Higher-Order Differential Equations CHAPTER 3 Higher-Order Differential Equations

Contents 3.1 Preliminary Theory: Linear Equations 3.2 Reduction of Order 3.3 Homogeneous Linear Equations with Constants Coefficients 3.4 Undetermined Coefficients 3.5 Variation of Parameters 3.6 Cauchy-Euler Equations 3.7 Nonlinear Equations 3.8 Linear Models: Initial-Value Problems 3.9 Linear Models: Boundary-Value Problems 3.10 Nonlinear Models 3.11 Solving Models of Linear Equations

3.1 Preliminary Theory: Linear Equ. Initial-value Problem An nth-order initial problem is Solve: Subject to: (1) with n initial conditions.

Let an(x), an-1(x), …, a0(x), and g(x) be continuous on I, an(x)  0 for all x on I. If x = x0 is any point in this interval, then a solution y(x) of (1) exists on the interval and is unique. THEOREM 3.1 Existence and Uniqueness

Example 1 The problem possesses the trivial solution y = 0. Since this DE with constant coefficients, from Theorem 3.1, hence y = 0 is the only one solution on any interval containing x = 1.

Example 2 Please verify y = 3e2x + e–2x – 3x, is a solution of This DE is linear and the coefficients and g(x) are all continuous, and a2(x)  0 on any I containing x = 0. This DE has an unique solution on I.

Boundary-Value Problem Solve: Subject to: is called a boundary-value problem (BVP). See Fig 3.1.

Fig 3.1

Example 3 In example 4 of Sec 1.1, we see the solution of is x = c1 cos 4t + c2 sin 4t (2) (a) Suppose x(0) = 0, then c1 = 0, x(t) = c2 sin 4t Furthermore, x(/2) = 0, we obtain 0 = 0, hence (3) has infinite many solutions. See Fig 3.2. (b) If (4) we have c1 = 0, c2 = 0, x = 0 is the only solution.

Example 3 (2) (c) If (5) we have c1 = 0, and 1 = 0 (contradiction). Hence (5) has no solutions.

Fig 3.2

The following DE. (6) is said to be homogeneous; The following DE (6) is said to be homogeneous; (7) with g(x) not identically zero, is nonhomogeneous.

Differential Operators Let dy/dx = Dy. This symbol D is called a differential operator. We define an nth-order differential operator as (8) In addition, we have (9) so the differential operator L is a linear operator. Differential Equations We can simply write the DEs as L(y) = 0 and L(y) = g(x)

Let y1, y2, …, yk be a solutions of the homogeneous Nth-order differential equation (6) on an interval I. Then the linear combination y = c1y1(x) + c2y2(x) + …+ ckyk(x) where the ci, i = 1, 2, …, k are arbitrary constants, is also a solution on the interval. THEOREM 3.2 Superposition Principles – Homogeneous Equations

(A) y = cy1 is also a solution if y1 is a solution. (B) A homogeneous linear DE always possesses the trivial solution y = 0. COROLLARY Corollaries to Theorem 3.2

Example 4 The function y1 = x2, y2 = x2 ln x are both solutions of Then y = x2 + x2 ln x is also a solution on (0, ). A set of f1(x), f2(x), …, fn(x) is linearly dependent on an interval I, if there exists constants c1, c2, …, cn, not all zero, such that c1f1(x) + c2f2(x) + … + cn fn(x) = 0 If not linearly dependent, it is linearly independent. DEFINITION 3.1 Linear Dependence and Linear Independence

In other words, if the set is linearly independent, when c1f1(x) + c2f2(x) + … + cn fn(x) = 0 then c1 = c2 = … = cn = 0 Referring to Fig 3.3, neither function is a constant multiple of the other, then these two functions are linearly independent.

Fig 3.3

Example 5 The functions f1 = cos2 x, f2 = sin2 x, f3 = sec2 x, f4 = tan2 x are linearly dependent on the interval (-/2, /2) since c1 cos2 x +c2 sin2 x +c3 sec2 x +c4 tan2 x = 0 when c1 = c2 = 1, c3 = -1, c4 = 1.

Example 6 The functions f1 = x½ + 5, f2 = x½ + 5x, f3 = x – 1, f4 = x2 are linearly dependent on the interval (0, ), since f2 = 1 f1 + 5 f3 + 0 f4

Suppose each of the functions f1(x), f2(x), …, fn(x) possesses at least n – 1 derivatives. The determinant is called the Wronskian of the functions. DEFINITION 3.2 Wronskian

Let y1(x), y2(x), …, yn(x) be solutions of the nth-order homogeneous DE (6) on an interval I. This set of solutions is linearly independent if and on if W(y1, y2, …, yn)  0 for every x in the interval. THEOREM 3.3 Criterion for Linear Independence Any set y1(x), y2(x), …, yn(x) of n linearly independent solutions is said to be a fundamental set of solutions. DEFINITION 3.3 Fundamental Set of a Solution

There exists a fundamental set of solutions for (6) on an interval I. THEOREM 3.4 Existence of a Fundamental Set Let y1(x), y2(x), …, yn(x) be a fundamental set of solutions of homogeneous DE (6) on an interval I. Then the general solution is y = c1y1(x) + c2y2(x) + … + cnyn(x) where ci are arbitrary constants. THEOREM 3.5 General Solution – Homogeneous Equations

Example 7 The functions y1 = e3x, y2 = e-3x are solutions of y” – 9y = 0 on (-, ) Now for every x. So y = c1y1 + c2y2 is the general solution.

Example 8 The functions y = 4 sinh 3x - 5e3x is a solution of example 7 (Verify it). Observer = 4 sinh 3x – 5e-3x

Example 9 The functions y1 = ex, y2 = e2x , y3 = e3x are solutions of y’’’ – 6y” + 11y’ – 6y = 0 on (-, ). Since for every real value of x. So y = c1ex + c2 e2x + c3e3x is the general solution on (-, ).

Any yp free of parameters satisfying (7) is called a particular solution. If y1(x), y2(x), …, yk(x) be a fundamental set of solutions of (6), then the general solution of (7) is y= c1y1 + c2y2 +… + ckyk + yp (10) THEOREM 3.6 General Solution – Nonhomogeneous Equations Complementary Function y = c1y1 + c2y2 +… + ckyk + yp = yc + yp = complementary + particular

Example 10 The function yp = -(11/12) – ½ x is a particular solution of (11) From previous discussions, the general solution of (11) is

DE (12) with gi(x), then (13) is a particular solution of (14) Given (12) where i = 1, 2, …, k. If ypi denotes a particular solution corresponding to the DE (12) with gi(x), then (13) is a particular solution of (14) THEOREM 3.7

Example 11 We find yp1 = -4x2 is a particular solution of yp2 = e2x is a particular solution of yp3 = xex is a particular solution of From Theorem 3.7, is a solution of

Note: If ypi is a particular solution of (12), then is also a particular solution of (12) when the right-hand member is

3.2 Reduction of Order Introduction: We know the general solution of (1) is y = c1y1 + c2y1. Suppose y1(x) denotes a known solution of (1). We assume the other solution y2 has the form y2 = uy1. Our goal is to find a u(x) and this method is called reduction of order.

Example 1 Given y1 = ex is a solution of y” – y = 0, find a second solution y2 by the method of reduction of order. Solution If y = uex, then And Since ex  0, we let w = u’, then

Example 1 (2) Thus (2) Choosing c1 = 0, c2 = -2, we have y2 = e-x. Because W(ex, e-x)  0 for every x, they are independent.

General Case Rewrite (1) as the standard form (3) Let y1(x) denotes a known solution of (3) and y1(x)  0 for every x in the interval. If we define y = uy1, then we have

This implies that or (4) where we let w = u’. Solving (4), we have or

then Let c1 = 1, c2 = 0, we find (5)

Example 2 The function y1= x2 is a solution of Find the general solution on (0, ). Solution: The standard form is From (5) The general solution is

3.3 Homogeneous Linear Equation with Constant Coefficients Introduction: (1) where ai are constants, an  0. Auxiliary Equation: For n = 2, (2) Try y = emx, then (3) is called an auxiliary equation.

From (3) the two roots are (1) From (3) the two roots are (1) b2 – 4ac > 0: two distinct real numbers. (2) b2 – 4ac = 0: two equal real numbers. (3) b2 – 4ac < 0: two conjugate complex numbers.

Case 1: Distinct real roots The general solution is (4) Case 2: Repeated real roots and from (5) of Sec 3.2, (5) The general solution is (6)

Case 3: Conjugate complex roots We write. , a general solution is Case 3: Conjugate complex roots We write , a general solution is From Euler’s formula: and (7) and

Since is a solution then set C1 = C1 = 1 and C1 = 1, C2 = -1 , we have two solutions: So, ex cos x and ex sin x are a fundamental set of solutions, that is, the general solution is (8)

Example 1 Solve the following DEs: (a) (b) (c)

Example 2 Solve Solution: See Fig 3.4.

Fig 3.4

Two Equations worth Knowing For the first equation: (9) For the second equation: (10) Let Then (11)

Higher-Order Equations Given (12) we have (13) as an auxiliary equation.

Example 3 Solve Solution:

Example 4 Solve Solution:

Repeated complex roots If m1 =  + i is a complex root of multiplicity k, then m2 =  − i is also a complex root of multiplicity k. The 2k linearly independent solutions:

3.4 Undetermined Coefficients Introduction If we want to solve (1) we have to find y = yc + yp. Thus we introduce the method of undetermined coefficients.

Example 1 Solve Solution: We can get yc as described in Sec 3.3. Now, we want to find yp. Since the right side of the DE is a polynomial, we set After substitution, 2A + 8Ax + 4B – 2Ax2 – 2Bx – 2C = 2x2 – 3x + 6

Example 1 (2) Then

Example 2 Find a particular solution of Solution: Let yp = A cos 3x + B sin 3x After substitution, Then

Example 3 Solve (3) Solution: We can find Let After substitution, Then

Example 4 Find yp of Solution: First let yp = Ae2x After substitution, 0 = 8e2x, (wrong guess) Let yp = Axe2x After substitution, -3Ae2x = 8e2x Then A = -8/3, yp = (−8/3)xe2x

Rule of Case 1: No function in the assumed yp is part of yc Table 3.1 shows the trial particular solutions.

Example 5 Find the form of yp of (a) Solution: We have and try There is no duplication between yp and yc . (b) y” + 4y = x cos x Solution: We try There is also no duplication between yp and yc .

Example 6 Find the form of yp of Solution: For 3x2: For -5 sin 2x: For 7xe6x: No term in duplicates a term in yc

Rule of Case 2: If any term in yp duplicates a term in yc, it should be multiplied by xn, where n is the smallest positive integer that eliminates that duplication.

Example 8 Solve Solution: First trial: yp = Ax + B + C cos x + E sin x (5) However, duplication occurs. Then we try yp = Ax + B + Cx cos x + Ex sin x After substitution and simplification, A = 4, B = 0, C = -5, E = 0 Then y = c1 cos x + c2 sin x + 4x – 5x cos x Using y() = 0, y’() = 2, we have y = 9 cos x + 7 sin x + 4x – 5x cos x

Example 9 Solve Solution: yc = c1e3x + c2xe3x After substitution and simplification, A = 2/3, B = 8/9, C = 2/3, E = -6 Then

Example 10 Solve Solution: m3 + m2 = 0, m = 0, 0, -1 yc = c1+ c2x + c3e-x yp = Aex cos x + Bex sin x After substitution and simplification, A = -1/10, B = 1/5 Then

Example 11 Find the form of yp of Solution: yc = c1+ c2x + c3x2 + c4e-x Normal trial: Multiply A by x3 and (Bx2e-x + Cxe-x + Ee-x) by x Then yp = Ax3 + Bx3e-x + Cx2e-x + Exe-x

3.5 Variation of Parameters Some Assumptions For the DE (1) we put (1) in the form (2) where P, Q, f are continuous on I.

Method of Variation of Parameters We try (3) After we obtain yp’, yp”, we put them into (2), then (4)

Making further assumptions:. y1u1’ + y2u2’ = 0, then from (4), Making further assumptions: y1u1’ + y2u2’ = 0, then from (4), y1’u1’ + y2’u2’ = f(x) Express the above in terms of determinants and (5) where (6)

Example 1 Solve Solution: m2 – 4m + 4 = 0, m = 2, 2 y1 = e2x, y2 = xe2x, Since f(x) = (x + 1)e2x, then

Example 1 (2) From (5), Then u1 = (-1/3)x3 – ½ x2, u2 = ½ x2 + x And

Example 2 Solve Solution: y” + 9y = (1/4) csc 3x m2 + 9 = 0, m = 3i, -3i y1 = cos 3x, y2 = sin 3x, f = (1/4) csc 3x Since

Example 2 (2) Then And

Example 3 Solve Solution: m2 – 1 = 0, m = 1, -1 y1 = ex, y2 = e-x, f = 1/x, and W(ex, e-x) = -2 Then The low and up bounds of the integral are x0 and x, respectively.

Example 3 (2)

Higher-Order Equations For the DEs of the form (8) then yp = u1y1 + u2y2 + … + unyn, where yi , i = 1, 2, …, n, are the elements of yc. Thus we have (9) and uk’ = Wk/W, k = 1, 2, …, n.

For the case n = 3, (10)

3.6 Cauchy-Eulaer Equation Form of Cauchy-Euler Equation Method of Solution We try y = xm, since

An Auxiliary Equation For n = 2, y = xm, then am(m – 1) + bm + c = 0, or am2 + (b – a)m + c = 0 (1) Case 1: Distinct Real Roots (2)

Example 1 Solve Solution: We have a = 1, b = -2 , c = -4 m2 – 3m – 4 = 0, m = -1, 4, y = c1x-1 + c2x4

Case 2: Repeated Real Roots Using (5) of Sec 3.2, we have Then (3)

Example 2 Solve Solution: We have a = 4, b = 8, c = 1 4m2 + 4m + 1 = 0, m = -½ , -½

Case 3: Conjugate Complex Roots Higher-Order: multiplicity Case 3: Conjugate Complex Roots m1 =  + i, m2 =  – i, y = C1x( + i) + C2x( - i) Since xi = (eln x)i = ei ln x = cos( ln x) + i sin( ln x) x-i = cos ( ln x) – i sin ( ln x) Then y = c1x cos( ln x) + c2x sin( ln x) = x [c1 cos( ln x) + c2 sin( ln x)] (4)

Example 3 Solve Solution: We have a = 4, b = 0 , c = 17 4m2 − 4m + 17 = 0, m = ½ + 2i Apply y(1) = -1, y’(1) = 0, then c1 = -1, c2 = 0, See Fig 3.15.

Fig 3.15

Example 4 Solve Solution: Let y = xm, Then we have xm(m + 2)(m2 + 4) = 0 m = -2, m = 2i, m = -2i y = c1x-2 + c2 cos(2 ln x) + c3 sin(2 ln x)

Example 5 Solve Solution: We have (m – 1)(m – 3) = 0, m = 1, 3 yc = c1x + c2x3 , use variation of parameters, yp = u1y1 + u2y2, where y1 = x, y2 = x3 Rewrite the DE as Then P = -3/x, Q = 3/x2, f = 2x2ex

Example 5 (2) Thus We find

Example 5 (3) Then

3.7 Nonlinear Equations Example 1 Solve Solution: This nonlinear equation misses y term. Let u(x) = y’, then du/dx = y”, or (This form is just for convenience) Since u-1 = 1/y’, So,

Example 2 Solve Solution: This nonlinear equation misses x term. Let u(x) = y’, then y” = du/dx = (du/dy)(dy/dx) = u du/dy or ln|u| = ln|y| + c1, u = c2y (where ) Since u = dy/dx = c2y, dy/y = c2 dx ln|y| = c2x + c3,

Example 3 Assume (1) exists. If we further assume y(x) possesses a Taylor series centered at 0: (2) Remember that y(0) = -1, y’(0) = 1. From the original DE, y”(0) = 0 + y(0) – y(0)2 = −2. Then (3)

Example 3 (2) (4) (5) and so on. So we can use the same method to obtain y(3)(0) = 4, y(4)(0) = −8, …… Then

Example 4 The DE in example 3 is equivalent to With the aid of a solver, Fig 3.16 shows the graph of this DE. For comparison, the curve of fifth-degree Taylor series is also shown.

Fig 3.16

3.8 Linear Models: IVP Newton’s Law See Fig 3.18, we have (1)

Fig 3.18

Fig3.19

Free Undamped Motion From (1), we have (2) where  = k/m. (2) is called a simple harmonic motion, or free undamped motion.

Solution and Equation of Motion From (2), the general solution is (3) Period T = 2/, frequency f = 1/T = /2.

Example 1 A mass weighing 2 pounds stretches a spring 6 inches. At t = 0, the mass is released from a 8 inches below the equilibrium position with an upward velocity 4/3 ft/s. Determine the equation of motion. Solution: Unit convert: 6 in = 1/2 ft; 8 in = 2/3 ft, m = W/g = 1/16 slug From Hooke’s Law, 2 = k(1/2), k = 4 lb/ft Hence (1) gives

Example 1 (2) together with x(0) = 2/3, x’(0) = -4/3. Since 2 = 64,  = 8, the solution is x(t) = c1 cos 8t + c2 sin 8t (4) Applying the initial condition, we have (5)

Alternate form of x(t) (4) can be written as x(t) = A sin(t + ) (6) where and  is a phase angle, (7) (8) (9)

Fig 3.20

Example 2 Solution (5) is x(t) = (2/3) cos 8t − (1/6) sin 8t = A sin(t + ) Then However it is not the solution, since we know tan-1 (+/−) will locate in the second quadrant Then so (9) The period is T = 2/8 = /4.

Fig 3.21 Fig 3.21 shows the motion.

Free Damped Motion If the DE is as (10) where  is a positive damping constant. Then x”(t) + (/m)x’ + (k/m)x = 0 can be written as (11) where 2 = /m, 2 = k/m (12) The auxiliary equation is m2 + 2m + 2 = 0, and the roots are

Case 1: 2 – 2 > 0. Let then (13) It is said to be overdamped. See Fig 3.23.

Fig3.23

Case 2: 2 – 2 = 0. then (14) It is said to be critically damped. See Fig 3.24.

Fig3.24

Case 3: 2 – 2 < 0. Let then (15) It is said to be underdamped. See Fig 3.25.

Fig 3.25

Example 3 The solution of is (16) See Fig 3.26.

Fig 3.26

Example 4 A mass weighing 8 pounds stretches a spring 2 feet. Assuming a damping force equal to 2 times the instantaneous velocity exists. At t = 0, the mass is released from the equilibrium position with an upward velocity 3 ft/s. Determine the equation of motion. Solution: From Hooke’s Law, 8 = k (2), k = 4 lb/ft, and m = W/g = 8/32 = ¼ slug, hence (17)

Example 4 (2) m2 + 8m + 16 = 0, m = −4, −4 x(t) = c1 e-4t + c2t e-4t (18) Initial conditions: x(0) = 0, x’(0) = −3, then x(t) = −3t e-4t (19) See Fig 3.27.

Fig 3.27

Example 5 A mass weighing 16 pounds stretches a spring from 5 feet to 8.2 feet. t. Assuming a damping force is equal to the instantaneous velocity exists. At t = 0, the mass is released from rest at a point 2 feet above the equilibrium position. Determine the equation of motion. Solution: From Hooke’s Law, 16 = k (3.2), k = 5 lb/ft, and m = W/g = 16/32 = ½ slug, hence (20) m2 + 2m + 10 = 0, m = −3 + 3i, −3 − 3i

Example 5 (2) (21) Initial conditions: x(0) = −2, x’(0) = 0, then (22)

Alternate form of x(t) (22) can be written as (23) where and

DE of Driven Motion with Damping As in Fig 3.28, (24) (25) where

Fig 3.28

Example 6 Interpret and solve (26) Solution: Interpret: m = 1/5, k = 2,  = 1.2, f(t) = 5 cos 4t release from rest at a point ½ below Sol:

Example 6 (2) Assuming xp(t) = A cos 4t + B sin 4t, we have A = −25/102, B = 50/51, then Using x(0) = 1/2, x’(0) = 0 c1 = 38/51, c2 = −86/51, (28)

Transient and Steady-State Graph of (28) is shown in Fig 3.29. xc(t) will vanish at t  : transient term xp(t) will still remain at t  : steady-state term

Fig 3.29

Example 7 The solution of is See Fig 3.30.

Fig 3.30

Example 8 Solve where F0 is a constant and   . Solution: xc = c1 cos t + c2 sin t Let xp = A cos t + B sin t, after substitution, A = 0, B = F0/(2− 2),

Example 8 (2) Since x(0) = 0, x’(0) = 0, then Thus (30)

Pure Resonance When  = , we consider the case   . (31)

When t  , the displacements become large In fact, |x(tn)|   when tn = n/, n = 1, 2, ….. As shown in Fig 3.31, it is said to be pure resonance.

Fig 3.31

LRC-Series Circuits The following equation is the DE of forced motion with damping: (32) If i(t) denotes the current shown in Fig 3.32, then (33) Since i = dq/dt, we have (34)

Fig 3.32

Example 9 Find q(t) in Fig 3.32, where L = 0.025 henry, R = 10 ohms, C = 0.001 farad, E(t) = 0, q(0) = q0 coulombs, and i(0) = 0 ampere. Solution: Using the given data: As described before, Using q(0) = q0, i(0) = q’(0) = 0, c1 = q0, c2 = q0/3

Example 10 Find the steady-state qp(t) and the steady-state current, when E(t) = E0 sin t . Solution: Let qp(t) = A sin t + B cos t,

Example 10 (2) If Using the similar method, we have So Note: X and Z are called the reactance and impedance, respectively.

3.9 Linear Models: BVP Deflection of a Beam The bending moment M(x) at a point x along the beam is related to the load per unit length w(x) by (1) In addition, M(x) is proportional to the curvature  of the elastic curve M(x) = EI (2) where E, I are constants.

From calculus, we have   y”, when the deflection y(x) is small From calculus, we have   y”, when the deflection y(x) is small. Finally we have (3) Then (4)

Terminology Ends of the beam Boundary Conditions embedded y = 0, y’ = 0 free y” = 0, y’’’ = 0 simply supported (hinged) y = 0, y” = 0 See Fig 3.41

Fig 3.41

Example 1 A beam of length L is embedded at both ends. Find the deflection of the beam if a constant load w0 is uniformly distributed aling its length, that is, w(x)= w0 , 0 < x < L Solution: From (4) we have Embedded ends means We have m4 = 0, yc(x) = c1 + c2x + c3x2 + c4x3, and

Example 1 (2) So Using the boundary conditions, we have c1 = 0, c2 = 0, c3 = w0L2/24EI, c4 = −w0L/12EI Choosing w0 = 24EI and L = 1, we have Fig 3.42.

Fig 3.42

Example 2 Solve Solution: Case 1 :  = 0 y = c1x + c2, y(0) = c2 = 0, y(L) = c1L = 0, c1 = 0 then y = 0, trivial solution. Case 2 :  < 0,  = −2,  > 0 Choose y = c1 cosh x + c2 sinh x y(0) = 0, c1 = 0; y(L) = 0, c2 = 0 then y = 0, trivial solution.

Example 2 (2) Case 3 :  > 0,  = 2,  > 0 Choose y = c1 cos x + c2 sin x y(0) = 0, c1 = 0; y(L) = 0, c2 sin L= 0 If c2 = 0, y = 0, trivial solution. So c2  0, sin L = 0, L = n,  = n/L Thus, y = c2 sin (nx/L) is a solution for each n.

Example 2 (3) Simply take c2 = 1, for each: the corresponding function: Note: n = (n/L)2, n = 1, 2, 3, … are known as characteristic values or eigenvalues. yn = sin (nx/L) are called characteristic functions or eigenfunctions.

Bulking of a Thin Vertical Column Referring to Fig 3.43, the DE is (5) where P is a constant vertical compressive force applied to the column’s top.

Fig 3.43

Example 3 Referring to Fig 3.43, when the column is hinged at both ends, find the deflection. Solution: The boundary-value problem is From the intuitive view, if the load P is not great enough, there is no deflection. The question is: For what values of P does the given BVP possess nontrivial solutions?

Example 3 (2) By writing  = P/EI, we see is identical to example 2. From Case 3, the deflection curves are yn = c2 sin (nx/L), corresponding to eigenvalues n = Pn/EI = n22/L2, n = 1, 2, 3, … Physically, only Pn = EIn22/L2, deflection occurs. We call these Pn the critical loads and the smallest P = P1 = EI2/L2 is called the Euler load, and y1 = c2 sin(x/L) is known as the first buckling mode. See Fig 3.44

Fig 3.44

Rotating String The simple DE y” + y = 0 (6) occurs again as a model of a rotating string. See Fig 3.45.

Fig 3.45

We have. F = T sin 2 – T sin 1. (7) When 1 and 2 are small, We have F = T sin 2 – T sin 1 (7) When 1 and 2 are small, sin 2  tan 2 , sin 1  tan 1 Since tan2, tan1 are slopes of the lines containing the vectors T1 and T2, then tan 2 = y’(x + x), tan 1 = y’(x) Thus (7) becomes (8) Because F = ma, m = x, a = r2. With x small, we take r = y.

Thus. (9) Letting (8) = (9), we have Thus (9) Letting (8) = (9), we have (10) For x close to zero, we have (11) And the boundary conditions are y(0) = y(L) = 0.

3.10 Nonlinear Models Nonlinerar Springs The model (1) when F(x) = kx is said to be linear. However, (2) is a nonlinear spring. Another model (3)

Hard and Soft Springs F(x) = kx + k1x3 is said to be hard if k1 > 0; and is soft, if k1 < 0. See Fig 3.50. Fig 3.50

Example 1 The DEs (4) and (5) are special cases of (2). Fig3.51 shows the graph from a numerical solver.

Fig 3.51

Nonlinear Pendulum The model of a simple pendulum is shown in Fig 3.52. From the figure, We have the angle acceleration a = s” = l”, the force Then (6)

Fig 3.52

Linearization Since If we use only the first two terms, If  is small, (7)

Example 2 Fig 3.53 shows some results with different initial conditions by a solver. We can see if the initial velocity is great enough, it will go out of bounds.

Fig 3.53

Telephone Wire Recalling from (17) in Sec 1.3 and Fig 1.26 dy/dx = W/T1, can be modified as (8) where  is the density and s is the arc length. Since the length s is (9)

then (10) Differentiating (8) w.s.t x and using (10), then (11)

Example 3 From Fig 1.26, we obtain From Fig 1.26, we obtain y(0) = a, y’(0) = 0. Let u = y’, equation (11) becomes Thus Now y’(0) = u(0) = 0, sinh-10 = 0 = c1 Since u = sinh(x/T1) = dy/dx, then Using y(0) = a, c2 = a − (T1/)

Rocket Motion From Fig 3.54, we have (12) when y = R, kMm/R2 = Mg, k = gR2/M, then (13)

Fig 3.54

Variable Mass Assuming the mass is variable, then F = ma should be modified as (14)

Example 4 A uniform 10-foot-long chain is coiled loosely on the ground. On end is pulled vertically by a force of 5 lb. The chain weigh 1 lb per foot. Determine the height of the end at time t. Solution: Let x(t) = the height v(t) = dx/dt (velocity) W = x1 = x (weight) m = W/g = x/32 (mass) F = 5 – W (net force)

Example 4 (2) Then (15) Since v = dx/dt (16) is of the form F(x, x’, x”) = 0 Since v = x’, and then (15) becomes (17)

Example 4 (3) Rewriting (17) as (v2+32x – 160) dx + xv = 0 (18) (18) can be multiplied by an integrating factor to become exact, where we can find the integrating factor is (x) = x (please verify). Then Use the method in Sec. 2.4 (19) Since x(0) = 0, then c1 = 0. By solving (19) = 0, for v = dx/dt > 0, we get

Example 4 (4) Thus please verify that (20) Using x(0) = 0 again, , we square both sides of (20) and solve for x (21)

3.11 Solving Systems of Linear Equations Coupled Spring/Mass System From Fig 3.58 and Newton’s Law (1)

Fig 3.58

Method of Solution Consider dx/dt = 3y, dy/dt = 2x or Dx – 3y = 0, 2x – Dy = 0 (2) Then, multiplying the first by D, the second by −3, and then eliminating y, gives D2x – 6x =0 (3) Similar method can give (4)

Return to the original equations, Return to the original equations, dx/dt = 3y then after simplification, we have (5)

Example 1 Solve Dx + (D + 2)y = 0 (D – 3)x – 2y = 0 (6) Solution: Multiplying the first by D – 3, the second by D, then subtracting, [(D – 3)(D + 2) + 2D]y = 0 (D2 + D – 6)y = 0 then y(t) = c1e2t + c2e-3t (7)

Example 1 (2) Using the similar method, x(t) = c3e2t + c4e-3t (8) Substituting (7) and (8) into the first equation of (6), (4c1 + 2c3)e2t + (−c2 – 3c4)e−3t = 0 Then 4c1 + 2c3 = 0 = −c2 – 3c4 c3 = –2c1, c4 = – ⅓c2

Example 2 Solve x’ – 4x + y” = t2 x’ + x + y’ = 0 (9) Solution: (D – 4)x + D2y = t2 (D + 1)x + Dy = 0 (10) By eliminating x, then and m = 0, 2i, −2i Let then we can get A = 1/12, B = ¼ , C = −1/8.

Example 2 (2) Thus (11) Similar method to get x(t) Then m= 2i, −2i, Let xp(t) = At2 + Bt + C, then we can get A = −1/4, B = 0, C = 1/8

Example 2 (3) Thus (12) By using the second equation of (9), we have

Example 3 In (3) of Sec. 2.9, we have Together with the given initial conditions, we can use the same method to solve x1 and x2, not mentioned here.

Example 4 Solve (13) with Solution: Then

Example 4 (2) Using the same method, we have (14)

Fig 3.59

Thank You !