Case III. Complex conjugate roots are of minor practical importance, and we discuss the derivation of real solutions from complex ones just in terms of a typical example
Case 2. Damped Forced Oscillations
3.2 Homogeneous Linear ODEs with Constant Coefficients
are linearly independent, so that W is not zero by Theorem 3. Hence (10) has a unique solution (by Cramer’s theorem in Sec. 7.7). With these values we obtain the particular solution
on I. Equation (10) shows that and its first derivatives agree at with Yand its corresponding derivatives. That is, and Ysatisfy, at , the same initial conditions.
The uniqueness theorem (Theorem 2) now implies that on I. This proves the theorem.
This completes our theory of the homogeneous linear ODE (2). Note that for it is identical with that in Sec. 2.6. This had to be expected.
n⫽2 䊏 y*⬅Y
x0 y*
x0
n⫺1 y*
y*(x)⫽C1y1(x)⫹ Á ⫹Cnyn(x) c1⫽C1,Á
, cn⫽Cn
SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients 111
1–6 BASES: TYPICAL EXAMPLES
To get a feel for higher order ODEs, show that the given functions are solutions and form a basis on any interval.
Use Wronskians. In Prob. 6, 1.
2.
3.
4.
5.
6.
7. TEAM PROJECT. General Properties of Solutions of Linear ODEs. These properties are important in obtaining new solutions from given ones. Therefore extend Team Project 38 in Sec. 2.2 to nth-order ODEs.
Explore statements on sums and multiples of solutions of (1) and (2) systematically and with proofs.
Recognize clearly that no new ideas are needed in this extension from to general n.
8–15 LINEAR INDEPENDENCE
Are the given functions linearly independent or dependent on the half-axis Give reason.
8. x2, 1>x2, 0 9. tan x, cot x, 1 xⱖ0?
n⫽2
1, x2, x4, x2yt⫺3xys⫹3yr⫽0
1, eⴚx cos 2x, eⴚx sin 2x, yt⫹2ys⫹5yr⫽0
eⴚ4x, xeⴚ4x, x2eⴚ4x, yt⫹12ys⫹48yr⫹64y⫽0
cos x, sin x, x cos x, x sin x, yiv⫹2ys⫹y⫽0
ex, eⴚx, e2x, yt⫺2ys⫺yr⫹2y⫽0
1, x, x2, x3, yiv⫽0
x⬎0,
P R O B L E M S E T 3 . 1
10. 11.
12. 13.
14. 15.
16. TEAM PROJECT. Linear Independence and Dependence. (a) Investigate the given question about a set Sof functions on an interval I. Give an example.
Prove your answer.
(1) If Scontains the zero function, can Sbe linearly independent?
(2) If Sis linearly independent on a subinterval Jof I, is it linearly independent on I?
(3) If Sis linearly dependent on a subinterval Jof I, is it linearly dependent on I?
(4) If S is linearly independent on I, is it linearly independent on a subinterval J?
(5) If S is linearly dependent on I, is it linearly independent on a subinterval J?
(6) If Sis linearly dependent on I, and if Tcontains S, is Tlinearly dependent on I?
(b) In what cases can you use the Wronskian for testing linear independence? By what other means can you perform such a test?
cosh 2x, sinh 2x, e2x cos2 x, sin2 x, 2p
sin x, cos x, sin 2x sin2 x, cos2 x, cos 2x
ex cos x, ex sin x, ex e2x, xe2x, x2e2x
3.2 Homogeneous Linear ODEs
where etc. As in Sec. 2.2, we substitute to obtain the characteristic equation
(2)
of (1). If is a root of (2), then is a solution of (1). To find these roots, you may need a numeric method, such as Newton’s in Sec. 19.2, also available on the usual CASs.
For general nthere are more cases than for We can have distinct real roots, simple complex roots, multiple roots, and multiple complex roots, respectively. This will be shown next and illustrated by examples.
Distinct Real Roots
If all the nroots of (2) are real and different, then the nsolutions (3)
constitute a basis for all x. The corresponding general solution of (1) is (4)
Indeed, the solutions in (3) are linearly independent, as we shall see after the example.
E X A M P L E 1 Distinct Real Roots Solve the ODE
Solution. The characteristic equation is It has the roots if you find one of them by inspection, you can obtain the other two roots by solving a quadratic equation (explain!). The corresponding general solution (4) is
Linear Independence of (3). Students familiar with nth-order determinants may verify that, by pulling out all exponential functions from the columns and denoting their product
by the Wronskian of the solutions in (3) becomes
(5)
⫽E 7
1 1 Á 1
l1 l2 Á ln l12 l22 Á ln2
# # Á #
l1nⴚ1 l2nⴚ1 Á lnnⴚ1
7.
W⫽7
el1x el2x Á elnx l1el1x l2el2x Á lnelnx l12el1x l22el2x Á ln2elnx
# # Á #
l1nⴚ1el1x l2nⴚ1el2x Á lnnⴚ1elnx 7 E⫽exp [l1⫹Á ⫹ln)x],
䊏
y⫽c1eⴚx⫹c2ex⫹c3e2x.
⫺1, 1, 2;
l3⫺2l2⫺l⫹2⫽0.
yt⫺2ys⫺yr⫹2y⫽0.
y⫽c1el1x⫹ Á ⫹cnelnx. y1⫽el1x, Á, yn⫽elnx. l1,Á, ln
n⫽2.
y⫽elx l
l(n)⫹anⴚ1l(nⴚ1)⫹ Á ⫹a1l⫹a0y⫽0 y⫽elx y(n)⫽dny>dxn,
The exponential function Eis never zero. Hence if and only if the determinant on the right is zero. This is a so-called Vandermondeor Cauchy determinant.1It can be shown that it equals
(6)
where Vis the product of all factors with for instance, when
we get This shows that the Wronskian is not zero
if and only if all the nroots of (2) are different and thus gives the following.
T H E O R E M 1 Basis
Solutions of (1) (with any real or complex ’s) form a basis of solutions of (1) on any open interval if and only if all n roots of (2) are different.
Actually, Theorem 1 is an important special case of our more general result obtained from (5) and (6):
T H E O R E M 2 Linear Independence
Any number of solutions of (1)of the form are linearly independent on an open interval I if and only if the corresponding are all different.
Simple Complex Roots
If complex roots occur, they must occur in conjugate pairs since the coefficients of (1) are real. Thus, if is a simple root of (2), so is the conjugate and two corresponding linearly independent solutions are (as in Sec. 2.2, except for notation)
E X A M P L E 2 Simple Complex Roots. Initial Value Problem Solve the initial value problem
Solution. The characteristic equation is It has the root 1, as can perhaps be seen by inspection. Then division by shows that the other roots are Hence a general solution and its derivatives (obtained by differentiation) are
ys⫽c1ex⫺100A cos 10x⫺100B sin 10x.
yr⫽c1ex⫺10A sin 10x⫹10B cos 10x, y⫽c1ex⫹A cos 10x⫹B sin 10x,
⫾10i.
l⫺1
l3⫺l2⫹100l⫺100⫽0.
ys(0)⫽ ⫺299.
yr(0)⫽11, y(0)⫽4,
yt⫺ys⫹100yr⫺100y⫽0,
y2⫽egx sin vx.
y1⫽egx cos vx,
l⫽g⫺iv, l⫽g⫹iv
l elx
lj y1⫽el1x,Á
, yn⫽elnx
⫺V⫽ ⫺(l1⫺l2)(l1⫺l3)(l2⫺l3).
n⫽3 j⬍k (⬉n);
lj⫺lk (⫺1)n(nⴚ1)>2V
W⫽0
SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients 113
1ALEXANDRE THÉOPHILE VANDERMONDE (1735–1796), French mathematician, who worked on solution of equations by determinants. For CAUCHY see footnote 4, in Sec. 2.5.
From this and the initial conditions we obtain, by setting ,
(a) (b) (c)
We solve this system for the unknowns A, B, Equation (a) minus Equation (c) gives Then from (a) and from (b). The solution is (Fig. 73)
This gives the solution curve, which oscillates about ex(dashed in Fig. 73). 䊏
y⫽ex⫹3 cos 10x⫹sin 10x.
B⫽1 c1⫽1
101A⫽303, A⫽3.
c1.
c1⫺100A⫽ ⫺299.
c1⫹10B⫽11, c1⫹A⫽4,
x⫽0
4 00 10
3 2
1 x
y 20
Fig. 73. Solution in Example 2
Multiple Real Roots
If a real double root occurs, say, then in (3), and we take and as corresponding linearly independent solutions. This is as in Sec. 2.2.
More generally, if is a real root of order m, then mcorresponding linearly independent solutions are
(7)
We derive these solutions after the next example and indicate how to prove their linear independence.
E X A M P L E 3 Real Double and Triple Roots Solve the ODE
Solution. The characteristic equation has the roots and and the answer is
(8)
Derivation of (7).We write the left side of (1) as
Let Then by performing the differentiations we have L[elx]⫽(ln⫹anⴚ1lnⴚ1⫹ Á ⫹a0)elx. y⫽elx.
L[y]⫽y(n)⫹anⴚ1y(nⴚ1)⫹ Á ⫹a0y.
䊏
y⫽c1⫹c2x⫹(c3⫹c4x⫹c5x2)ex. l5⫽1,
l3⫽l4⫽ l1⫽l2⫽0,
l5⫺3l4⫹3l3⫺l2⫽0 yv⫺3yiv⫹3yt⫺ys⫽0.
elx, xelx, x2elx, Á, xmⴚ1elx. l
xy1
y1
y1⫽y2
l1⫽l2,
Now let be a root of mth order of the polynomial on the right, where For let be the other roots, all different from Writing the polynomial in product form, we then have
with if and if Now comes the
key idea: We differentiate on both sides with respect to (9)
The differentiations with respect to xand are independent and the resulting derivatives are continuous, so that we can interchange their order on the left:
(10)
The right side of (9) is zero for because of the factors (and since we have a multiple root!). Hence by (9) and (10). This proves that is a solution of (1).
We can repeat this step and produce by another such
differentiations with respect to Going one step further would no longer give zero on the right because the lowest power of would then be multiplied by and because has no factors so we get preciselythe solutions in (7).
We finally show that the solutions (7) are linearly independent. For a specific nthis can be seen by calculating their Wronskian, which turns out to be nonzero. For arbitrary mwe can pull out the exponential functions from the Wronskian. This gives
times a determinant which by “row operations” can be reduced to the Wronskian of 1, The latter is constant and different from zero (equal to
These functions are solutions of the ODE so that linear independence follows from Theroem 3 in Sec. 3.1.
Multiple Complex Roots
In this case, real solutions are obtained as for complex simple roots above. Consequently, if is a complex double root, so is the conjugate Corresponding linearly independent solutions are
(11)
The first two of these result from and as before, and the second two from and in the same fashion. Obviously, the corresponding general solution is (12)
For complex triple roots(which hardly ever occur in applications), one would obtain two more solutions x2egx cos vx, x2egx sin vx,and so on.
y⫽egx[(A1⫹A2x) cos vx⫹(B1⫹B2x) sin vx].
xelx
xelx elx
elx
egx cos vx, egx sin vx, xegx cos vx, xegx sin vx.
l⫽g⫺iv.
l⫽g⫹iv
y(m)⫽0,
1!2!Á
(m⫺1)!).
x,Á , xmⴚ1.
(elx)m⫽elmx l⫺l1;
h(l) h(l1)⫽0
m!h(l) (l⫺l1)0,
l⫺l1 l.
m⫺2 x2el1x,Á
, xmⴚ1el1x
xel1x L[xel1x]⫽0
m⭌2 l⫺l1
l⫽l1 0
0l L[elx]⫽Lc 0
0l elxd ⫽L[xelx].
l 0
0l L[elx]⫽m(l⫺l1)mⴚ1h(l)elx⫹(l⫺l1)m 0
0l [h(l)elx].
l,
m⬍n.
h(l)⫽(l⫺lm⫹1)Á(l⫺ln) m⫽n,
h(l)⫽1
L[elx]⫽(l⫺l1)mh(l)elx l1. lmⴙ1,Á, ln
m⬍n m⬉n.
l1
SEC. 3.2 Homogeneous Linear ODEs with Constant Coefficients 115