4xe"' + 3sinx
(i) y" + y = 3 sin 2x- 5 + 2x 2
(j) y" + y' - 2y = x 3 - e-x
11
(!{) y + y = 6 COS X + 2
(!!)
(b)
(c)
(Q)
(e)
=
(I) y" + 2y'
x2 + 4e 2 "'
(m) y"- 2y' + y = x 2 e"'
(n) y"- 4y = 5(cosh 2x- x)
(Q) y" - y' = 2xe"'
(p) y"' - y' = 25 cos 2x
(q) y"' - y" = 6x + 2 cosh x
(r) y"" + y" - 2y = 3x 2 - 1
(s) y""- y
=
(j) y"
(k) y"
+ y = 4sinx
+ 4y' + 4y = 2e- 2"'
6y"-5y'+y=x 2
x 3 y" + x 2 y'- 4xy = 1 (x > 0)
x 2 y" - xy' - 3y = 4x (x < 0)
(Q) y"' + y" - y' - y = X
(p) y"'- 6y" + lly'- 6y = e 4 "'
5. (a)-(p) Use computer software to solve the corresponding
problem in Exercise 4.
6. In the method of variation of parameters we used indefinite
integrals [in (49) and (78)]. However, we could use definite
integrals in those formulas, instead, if we choose. Specifically,
show that, in place of (49),
= 5(x + cosx)
3. (a)-(s) Use computer software to solve the corresponding
problem in Exercise 2.
(
y' + 2y = 4e 2x
(b) y' - y = xe"' + 1
(c) xy' - y
x3
(!!) xy' + y = 1/x (x > 0)
(g) e9 x
(h) (x- l)j(x + 2)
(e) x 3 y' + x 2 y = 1 (x > 0)
(i) tan x
(j) e"' cos 3x
~y"- y = 8x
(k) x 3 e-"' sinhx
(I) cosxcos 2x
«tJ>y"- y = 8e"'
(m) sin x sin 2x sin 3x
(n) e"' j(x + 1)
(h) y" - 2y' + y = 6x 2
2. Obtain a general solution using the method of undetermined
(i) y" - 2y' + y = 2e"'
coefficients.
(
is also correct, for any choice of the constants a 1 , a 2 (although
normally one would choose a1 and a2 to be the same).
1
I