Homogeneous Linear ODEs

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

3.1 Homogeneous Linear ODEs

105

C H A P T E R 3

Higher Order Linear ODEs

The concepts and methods of solving linear ODEs of order extend nicely to linear ODEs of higher order n, that is, etc. This shows that the theory explained in Chap. 2 for second-order linear ODEs is attractive, since it can be extended in a straightforward way to arbitrary n. We do so in this chapter and notice that the formulas become more involved, the variety of roots of the characteristic equation (in Sec. 3.2) becomes much larger with increasing n, and the Wronskian plays a more prominent role.

The concepts and methods of solving second-order linear ODEs extend readily to linear ODEs of higher order.

This chapter follows Chap. 2 naturally, since the results of Chap. 2 can be readily extended to that of Chap. 3.

Prerequisite:Secs. 2.1, 2.2, 2.6, 2.7, 2.10.

References and Answers to Problems:App. 1 Part A, and App. 2.

and is called homogeneous. If is not identically zero, then the ODE is called nonhomogeneous. This is as in Sec. 2.1.

A solution of an nth-order (linear or nonlinear) ODE on some open interval I is a function that is defined and ntimes differentiable on Iand is such that the ODE becomes an identity if we replace the unknown function yand its derivatives by hand its corresponding derivatives.

Sections 3.1–3.2 will be devoted to homogeneous linear ODEs and Section 3.3 to nonhomogeneous linear ODEs.

Homogeneous Linear ODE: Superposition Principle, General Solution

The basic superposition or linearity principle of Sec. 2.1 extends to nth order homogeneous linear ODEs as follows.

T H E O R E M 1 Fundamental Theorem for the Homogeneous Linear ODE (2)

For a homogeneous linear ODE (2), sums and constant multiples of solutions on some open interval I are again solutions on I. (This does not hold for a nonhomogeneous or nonlinear ODE!)

The proof is a simple generalization of that in Sec. 2.1 and we leave it to the student.

Our further discussion parallels and extends that for second-order ODEs in Sec. 2.1.

So we next define a general solution of (2), which will require an extension of linear independence from 2 to nfunctions.

D E F I N I T I O N General Solution, Basis, Particular Solution

A general solutionof (2) on an open interval Iis a solution of (2) on Iof the form (3)

where is a basis(or fundamental system) of solutions of (2) on I;that is, these solutions are linearly independent on I, as defined below.

A particular solutionof (2) on Iis obtained if we assign specific values to the

nconstants in (3).

D E F I N I T I O N Linear Independence and Dependence

Consider nfunctions defined on some interval I.

These functions are called linearly independentonIif the equation (4)

implies that all are zero. These functions are called linearly dependent on Iif this equation also holds on Ifor some k₁,Á, k_nnot all zero.

k₁,Á, k_n

k₁y₁(x)⫹ Á ⫹k_ny_n(x)⫽0 on I y1(x),Á

, yn(x) c1,Á

, cn

y1,Á , yn

(c₁,Á, c_n arbitrary) y(x)⫽c₁y₁(x)⫹ Á ⫹c_ny_n(x)

y⫽h(x)

r(x)

If and only if are linearly dependent on I, we can express (at least) one of these functions on Ias a “linear combination” of the other functions, that is, as a sum of those functions, each multiplied by a constant (zero or not). This motivates the term “linearly dependent.” For instance, if (4) holds with we can divide by and express as the linear combination

Note that when these concepts reduce to those defined in Sec. 2.1.

E X A M P L E 1 Linear Dependence

Show that the functions are linearly dependent on any interval.

Solution. . This proves linear dependence on any interval.

E X A M P L E 2 Linear Independence

Show that are linearly independent on any interval, for instance, on

Solution. Equation (4) is Taking (a) (b) (c) we get

(a) (b) (c)

from Then from (c) (b). Then from (b). This proves linear independence.

A better method for testing linear independence of solutions of ODEs will soon be explained.

E X A M P L E 3 General Solution. Basis Solve the fourth-order ODE

(where ).

Solution. As in Sec. 2.2 we substitute . Omitting the common factor we obtain the characteristic equation

This is a quadratic equation in namely,

The roots are and 4. Hence This gives four solutions. A general solution on any interval is

provided those four solutions are linearly independent. This is true but will be shown later.

Initial Value Problem. Existence and Uniqueness

An initial value problemfor the ODE (2) consists of (2) and ninitial conditions

(5) ,

with given in the open interval Iconsidered, and given K₀,Á . , K_nⴚ1 x0

y^(nⴚ1)(x₀)⫽K_nⴚ1 y

r

_(x0)⫽K1, Á

y(x₀)⫽K₀,

䊏

y⫽c1e^ⴚ2x⫹c2e^ⴚx⫹c3e^x⫹c4e^2x l⫽ ⫺2, ⫺1, 1, 2.

␮⫽1

␮²⫺5␮⫹4⫽(␮⫺1)(␮⫺4)⫽0.

␮⫽l²,

l⁴⫺5l²⫹4⫽0.

e^lx, y⫽e^lx

y^iv⫽d⁴y>dx⁴ y^iv⫺5ys_⫹4y⫽0

k1⫽0 䊏

⫺2 k3⫽0

(a)⫹(b).

k2⫽0

2k₁⫹4k₂⫹8k₃⫽0.

k₁⫹k₂⫹k₃⫽0,

⫺k₁⫹k₂⫺k₃⫽0,

x⫽2, x⫽1,

x⫽ ⫺1, k1x⫹k2x²⫹k3x³⫽0.

⫺1⬉x⬉2.

y₁⫽x, y₂⫽x², y₃⫽x³

䊏

y2⫽0y1⫹2.5y3

y₁⫽x², y₂⫽5x, y₃⫽2x

n⫽2,

y₁⫽ ⫺ 1

k₁(k₂y₂⫹ Á ⫹k_ny_n).

y1

k1

k1⫽0, n⫺1 y1,Á

, yn

SEC. 3.1 Homogeneous Linear ODEs 107

In extension of the existence and uniqueness theorem in Sec. 2.6 we now have the following.

T H E O R E M 2 Existence and Uniqueness Theorem for Initial Value Problems

If the coefficients of (2) are continuous on some open interval I and is in I, then the initial value problem (2), (5)has a unique solution on I.

Existence is proved in Ref. [A11] in App. 1. Uniqueness can be proved by a slight generalization of the uniqueness proof at the beginning of App. 4.

E X A M P L E 4 Initial Value Problem for a Third-Order Euler–Cauchy Equation

Solve the following initial value problem on any open interval Ion the positive x-axis containing

Solution. Step 1. General solution.As in Sec. 2.5 we try By differentiation and substitution,

Dropping and ordering gives If we can guess the root We can divide

by and find the other roots 2 and 3, thus obtaining the solutions which are linearly independent on I(see Example 2). [In general one shall need a root-finding method, such as Newton’s (Sec. 19.2), also available in a CAS (Computer Algebra System).] Hence a general solution is

valid on any interval I, even when it includes where the coefficients of the ODE divided by (to have the standard form) are not continuous.

Step 2. Particular solution.The derivatives are and From this, and yand the initial conditions, we get by setting

(a) (b) (c)

This is solved by Cramer’s rule (Sec. 7.6), or by elimination, which is simple, as follows. gives

(d) Then (c) (d) gives Then (c) gives Finally from (a).

Answer:

Linear Independence of Solutions. Wronskian

Linear independence of solutions is crucial for obtaining general solutions. Although it can often be seen by inspection, it would be good to have a criterion for it. Now Theorem 2 in Sec. 2.6 extends from order to any n. This extended criterion uses the Wronskian Wof nsolutions defined as the nth-order determinant

(6) W(y₁,Á, y_n)⫽ 5

y₁ y₂ Á y_n

y₁

r

_y2

r

^Á _yn

r

# # Á #

y₁^(nⴚ1) y₂^(nⴚ1) Á y_n^(nⴚ1) 5 . y₁,Á, y_n

n⫽2

䊏

y⫽2x⫹x²⫺x³.

c1⫽2 c2⫽1.

c3⫽ ⫺1.

⫺2 c2⫹2c3⫽ ⫺1.

(b)⫺(a) ys_{(1)⫽ 2c2⫹6c3⫽ ⫺4.}

yr_{(1)⫽c1⫹2c2⫹3c3⫽ 1} y(1) ⫽c₁⫹ c₂⫹ c₃⫽ 2

x⫽1

ys_{⫽2c2⫹6c3x.}

yr_{⫽c1⫹2c2x⫹3c3x}²

x³ x⫽0

y⫽c1x⫹c2x²⫹c3x³

x, x², x³, m⫺1

m⫽1.

m³⫺6m²⫹11m⫺6⫽0.

x^m

m(m⫺1)(m⫺2)x^m⫺3m(m⫺1)x^m⫹6mx^m⫺6x^m⫽0.

y⫽x^m.

ys_{(1)⫽ ⫺4.}

yr_(1)⫽1, y(1)⫽2,

x³yt_⫺3x²_ys_⫹6xyr_⫺6y⫽0,

x⫽1.

y(x) x₀

p₀(x),Á, p_nⴚ1(x)

Note that W depends on xsince do. The criterion states that these solutions form a basis if and only if Wis not zero; more precisely:

T H E O R E M 3 Linear Dependence and Independence of Solutions

Let the ODE(2)have continuous coefficients on an open interval I. Then n solutions of (2)on I are linearly dependent on I if and only if their Wronskian is zero for some in I. Furthermore, if W is zero for then W is identically zero on I. Hence if there is an in I at which W is not zero, then

are linearly independent on I, so that they form a basis of solutions of(2)on I.

P R O O F (a)Let be linearly dependent solutions of (2) on I. Then, by definition, there are constants not all zero, such that for all xin I,

(7)

By differentiations of (7) we obtain for all xin I

(8)

(7), (8) is a homogeneous linear system of algebraic equations with a nontrivial solution Hence its coefficient determinant must be zero for every xon I, by Cramer’s theorem (Sec. 7.7). But that determinant is the Wronskian W, as we see from (6). Hence Wis zero for every xon I.

(b) Conversely, if W is zero at an in I, then the system (7), (8) with has a solution not all zero, by the same theorem. With these constants we define the solution of (2) on I. By (7), (8) this solution satisfies the

initial conditions But another solution satisfying the

same conditions is Hence by Theorem 2, which applies since the coefficients of (2) are continuous. Together, on I. This means linear dependence of on I.

(c)If Wis zero at an in I, we have linear dependence by (b) and then by (a).

Hence if Wis not zero at an in I, the solutions must be linearly independent on I.

E X A M P L E 5 Basis, Wronskian

We can now prove that in Example 3 we do have a basis. In evaluating W, pull out the exponential functions columnwise. In the result, subtract Column 1 from Columns 2, 3, 4 (without changing Column 1). Then expand by Row 1. In the resulting third-order determinant, subtract Column 1 from Column 2 and expand the result by Row 2:

䊏

W⫽⁶

e^ⴚ2x e^ⴚx e^x e^2x

⫺2e^ⴚ2x ⫺e^ⴚx e^x 2e^2x 4e^ⴚ2x e^ⴚx e^x 4e^2x

⫺8e^ⴚ2x ⫺e^ⴚx e^x 8e^2x

6⫽⁶

1 1 1 1

⫺2 ⫺1 1 2

4 1 1 4

⫺8 ⫺1 1 8

6⫽3

1 3 4

⫺3 ⫺3 0

7 9 16

3_⫽72.

䊏 y₁,Á, y_n

x₁

W⬅0 x₀

y₁,Á, y_n

y*⫽k₁*y₁⫹ Á ⫹k_n*y_n⬅0 y*⬅y

y⬅0.

y*(x₀)⫽0,Á, y*^(nⴚ1)(x₀)⫽0.

y*⫽k₁*y₁⫹ Á ⫹k_n*y_n k₁*,Á, k_n*,

x⫽x₀ x₀

k₁,Á, k_n.

k₁y₁^(nⴚ1)⫹ Á ⫹k_ny_n^(nⴚ1) ⫽0.

.. .

k1y1

r

_⫹ ^Á _⫹knyn

r

_⫽0

n⫺1

k1y1⫹ Á ⫹knyn⫽0.

k1,Á , kn

y1,Á , yn

y1,Á , yn

x1

x⫽x0, x⫽x0

y1,Á , yn

p0(x),Á

, p_nⴚ1(x) y1,Á

, yn

SEC. 3.1 Homogeneous Linear ODEs 109

A General Solution of (2) Includes All Solutions

Let us first show that general solutions always exist. Indeed, Theorem 3 in Sec. 2.6 extends as follows.

T H E O R E M 4 Existence of a General Solution

If the coefficients of (2)are continuous on some open interval I, then (2)has a general solution on I.

P R O O F We choose any fixed in I. By Theorem 2 the ODE (2) has nsolutions where satisfies initial conditions (5) with and all other K’s equal to zero. Their Wronskian at equals 1. For instance, when then

and the other initial values are zero. Thus, as claimed,

Hence for any nthose solutions are linearly independent on I, by Theorem 3.

They form a basis on I, and is a general solution of (2) on I.

We can now prove the basic property that, from a general solution of (2), every solution of (2) can be obtained by choosing suitable values of the arbitrary constants. Hence an nth-order linearODE has no singular solutions, that is, solutions that cannot be obtained from a general solution.

T H E O R E M 5 General Solution Includes All Solutions

If the ODE (2)has continuous coefficients on some open interval I, then every solution of (2)on I is of the form

(9)

where is a basis of solutions of (2)on I and are suitable constants.

P R O O F Let Ybe a given solution and a general solution of (2) on I. We choose any fixed in Iand show that we can find constants for which yand its first derivatives agree with Y and its corresponding derivatives at That is, we should have at

(10)

But this is a linear system of equations in the unknowns Its coefficient determinant is the Wronskian Wof y₁,Á, y_nat x₀.Since y₁,Á, y_nform a basis, they

c₁,Á, c_n. c₁y₁^(nⴚ1)⫹ Á ⫹c_ny_n^(nⴚ1) ⫽Y^(nⴚ1).

.. . c₁y₁

r

_⫹ ^Á _{⫹ c}

ny_n

r

_⫽Y

r

c₁y₁⫹ Á ⫹ c_ny_n ⫽Y x⫽x₀

x₀. n⫺1

c₁,Á, c_n x₀

y⫽c₁y₁⫹ Á ⫹c_ny_n

C₁,Á, C_n y₁,Á, y_n

Y(x)⫽C₁y₁(x)⫹ Á ⫹C_ny_n(x) y⫽Y(x)

p₀(x),Á, p_nⴚ1(x)

䊏 y⫽c₁y₁⫹ Á ⫹c_ny_n

y₁,Á, y_n W(y1(x0), y2(x0), y3(x0))⫽⁴

y1(x0) y2(x0) y3(x0) y1

r

_{(x0) y2}

r

_{(x0) y3}

r

_(x0)

y1

s

_{(x0) y2}

s

_{(x0) y3}

s

_(x0)

4⫽⁴

1 0 0

0 1 0

0 0 1

4⫽1.

y₃

s

_(x0)⫽1,

y₁(x₀)⫽1, y₂

r

_(x0)⫽1,

n⫽3, x₀

K_jⴚ1⫽1 y_j

y₁,Á, y_n, x₀

p₀(x),Á, p_nⴚ1(x)

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, e^ⴚx cos 2x, e^ⴚx sin 2x, yt_⫹2ys_⫹5yr_⫽0

e^ⴚ4x, xe^ⴚ4x, x²e^ⴚ4x, yt_⫹12ys_⫹48yr_⫹64y⫽0

cos x, sin x, x cos x, x sin x, y^iv⫹2ys_⫹y⫽0

e^x, e^ⴚx, e^2x, yt_⫺2ys_⫺yr_⫹2y⫽0

1, x, x², x³, y^iv⫽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, e^2x cos² x, sin² x, 2p

sin x, cos x, sin 2x sin² x, cos² x, cos 2x

e^x cos x, e^x sin x, e^x e^2x, xe^2x, x²e^2x

3.2 Homogeneous Linear ODEs