Homogeneous Linear ODEs of Second Order

46

C H A P T E R 2

Second-Order Linear ODEs

Many important applications in mechanical and electrical engineering, as shown in Secs.

2.4, 2.8, and 2.9, are modeled by linear ordinary differential equations (linear ODEs) of the second order. Their theory is representative of all linear ODEs as is seen when compared to linear ODEs of third and higher order, respectively. However, the solution formulas for second-order linear ODEs are simpler than those of higher order, so it is a natural progression to study ODEs of second order first in this chapter and then of higher order in Chap. 3.

Although ordinary differential equations (ODEs) can be grouped into linear and nonlinear ODEs, nonlinear ODEs are difficult to solve in contrast to linear ODEs for which many beautiful standard methods exist.

Chapter 2 includes the derivation of general and particular solutions, the latter in connection with initial value problems.

For those interested in solution methods for Legendre’s, Bessel’s, and the hypergeometric equations consult Chap. 5 and for Sturm–Liouville problems Chap. 11.

C O M M E N T . Numerics for second-order ODEs can be studied immediately after this chapter.See Sec. 21.3, which is independent of other sections in Chaps. 19–21.

Prerequisite:Chap. 1, in particular, Sec. 1.5.

Sections that may be omitted in a shorter course:2.3, 2.9, 2.10.

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

The definitions of homogeneous and nonhomogenous second-order linear ODEs are very similar to those of first-order ODEs discussed in Sec. 1.5. Indeed, if (that is, for all x considered; read “ is identically zero”), then (1) reduces to (2)

and is called homogeneous. If then (1) is called nonhomogeneous. This is similar to Sec. 1.5.

An example of a nonhomogeneous linear ODE is

and a homogeneous linear ODE is

written in standard form .

Finally, an example of a nonlinear ODE is

.

The functions pand qin (1) and (2) are called the coefficientsof the ODEs.

Solutionsare defined similarly as for first-order ODEs in Chap. 1. A function

is called a solutionof a (linear or nonlinear) second-order ODE on some open interval I if his defined and twice differentiable throughout that interval and is such that the ODE becomes an identity if we replace the unknown yby h, the derivative by , and the second derivative by . Examples are given below.

Homogeneous Linear ODEs: Superposition Principle

Sections 2.1–2.6 will be devoted to homogeneous linear ODEs (2) and the remaining sections of the chapter to nonhomogeneous linear ODEs.

Linear ODEs have a rich solution structure. For the homogeneous equation the backbone of this structure is the superposition principleor linearity principle,which says that we can obtain further solutions from given ones by adding them or by multiplying them with any constants. Of course, this is a great advantage of homogeneous linear ODEs. Let us first discuss an example.

E X A M P L E 1 Homogeneous Linear ODEs: Superposition of Solutions

The functions and are solutions of the homogeneous linear ODE

for all x.We verify this by differentiation and substitution. We obtain ; hence ys_{y(cos x)}s_{cos x cos xcos x0.}

(cos x)s _{cos x} ys_y0

ysin x ycos x

h

s

y

s

h

r

y

r

yh(x) y

s

_yy

r

²₀

y

s

¹

x y

r

_y0

xy

s

_y

r

_xy0,

y

s

_25ye^ⴚx _{cos x,}

r(x) [ 0,

y

s

_p(x)y

r

_q(x)y0

r(x) r(x)0

r(x)⬅0

SEC. 2.1 Homogeneous Linear ODEs of Second Order 47

Similarly for (verify!). We can go an important step further. We multiply by any constant, for instance, 4.7, and by, say, , and take the sum of the results, claiming that it is a solution. Indeed, differentiation and substitution gives

In this example we have obtained from and a function of the form

(3) ( arbitrary constants).

This is called a linear combinationof and . In terms of this concept we can now formulate the result suggested by our example, often called the superposition principle or linearity principle.

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

For a homogeneous linear ODE (2),any linear combination of two solutions on an open interval I is again a solution of (2)on I.In particular, for such an equation, sums and constant multiples of solutions are again solutions.

P R O O F Let and be solutions of (2) on I. Then by substituting and its derivatives into (2), and using the familiar rule , etc., we get

since in the last line, because and are solutions, by assumption. This shows that yis a solution of (2) on I.

CAUTION! Don’t forget that this highly important theorem holds for homogeneous linear ODEs only but does not holdfor nonhomogeneous linear or nonlinear ODEs, as the following two examples illustrate.

E X A M P L E 2 A Nonhomogeneous Linear ODE

Verify by substitution that the functions and are solutions of the nonhomogeneous linear ODE

but their sum is not a solution. Neither is, for instance, or . E X A M P L E 3 A Nonlinear ODE

Verify by substitution that the functions and are solutions of the nonlinear ODE

but their sum is not a solution. Neither is ^x2, so you cannot even multiply by ¹! 䊏

ys_yxyr_0, y1 yx²

䊏

5(1sin x) 2(1cos x)

ys_y1,

y1sin x y1cos x

䊏 y2

y1

(Á )0

c1(y

s

_1py1

r

_qy1)c2(y2

s

_py

r

_2qy2)0,

c1y₁

s

_c2y

s

_2p(c1y1

r

_c2y2

r

_)q(c1y1c2y2)

y

s

_py

r

_qy(c1y1c2y2)

s

_p(c1y1c2y2)

r

_q(c1y1c2y2)

(c1y1c2y2)

r

_c1y1

r

_c2y

r

₂

yc1y1c2y2

y2

y1

y2

y1

c1, c2

yc1y1c2y2

y2 ( sin x) y1 ( cos x)

䊏

(4.7 cos x2 sin x)s_{(4.7 cos x2 sin x) 4.7 cos x2 sin x4.7 cos x2 sin x0.}

2 sin x

cos x ysin x

Initial Value Problem. Basis. General Solution

Recall from Chap. 1 that for a first-order ODE, an initial value problemconsists of the ODE and one initial condition . The initial condition is used to determine the arbitrary constant cin the general solutionof the ODE. This results in a unique solution, as we need it in most applications. That solution is called a particular solution of the ODE. These ideas extend to second-order ODEs as follows.

For a second-order homogeneous linear ODE (2) an initial value problemconsists of (2) and two initial conditions

(4)

These conditions prescribe given values and of the solution and its first derivative (the slope of its curve) at the same given in the open interval considered.

The conditions (4) are used to determine the two arbitrary constants and in a general solution

(5)

of the ODE; here, and are suitable solutions of the ODE, with “suitable” to be explained after the next example. This results in a unique solution, passing through the point with as the tangent direction (the slope) at that point. That solution is called a particular solutionof the ODE (2).

E X A M P L E 4 Initial Value Problem Solve the initial value problem

Solution. Step 1. General solution.The functions and are solutions of the ODE (by Example 1), and we take

This will turn out to be a general solution as defined below.

Step 2. Particular solution.We need the derivative . From this and the initial values we obtain, since and ,

This gives as the solution of our initial value problem the particular solution

Figure 29 shows that at it has the value 3.0 and the slope , so that its tangent intersects the x-axis at . (The scales on the axes differ!)

Observation. Our choice of and was general enough to satisfy both initial conditions. Now let us take instead two proportional solutions and

so that . Then we can write in the form

. yc1 cos xc2(k cos x)C cos x where Cc1c2k

yc1y1c2y2

y1/y21/kconst

y2k cos x, y1cos x

y2

y1

䊏

x3.0>0.56.0

0.5 x0

y3.0 cos x0.5 sin x.

y(0)c₁3.0 and yr_{(0)c2 0.5.}

sin 00 cos 01

yr _{c1 sin xc2 cos x} yc1 cos xc2 sin x.

sin x cos x

ys_{y0, y(0)3.0, y}r_{(0) 0.5.}

K₁ (x₀, K₀)

y₂ y₁

yc₁y₁c₂y₂

c₂ c₁ xx₀

K₁ K₀ y(x₀)K₀, y

r

_(x

0)K₁. y(x₀)y₀

SEC. 2.1 Homogeneous Linear ODEs of Second Order 49

2 4 6 8 10 x

–3 –2 –1 0 1 2 3 y

Fig. 29. Particular solution and initial tangent in

Example 4

Hence we are no longer able to satisfy two initial conditions with only one arbitrary constant C. Consequently, in defining the concept of a general solution, we must exclude proportionality. And we see at the same time why the concept of a general solution is of importance in connection with initial value problems.

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

A general solutionof an ODE (2) on an open interval Iis a solution (5) in which and are solutions of (2) on Ithat are not proportional, and and are arbitrary constants. These , are called a basis(or a fundamental system) of solutions of (2) on I.

A particular solutionof (2) on I is obtained if we assign specific values to and in (5).

For the definition of an intervalsee Sec. 1.1. Furthermore, as usual, and are called proportionalon Iif for all xon I,

(6) (a) or (b)

where kand lare numbers, zero or not. (Note that (a) implies (b) if and only if ).

Actually, we can reformulate our definition of a basis by using a concept of general importance. Namely, two functions and are called linearly independent on an interval Iwhere they are defined if

(7) everywhere on Iimplies .

And and are called linearly dependenton Iif (7) also holds for some constants , not both zero. Then, if , we can divide and see that and are proportional,

or

In contrast, in the case of linear independencethese functions are not proportional because then we cannot divide in (7). This gives the following

D E F I N I T I O N Basis (Reformulated)

A basisof solutions of (2) on an open interval Iis a pair of linearly independent solutions of (2) on I.

If the coefficients pand qof (2) are continuous on some open interval I, then (2) has a general solution. It yields the unique solution of any initial value problem (2), (4). It includes all solutions of (2) on I; hence (2) has no singular solutions (solutions not obtainable from of a general solution; see also Problem Set 1.1). All this will be shown in Sec. 2.6.

y2

k1

k2

y1. y1

k2

k1

y2

y₂ y₁ k₁0 or k₂0

k₂

k₁ y₂

y₁

k₁0 and k₂0 k₁y₁(x) k₂y₂(x)0

y₂ y₁

k0 y₂ly₁

y₁ky₂

y₂ y₁ c2

c1

y2

y1

c2

c1

y2

y1

E X A M P L E 5 Basis, General Solution, Particular Solution

and in Example 4 form a basis of solutions of the ODE for all x because their

quotient is (or ). Hence is a general solution. The solution

of the initial value problem is a particular solution.

E X A M P L E 6 Basis, General Solution, Particular Solution

Verify by substitution that and are solutions of the ODE . Then solve the initial value problem

.

Solution. and show that and are solutions. They are not proportional, .Hence , form a basis for all x. We now write down the corresponding general solution and its derivative and equate their values at 0 to the given initial conditions,

. By addition and subtraction, , so that the answeris . This is the particular solution satisfying the two initial conditions.

Find a Basis if One Solution Is Known.

Reduction of Order

It happens quite often that one solution can be found by inspection or in some other way.

Then a second linearly independent solution can be obtained by solving a first-order ODE.

This is called the method of reduction of order.¹We first show how this method works in an example and then in general.

E X A M P L E 7 Reduction of Order if a Solution Is Known. Basis Find a basis of solutions of the ODE

.

Solution. Inspection shows that is a solution because and , so that the first term vanishes identically and the second and third terms cancel. The idea of the method is to substitute

into the ODE. This gives

uxand –xucancel and we are left with the following ODE, which we divide by x, order, and simplify, ,

This ODE is of first order in , namely, . Separation of variables and integration gives

, ln ƒvƒln ƒx1ƒ2 ln ƒxƒln ƒx1ƒ. x² dv

v

x2

x²x dxa 1 x12

xb dx

(x²x)vr_(x2)v0 vur

(x²x)us_(x2)ur_0.

(x²x)(us_x2ur_)x²_ur₀

(x²x)(us_x2ur_)x(ur_xu)ux0.

yuy1ux, yr_ur_{xu, y}s_us_x2ur ys₁₀ yr₁₁

y₁x

(x²x)ys_xyr_y0

y2e^x4e^ⴚx 䊏

c₁2, c₂4

y c1e^xc2e^ⴚx, yr_c1e^x _c2e^ⴚx_{, y(0)c1c26, y}r_{(0)c1c2 2} e^ⴚx

e^x e^x/e^ⴚx e^2xconst

e^ⴚx e^x (e^ⴚx)s _e^ⴚx₀

(e^x)s _e^x₀

ys_{y0, y(0)6, y}r_{(0) 2}

ys_y0 y₂e^ⴚx

y₁e^x

䊏

y3.0 cos x0.5 sin x

yc₁ cos xc₂ sin x tan xconst

cot xconst

ys_y0 sin x

cos x

SEC. 2.1 Homogeneous Linear ODEs of Second Order 51

1Credited to the great mathematician JOSEPH LOUIS LAGRANGE (1736–1813), who was born in Turin, of French extraction, got his first professorship when he was 19 (at the Military Academy of Turin), became director of the mathematical section of the Berlin Academy in 1766, and moved to Paris in 1787. His important major work was in the calculus of variations, celestial mechanics, general mechanics (Mécanique analytique, Paris, 1788), differential equations, approximation theory, algebra, and number theory.

We need no constant of integration because we want to obtain a particular solution; similarly in the next integration. Taking exponents and integrating again, we obtain

, , hence .

Since are linearly independent (their quotient is not constant), we have obtained a basis of solutions, valid for all positive x.

In this example we applied reduction of orderto a homogeneous linear ODE [see (2)]

.

Note that we now take the ODE in standard form, with not —this is essential in applying our subsequent formulas. We assume a solution of (2), on an open interval I, to be known and want to find a basis. For this we need a second linearly independent solution of (2) on I. To get , we substitute

, ,

into (2). This gives (8)

Collecting terms in and u, we have

.

Now comes the main point. Since is a solution of (2), the expression in the last parentheses is zero. Hence uis gone, and we are left with an ODE in and . We divide this remaining ODE by and set

, thus .

This is the desired first-order ODE, the reduced ODE. Separation of variables and integration gives

and .

By taking exponents we finally obtain

(9) .

Here so that . Hence the desired second solution is .

The quotient cannot be constant , so that and form

a basis of solutions.

y₂ y₁ (since U0)

y₂/y₁u 兰U dx

y₂y₁uy₁

冮

^{U dx}

u 兰U dx Uu

r

_,

U 1

y²₁ e^{ⴚ兰p dx}

ln ƒUƒ 2 ln ƒy1ƒ

冮

^{p dx}

dU

U a2y₁

r

y₁ pb dx

U

r

_a^2y₁

r

y₁ pb U0 u

s

_u

r

^2y₁

r

_py1

y₁ 0

u

r

_{U, u}

s

_U

r

_,

y₁

u

s

u

r

y₁

u

s

_y1u

r

_(2y1

r

_py1)u(y1

s

_py1

r

_qy1)0

u

s

_{, u}

r

_,

u

s

_y12u

r

_y1

r

_uy

s

_1p(u

r

_y1uy1

r

_)quy10.

y

s

_y2

s

_u

s

_y12u

r

_y1

r

_uy

s

₁

y

r

_y2

r

_u

r

_y1uy1

r

yy₂uy₁

y₂ y₂

y₁

f(x)y

s

y

s

_,

y

s

_p(x)y

r

_q(x)y0

y1x and y2x ln ƒxƒ1 䊏

y₂uxx ln ƒxƒ1 u

冮

^{v dxln ƒxƒ1}x

vx1 x² 1

x 1 x²

SEC. 2.2 Homogeneous Linear ODEs with Constant Coefficients 53

REDUCTION OF ORDER is important because it gives a simpler ODE. A general second-order ODE , linear or not, can be reduced to first order if ydoes not occur explicitly (Prob. 1) or if xdoes not occur explicitly (Prob. 2) or if the ODE is homogeneous linear and we know a solution (see the text).

1. Reduction. Show that can be

reduced to first order in (from which yfollows by integration). Give two examples of your own.

2. Reduction. Show that can be

reduced to a first-order ODE with yas the independent variable and , where derive this by the chain rule. Give two examples.

3–10 REDUCTION OF ORDER

Reduce to first order and solve, showing each step in detail.

3.

4.

5.

6. ,

7.

8.

9.

10.

11–14 APPLICATIONS OF REDUCIBLE ODEs 11. Curve. Find the curve through the origin in the

xy-plane which satisfies and whose tangent at the origin has slope 1.

12. Hanging cable. It can be shown that the curve of an inextensible flexible homogeneous cable hanging between two fixed points is obtained by solving y(x) ys_⫽2yr

ys_{⫹(1⫹1/y)y}r²_⫽0

x²ys_⫺5xyr_{⫹9y⫽0, y1⫽x}³

ys_⫽1⫹yr²

ys_⫹yr³ _{sin y⫽0}

y₁⫽(cos x)/x xys_⫹2yr_⫹xy⫽0

yys_⫽3yr²

2xys_⫽3yr

ys_⫹yr_⫽0

z⫽yr_;

ys_⫽(dz/dy)z

F(y, yr_{, y}s_)⫽0

z⫽yr

F(x, yr_{, y}s_)⫽0

F(x, y, yr_{, y}s_)⫽0

, where the constant kdepends on the weight. This curve is called catenary (from Latin catena = the chain). Find and graph , assuming that and those fixed points are and in a vertical xy-plane.

13. Motion. If, in the motion of a small body on a straight line, the sum of velocity and acceleration equals a positive constant, how will the distance depend on the initial velocity and position?

14. Motion. In a straight-line motion, let the velocity be the reciprocal of the acceleration. Find the distance for arbitrary initial position and velocity.

15–19 GENERAL SOLUTION. INITIAL VALUE PROBLEM (IVP)

(More in the next set.) (a)Verify that the given functions are linearly independent and form a basis of solutions of the given ODE. (b) Solve the IVP. Graph or sketch the solution.

15.

16.

17.

18.

19.

20. CAS PROJECT. Linear Independence. Write a program for testing linear independence and depen- dence. Try it out on some of the problems in this and the next problem set and on examples of your own.

e^ⴚx sin x e^ⴚx cos x,

ys _⫹2yr_{⫹2y⫽0, y(0)⫽0, y}r_(0)⫽15,

x, x ln x

x²ys _⫺xyr_{⫹y⫽0, y(1)⫽4.3, y}r_(1)⫽0.5,

x^3>2, x^ⴚ1>2

4x²ys_{⫺3y⫽0, y(1) ⫽ ⫺3, y}r_(1)⫽0,

e^ⴚ0.3x, xe^ⴚ0.3x

yr_(0)⫽0.14,

ys _⫹0.6yr_{⫹0.09y⫽0, y(0)⫽2.2,}

cos 2.5x, sin 2.5x

4ys_{⫹25y⫽0, y(0)⫽3.0, y}r_{(0)⫽ ⫺2.5,}

y(t) y(t)

(1, 0) (⫺1, 0)

k⫽1

y(x) ys _⫽k21⫹yr²

P R O B L E M S E T 2 . 1

2.2 Homogeneous Linear ODEs