Nonhomogeneous 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

3.3 Nonhomogeneous Linear ODEs

An initial value problemfor (1) consists of (1) and ninitial conditions (4)

with in I. Under those continuity assumptions it has a unique solution. The ideas of proof are the same as those for in Sec. 2.7.

Method of Undetermined Coefficients

Equation (2) shows that for solving (1) we have to determine a particular solution of (1).

For a constant-coefficient equation (5)

( constant) and special as in Sec. 2.7, such a can be determined by the method of undetermined coefficients,as in Sec. 2.7, using the following rules.

(A) Basic Ruleas in Sec. 2.7.

(B) Modification Rule. If a term in your choice for is a solution of the homogeneous equation (3), then multiply this term by where k is the smallest positive integer such that this term times is not a solution of (3).

(C) Sum Ruleas in Sec. 2.7.

The practical application of the method is the same as that in Sec. 2.7. It suffices to illustrate the typical steps of solving an initial value problem and, in particular, the new Modification Rule, which includes the old Modification Rule as a particular case (with or 2). We shall see that the technicalities are the same as for except perhaps for the more involved determination of the constants.

E X A M P L E 1 Initial Value Problem. Modification Rule Solve the initial value problem

(6)

Solution. Step 1. The characteristic equation is It has the triple root Hence a general solution of the homogeneous ODE is

Step 2. If we try we get which has no solution. Try and

The Modification Rule calls for

Then

ypt_{⫽C(6⫺18x⫹9x}²_⫺x³_)e^ⴚx_. yps_⫽C(6x⫺6x²_⫹x³_)e^ⴚx_,

ypr_⫽C(3x²_⫺x³_)e^ⴚx_, yp⫽Cx³e^ⴚx. Cx²e^ⴚx.

Cxe^ⴚx

⫺C⫹3C⫺3C⫹C⫽30, y_p⫽Ce^ⴚx,

⫽(c1⫹c2x⫹c3x²)e^ⴚx. yh⫽c1e^ⴚx⫹c2xe^ⴚx⫹c3x²e^ⴚx l⫽ ⫺1.

l³⫹3l²⫹3l⫹1⫽(l⫹1)³⫽0.

yt_⫹3ys_⫹3yr_⫹y⫽30e^ⴚx_{, y(0)⫽3, y}r_{(0)⫽ ⫺3, y}s_{(0)⫽ ⫺47.}

n⫽2, k⫽1

x^k

x^k, yp(x)

y_p(x) r(x)

a₀,Á, a_nⴚ1

y⁽ⁿ⁾⫹anⴚ1y^(nⴚ1)⫹ Á ⫹a1y

r

_{⫹a0y⫽r(x)}

n⫽2 x0

y(x₀)⫽K₀, y

r

_(x0)⫽K1, ^Á_{, y}^(nⴚ1)_{(x0)⫽Knⴚ1}

SEC. 3.3 Nonhomogeneous Linear ODEs 117

Substitution of these expressions into (6) and omission of the common factor gives

The linear, quadratic, and cubic terms drop out, and Hence This gives

Step 3.We now write down the general solution of the given ODE. From it we find by the first initial condition. We insert the value, differentiate, and determine from the second initial condition, insert the value, and finally determine from and the third initial condition:

Hence the answerto our problem is (Fig. 73)

The curve of ybegins at (0, 3) with a negative slope, as expected from the initial values, and approaches zero

as ^x:⬁.The dashed curve in Fig. 74 is ^yp. 䊏

y⫽(3⫺25x²)e^ⴚx⫹5x³e^ⴚx.

ys_{⫽[3⫹2c3⫹(30⫺4c3)x⫹(⫺30⫹c3)x}²_⫹5x³_]e^ⴚx_{, y}s_{(0)⫽3⫹2c3⫽ ⫺47, c3⫽ ⫺25.}

yr_{⫽[⫺3⫹c2⫹(⫺c2⫹2c3)x⫹(15⫺c3)x}²_⫺5x³_]e^ⴚx_{, y}r_{(0)⫽ ⫺3⫹c2⫽ ⫺3, c2⫽0} y⫽yh⫹yp⫽(c1⫹c2x⫹c3x²)e^ⴚx⫹5x³e^ⴚx, y(0)⫽c1⫽3

ys₍₀₎ c₃

c₂

c₁ y⫽y_h⫹y_p,

yp⫽5x³e^ⴚx. C⫽5.

6C⫽30.

C(6⫺18x⫹9x²⫺x³)⫹3C(6x⫺6x²⫹x³)⫹3C(3x²⫺x³)⫹Cx³⫽30.

e^ⴚx

– 5 5

0 5 x

y

10

Fig. 74. yand yp(dashed) in Example 1

Method of Variation of Parameters

The method of variation of parameters (see Sec. 2.10) also extends to arbitrary order n.

It gives a particular solution for the nonhomogeneous equation (1) (in standard form with as the first term!) by the formula

(7)

on an open interval Ion which the coefficients of (1) and are continuous. In (7) the functions form a basis of the homogeneous ODE (3), with Wronskian W, and is obtained from W by replacing the jth column of W by the column

Thus, when this becomes identical with (2) in Sec. 2.10, W⫽ `y₁ y₂

y₁

r

_y

2

r

^{`, W1⫽ `}

0 y₂

1 y₂

r

^{` ⫽ ⫺y2, W2⫽ `}

y₁ 0 y₁

r

₁^{` ⫽y1.}

n⫽2, [0 0 Á 0 1]^T.

W_j ( j⫽1,Á, n) y₁,Á, y_n

r(x)

⫽y₁(x)

冮

^{WW(x)1(x) r(x) dx⫹ Á ⫹yn(x)}

冮

^{WW(x)n(x) r(x) dx}

y_p(x)⫽ a

n

k⫽1 y_k(x)

冮

^{WW(x)k(x) r(x) dx}

y⁽ⁿ⁾

y_p

The proof of (7) uses an extension of the idea of the proof of (2) in Sec. 2.10 and can be found in Ref [A11] listed in App. 1.

E X A M P L E 2 Variation of Parameters. Nonhomogeneous Euler–Cauchy Equation Solve the nonhomogeneous Euler–Cauchy equation

Solution. Step 1. General solution of the homogeneous ODE.Substitution of and the derivatives into the homogeneous ODE and deletion of the factor gives

The roots are 1, 2, 3 and give as a basis

Hence the corresponding general solution of the homogeneous ODE is

Step 2. Determinants needed in(7).These are

Step 3. Integration.In (7) we also need the right side of our ODE in standard form, obtained by division of the given equation by the coefficient of thus, In (7) we have the simple

quotients Hence (7) becomes

Simplification gives Hence the answer is

Figure 75 shows Can you explain the shape of this curve? Its behavior near The occurrence of a minimum?

Its rapid increase? Why would the method of undetermined coefficients not have given the solution?^{yp. x⫽0?} 䊏

y⫽y_h⫹y_p⫽c₁x⫹c₂x²⫹c₃x³⫹¹6 x⁴ (ln x⫺¹¹6).

yp⫽¹6 x⁴ (ln x⫺¹¹6).

⫽x 2ax³

3 ln x⫺x³

9b⫺x² ax² 2 ln x⫺x²

4b⫹x³

2 (x ln x⫺x).

y_p⫽x

冮

^x2 x ln x dx⫺x²

冮

^{x ln x dx⫹x3}

冮

2x¹ x ln x dx W₁>W⫽x>2, W₂>W⫽ ⫺1, W₃>W⫽1>(2x).

r(x)⫽(x⁴ ln x)>x³⫽x ln x.

yt_; x³

r(x) W3⫽⁴

x x² 0

1 2x 0

0 2 1

4⫽x².

W2⫽⁴

x 0 x³ 1 0 3x²

0 1 6x

4⫽ ⫺2x³

W1⫽⁴

0 x² x³ 0 2x 3x²

1 2 6x

4⫽x⁴

W⫽3

x x² x³ 1 2x 3x²

0 2 6x

3_⫽2x³

yh⫽c1x⫹c2x²⫹c3x³. y1⫽x, y2⫽x², y3⫽x³. m(m⫺1)(m⫺2)⫺3m(m⫺1)⫹6m⫺6⫽0.

x^m

y⫽x^m

(x⬎0).

x³yt_⫺3x²_ys_⫹6xyr_⫺6y⫽x⁴ _{ln x}

SEC. 3.3 Nonhomogeneous Linear ODEs 119

Application: Elastic Beams

Whereas second-order ODEs have various applications, of which we have discussed some of the more important ones, higher order ODEs have much fewer engineering applications.

An important fourth-order ODE governs the bending of elastic beams, such as wooden or iron girders in a building or a bridge.

A related application of vibration of beams does not fit in here since it leads to PDEs and will therefore be discussed in Sec. 12.3.

E X A M P L E 3 Bending of an Elastic Beam under a Load

We consider a beam Bof length Land constant (e.g., rectangular) cross section and homogeneous elastic material (e.g., steel); see Fig. 76. We assume that under its own weight the beam is bent so little that it is practically straight. If we apply a load to Bin a vertical plane through the axis of symmetry (the x-axis in Fig. 76), Bis bent. Its axis is curved into the so-called elastic curveC(or deflection curve). It is shown in elasticity theory that the bending moment is proportional to the curvature of C. We assume the bending to be small, so that the deflection and its derivative (determining the tangent direction of C) are small.

Then, by calculus, Hence

EIis the constant of proportionality. Eis Young’s modulus of elasticityof the material of the beam. Iis the moment of inertia of the cross section about the (horizontal) z-axis in Fig. 76.

Elasticity theory shows further that where is the load per unit length. Together,

(8) EIy^iv⫽f(x).

f(x) Ms_(x)⫽f(x),

M(x)⫽EIys_(x).

k⫽ys_>(1⫹yr²₎^3>2_⬇ys_.

yr_(x) y(x)

k(x) M(x)

– 20 20

5 x

y 30

10 –10

10 0

Fig. 75. Particular solution of the nonhomogeneous Euler–Cauchy equation in Example 2

yp

L Undeformed beam

Deformed beam under uniform load (simply supported) x

y z

x

y z

Fig. 76. Elastic beam

In applications the most important supports and corresponding boundary conditions are as follows and shown in Fig. 77.

(A) Simply supported at and L

(B) Clamped at both ends at and L

(C) Clamped at , free at

The boundary condition means no displacement at that point, means a horizontal tangent, means no bending moment, and means no shear force.

Let us apply this to the uniformly loaded simply supported beam in Fig. 76. The load is Then (8) is

(9)

This can be solved simply by calculus. Two integrations give

gives Then (since ). Hence

Integrating this twice, we obtain

with from Then

Inserting the expression for k, we obtain as our solution

Since the boundary conditions at both ends are the same, we expect the deflection to be “symmetric” with respect to that is, Verify this directly or set and show that ybecomes an even function of u,

From this we can see that the maximum deflection in the middle at is Recall

that the positive direction points downward. ^5f0L 䊏

4>(16#_24EI).

u⫽0(x⫽L>2) y⫽ f₀

24EIau²⫺1

4 L²bau²⫺5 4 L²b. x⫽u⫹L>2 y(x)⫽y(L⫺x).

L>2,

y(x) y⫽ f0

24EI (x⁴⫺2Lx³⫹L³x).

y(L)⫽kL 2 aL³

12⫺L³

6 ⫹c3b⫽0, c3⫽L³ 12. y(0)⫽0.

c₄⫽0

y⫽k 2a1

12 x⁴⫺L

6 x³⫹c₃x⫹c₄b ys_⫽^k

2 (x²⫺Lx).

L⫽0 ys_(L)⫽L(¹_{2kL⫹c1)⫽0, c1⫽ ⫺kL>2}

c₂⫽0.

ys_(0)⫽0

ys_⫽^k

2 x²⫹c₁x⫹c₂. y^iv⫽k, k⫽ f₀

EI.

f(x)⬅f₀⫽const.

yt_⫽0

ys_⫽0 yr_⫽0

y⫽0

y(0)⫽yr_{(0)⫽0, y}s_(L)⫽yt_(L)⫽0.

x⫽L x⫽0

x⫽0 y⫽yr_⫽0

x⫽0 y⫽ys_⫽0

SEC. 3.3 Nonhomogeneous Linear ODEs 121

x

x = 0 x = L

x = 0 x = L

x = 0 x = L

(A) Simply supported

(B) Clamped at both ends

(C) Clamped at the left end, free at the right end

Fig. 77. Supports of a beam

1–7 GENERAL SOLUTION

Solve the following ODEs, showing the details of your work.

1.

2.

3.

4.

5.

6.

7.

8–13 INITIAL VALUE PROBLEM

Solve the given IVP, showing the details of your work.

8.

9.

10.

11.

12.

ys_(0)⫽17.2

yr_(0)⫽8.8,

y(0)⫽4.5, (D³⫺2D²⫺9D⫹18I)y⫽e^2x,

ys_{(0)⫽ ⫺5.2}

yr_(0)⫽3.2,

y(0)⫽ ⫺1.4, (D³⫺2D²⫺3D)y⫽74e^ⴚ3x sin x,

ys_(1)⫽14

yr_(1)⫽3,

y(1)⫽1, x³yt_⫹xyr_⫺y⫽x²_,

yt_{(0)⫽ ⫺32}

ys_{(0)⫽ ⫺1,}

yr_(0)⫽2,

y(0)⫽1, y^iv⫹5ys_{⫹4y⫽90 sin 4x,}

yt_(0)⫽0

ys_(0)⫽0,

yr_(0)⫽0,

y(0)⫽1, y^iv⫺5ys_⫹4y⫽10e^ⴚ3x_,

(D³⫺9D²⫹27D⫺27I)y⫽27 sin 3x (D³⫹4D)y⫽sin x

(x³D³⫹x²D²⫺2xD⫹2I)y⫽x^ⴚ2 (D³⫹3D²⫺5D⫺39I)y⫽ ⫺300 cos x (D⁴⫹10D²⫹9I)y⫽6.5 sinh 2x yt_⫹2ys_⫺yr_{⫺2y⫽1⫺4x}³

yt_⫹3ys_⫹3yr_⫹y⫽e^x_⫺x⫺1

P R O B L E M S E T 3 . 3

13.

14. CAS EXPERIMENT. Undetermined Coefficients.

Since variation of parameters is generally complicated, it seems worthwhile to try to extend the other method.

Find out experimentally for what ODEs this is possible and for what not. Hint:Work backward, solving ODEs with a CAS and then looking whether the solution could be obtained by undetermined coefficients. For example, consider

and

15. WRITING REPORT. Comparison of Methods. Write a report on the method of undetermined coefficients and the method of variation of parameters, discussing and comparing the advantages and disadvantages of each method. Illustrate your findings with typical examples.

Try to show that the method of undetermined coefficients, say, for a third-order ODE with constant coefficients and an exponential function on the right, can be derived from the method of variation of parameters.

x³yt_⫹x²_ys_⫺2xyr_⫹2y⫽x³ _{ln x.}

yt_⫺3ys_⫹3yr_⫺y⫽x^1>2_e^x

ys_{(0)⫽ ⫺1}

yr_{(0)⫽ ⫺2,}

y(0)⫽3, (D³⫺4D)y⫽10 cos x⫹5 sin x,

1. What is the superposition or linearity principle? For what nth-order ODEs does it hold?

2. List some other basic theorems that extend from second-order to nth-order ODEs.

3. If you know a general solution of a homogeneous linear ODE, what do you need to obtain from it a general solution of a corresponding nonhomogeneous linear ODE?

4. What form does an initial value problem for an nth- order linear ODE have?

5. What is the Wronskian? What is it used for?

6–15 GENERAL SOLUTION

Solve the given ODE. Show the details of your work.

6.

7.

8.

9.

10. x²yt_⫹3xys_⫺2yr_⫽0

(D⁴⫺16I)y⫽ ⫺15 cosh x yt_⫺4ys_⫺yr_⫹4y⫽30e^2x

yt_⫹4ys_⫹13yr_⫽0

y^iv⫺3ys_⫺4y⫽0

C H A P T E R 3 R E V I E W Q U E S T I O N S A N D P R O B L E M S

11.

12.

13.

14.

15.

16–20 INITIAL VALUE PROBLEM Solve the IVP. Show the details of your work.

16.

17.

18.

19.

20.

ys_(0)⫽5

yr_{(0)⫽ ⫺3,}

y(0)⫽ ⫺1, (D³⫹3D²⫹3D⫹I)y⫽8 sin x,

D²y(0)⫽189 Dy(0)⫽ ⫺41,

y(0)⫽9,

(D³⫹9D²⫹23D⫹15I)y⫽12exp(⫺4x), D³y(0)⫽ ⫺130 D²y(0)⫽34,

Dy(0)⫽ ⫺6,

y(0)⫽12.16, (D⁴⫺26D²⫹25I)y⫽50(x⫹1)²,

ys_{⫽ ⫺24}

yr_{(0)⫽ ⫺3.95,}

y(0)⫽1.94, yt_⫹5ys_⫹24yr_⫹20y⫽x,

D²y(0)⫽0

Dy(0)⫽1, y(0)⫽0,

(D³⫺D²⫺D⫹I)y⫽0, 4x³yt_⫹3xyr_⫺3y⫽10

(D⁴⫺13D²⫹36I)y⫽12e^x (D³⫹6D²⫹12D⫹8I)y⫽8x² (D³⫺D)y⫽sinh 0.8x

yt_⫹4.5ys_⫹6.75yr_{⫹3.375y⫽0}

Summary of Chapter 3 123

Compare with the similar Summary of Chap. 2 (the case ).

Chapter 3 extends Chap. 2 from order to arbitrary order n. An nth-order linear ODEis an ODE that can be written

(1)

with as the first term; we again call this the standard form. Equation (1) is called homogeneous if on a given open interval I considered, nonhomogeneousif on I. For the homogeneous ODE

(2)

the superposition principle (Sec. 3.1) holds, just as in the case A basisor fundamental system of solutions of (2) on I consists of n linearly independent solutions of (2) on I. A general solutionof (2) on Iis a linear combination of these,

(3) ( arbitrary constants).

A general solutionof the nonhomogeneous ODE (1) on Iis of the form

(4) (Sec. 3.3).

Here, is a particular solution of (1) and is obtained by two methods (undetermined coefficientsor variation of parameters) explained in Sec. 3.3.

An initial value problem for (1) or (2) consists of one of these ODEs and n initial conditions (Secs. 3.1, 3.3)

(5)

with given in Iand given If are continuous on I,

then general solutions of (1) and (2) on Iexist, and initial value problems (1), (5) or (2), (5) have a unique solution.

p₀,Á, p_nⴚ1, r K₀,Á, K_nⴚ1.

x₀

y(x₀)⫽K₀, y

r

_(x0)⫽K1, ^Á_{, y}^(nⴚ1)_{(x0)⫽Knⴚ1}

y_p

y⫽y_h⫹y_p

c₁,Á, c_n y⫽c₁y₁⫹ Á ⫹c_ny_n

y₁,Á, y_n

n⫽2.

y⁽ⁿ⁾⫹p_nⴚ1(x)y^(nⴚ1)⫹ Á ⫹p₁(x)y

r

_{⫹p0(x)y⫽0}

r(x) [ 0

r(x)⬅0 y⁽ⁿ⁾⫽dⁿy>dxⁿ

y⁽ⁿ⁾⫹p_nⴚ1(x)y^(nⴚ1)⫹ Á ⫹p₁(x)y

r

_{⫹p0(x)y⫽r(x)}

n⫽2

nⴝ2

S U M M A R Y O F C H A P T E R 3

Higher Order Linear ODEs

124

C H A P T E R 4

Systems of ODEs. Phase Plane.

Qualitative Methods

Tying in with Chap. 3, we present another method of solving higher order ODEs in Sec. 4.1. This converts any nth-order ODE into a system of nfirst-order ODEs. We also show some applications. Moreover, in the same section we solve systems of first-order ODEs that occur directly in applications, that is, not derived from an nth-order ODE but dictated by the application such as two tanks in mixing problems and two circuits in electrical networks. (The elementary aspects of vectors and matrices needed in this chapter are reviewed in Sec. 4.0 and are probably familiar to most students.)

In Sec. 4.3 we introduce a totally different way of looking at systems of ODEs. The method consists of examining the general behavior of whole families of solutions of ODEs in the phase plane, and aptly is called the phase plane method. It gives information on the stability of solutions. (Stability of a physical systemis desirable and means roughly that a small change at some instant causes only a small change in the behavior of the system at later times.) This approach to systems of ODEs is a qualitative methodbecause it depends only on the nature of the ODEs and does not require the actual solutions. This can be very useful because it is often difficult or even impossible to solve systems of ODEs. In contrast, the approach of actually solving a system is known as a quantitative method.

The phase plane method has many applications in control theory, circuit theory, population dynamics and so on. Its use in linear systems is discussed in Secs. 4.3, 4.4, and 4.6 and its even more important use in nonlinear systems is discussed in Sec. 4.5 with applications to the pendulum equation and the Lokta–Volterra population model. The chapter closes with a discussion of nonhomogeneous linear systems of ODEs.

N O T A T I O N . We continue to denote unknown functions by y; thus, — analogous to Chaps. 1–3. (Note that some authors use xfor functions, when dealing with systems of ODEs.)

Prerequisite: Chap. 2.

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