- - - - - - --·

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(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(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(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 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