Higher-Order Differential Equations

1. CHAPTER 3 Higher-Order Differential Equations

2. 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. 3.1 Preliminary Theory: Linear Equ. • Initial-value ProblemAn initial value problem for nth-order linear DE iswith (1)as n initial conditions.

4. 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.

5. 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.

6. The following DE (6)is homogeneous; (7)with g(x)0, is nonhomogeneous.

7. 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)

8. 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

9. 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.

10. 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

11. 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.

12. Fig 3.3

13. 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.

14. 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

15. 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

16. 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. . . .

17. 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

18. 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

19. 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.

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

21. 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 (-, ).

22. 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

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

24. 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)

25. 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

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

27. 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.

28. 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.

29. Case 1: Distinct real rootsThe general solution is (why?) (4) • Case 2: Repeated real roots, (why?) (5)The general solution is (why?) (6)

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

31. Since is a solution then setC1= C1 = 1and C1 = 1, C2 = -1 , we have two solutions: So, ex cos x and exsin x are a fundamental set of solutions, that is, the general solution is(8)

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

33. Example 2 Solve Solution:See Fig 3.4.

34. Fig 3.4

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

36. Example 3 Solve Solution:

37. Example 4 Solve Solution:

38. 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:

39. 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.

40. 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

41. Example 1 (2) Then

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

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

44. 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

45. Rule of Case 1: • No function in the assumed yp is part of ycTable 3.1 shows the trial particular solutions.

46. 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 .

47. Example 6 Find the form of ypof Solution: For 3x2: For -5sin2x: For 7xe6x: No term in duplicates a term in yc

48. 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.

49. 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 = 9cos x + 7sin x + 4x – 5x cos x

50. Example 9 Solve Solution: yc= c1e3x + c2xe3xAfter substitution and simplification,A = 2/3, B = 8/9, C = 2/3, E = -6 Then