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## Higher-Order Differential Equations

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**CHAPTER 3**Higher-Order Differential Equations**Contents**• 3.1 Preliminary Theory: Linear Equations • 3.3 Homogeneous Linear Equations with Constant Coefficients • 3.4 Undetermined Coefficients • 3.5 Variation of Parameters • 3.8 Linear Models: Initial-Value Problems**3.1 Preliminary Theory: Linear Equ.**• Initial-value ProblemAn initial value problem for nth-order linear DE iswith (1)as n initial conditions.**THEOREM 3.1**Existence and Uniqueness Let an(x), an-1(x), …, a0(x),and g(x)be continuous on I, an(x) 0for all x on I. If x = x0 is a point in this interval, then a solution y(x)of (1) exists on the interval and is unique.**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.**The following DE (6)is homogeneous;**(7)with g(x)0, 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 EquationsWe can simply write the n-th order linear DEs as L(y) = 0andL(y) = g(x)**THEOREM 3.2**Let y1, y2, …, ynbe n solutions of the homogeneous nth-order differential equation (6) on an interval I. Then the linear combinationy = c1y1(x) + c2y2(x)+ …+ cnyn(x)where the ci, i = 1, 2, …, n are arbitrary constants, is also a solution on the interval. Superposition Principles – Homogeneous Equations**COROLLARY**Corollaries to Theorem 3.2 (A) y = cy1 is also a solution if y1is a solution. (B) A homogeneous linear DE always possesses the trivial solution y = 0.**DEFINITION 3.1**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 thatc1f1(x) + c2f2(x) + … + cn fn(x) = 0If not linearly dependent, it is linearly independent. Linear Dependence and Linear Independence Linear Dependence and Independence**In other words, if the set islinearly independent, then**c1f1(x) + c2f2(x) + … + cn fn(x) = 0implies 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.**Example 5**• The functions f1 = cos2 x, f2 = sin2x, f3 = sec2x, f4 =tan2x are linearly dependent on the interval (-/2, /2) sincec1 cos2 x +c2 sin2x +c3 sec2x +c4 tan2x = 0when 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**DEFINITION 3.2**Suppose each of the functions f1(x), f2(x), …, fn(x) possesses at least n – 1 derivatives. The determinantis called the Wronskian of the functions. Wronskian**THEOREM 3.3**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. Criterion for Linear Independence Corollary 3.3 IfW(y1, y2, …, yn) 0 for some x in the interval then y1, y2, …, yn are linearly independent in the interval. If W(y1, y2, …, yn) = 0 for some x in the interval then y1, y2, …, yn are linearly dependent in the interval. . . .**DEFINITION 3.3**Any set y1(x), y2(x), …, yn(x) of n linearly independent solutions is said to be a fundamental set of solutions. Fundamental Set of a Solution**THEOREM 3.4**There exists a fundamental set of solutions for (6) on an interval I. Existence of a Fundamental Set THEOREM 3.5 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 isy = c1y1(x) + c2y2(x) + … + cnyn(x) where ci are arbitrary constants. General Solution – Homogeneous Equations**Example 7**• The functions y1 = e3x, y2 = e-3xare solutions of y” –9y = 0on (-, )Now for every x. So y = c1y1 + c2y2is the general solution.**Example 8**• The functions y = 4sinh 3x - 5e3x is a solution of example 7 (Verify it). Observe = 4 sinh 3x – 5e-3x**Example 9**• The functions y1= ex,y2 = e2x , y3 = e3xare solutions of y’’’ –6y” + 11y’ –6y = 0on (-, ). Sincefor everyrealvalue of x. So y = c1ex + c2e2x + c3e3xis the general solution on (-, ).**THEOREM 3.6**Any ypfree of parameters satisfying (7) is called a particular solution. If y1(x), y2(x), …, yn(x) be a fundamental set of solutions of (6), then the general solution of (7) isy= c1y1 + c2y2 +… +cnyn + yp (10) General Solution – Nonhomogeneous Equations • Complementary Function: yy = c1y1 + c2y2 +… +cnyn +yp = yc + yp = complementary + particular**Example 10**• The function yp= -(11/12) –½ xis a particular solution of (11)From previous discussions, the general solution of (11) is**THEOREM 3.7**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)**Example 11**• We findyp1 = -4x2 is a particular solution ofyp2 = e2xis a particular solution ofyp3 = xexis 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**3.3 Homogeneous Linear Equation with Constant Coefficients**• Introduction: (1)where ai are constants, an 0. • Auxiliary Equation (Characteristic Equation):For n = 2, (2)Try y = emx, then (3)is called an auxiliary equation or characteristic equation.**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 rootsThe general solution is**(why?) (4) • Case 2: Repeated real roots, (why?) (5)The general solution is (why?) (6)**Case 3: Conjugate complex rootsWe write , a general solution**is From Euler’s formula: and(7) and**Since is a solution then setC1= C1 = 1and C1 = 1, C2 = -1 ,**we have two solutions: So, ex cos x and exsin 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.**Higher-Order Equations**• Given (12)we have (13)as an auxiliary equation (or characteristic 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**• IntroductionIf we want to solve the nonhomogeneous linear DE (1)we have to find y = yc + yp. Thus we introduce the method of undetermined coefficients to solve for yp.**Example 1**Solve Solution: We can get ycas described in Sec 3.3. Now, we want to find yp. Since the right side of the DE is a polynomial, we setAfter 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 3xAfter substitution, Then**Example 3**Solve (3) Solution: We can find Let After substitution, Then**Example 4**Find ypof Solution: First let yp = AexAfter substitution, 0 = 8ex, (wrong guess) Let yp = AxexAfter substitution, -3Aex= 8exThen A = -8/3,yp= (−8/3)xex**Rule of Case 1:**• No function in the assumed yp is part of ycTable 3.1 shows the trial particular solutions.**Example 5**Find the form of ypof (a) Solution:We have and tryThere is no duplication between ypand yc . (b) y” + 4y = x cos x Solution: We try There is also no duplication between ypand yc .**Example 6**Find the form of ypof Solution: For 3x2: For -5sin2x: 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 tryyp= Ax + B + Cx cos x + Ex sin x After substitution and simplification,A = 4, B = 0, C = -5, E = 0Then y = c1 cos x + c2 sin x + 4x –5x cos xUsing y()= 0, y’() =2,we have y = 9cos x + 7sin x + 4x – 5x cos x**Example 9**Solve Solution: yc= c1e3x + c2xe3xAfter substitution and simplification,A = 2/3, B = 8/9, C = 2/3, E = -6 Then