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Chapter 2: Second-Order Differential Equations 2.1. Preliminary Concepts ○ Second-order differential equation e.g., Solution: A function satisfies, (I.

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Presentation on theme: "Chapter 2: Second-Order Differential Equations 2.1. Preliminary Concepts ○ Second-order differential equation e.g., Solution: A function satisfies, (I."— Presentation transcript:

1 Chapter 2: Second-Order Differential Equations 2.1. Preliminary Concepts ○ Second-order differential equation e.g., Solution: A function satisfies, (I : an interval) 1

2 ○ Linear second-order differential equation Nonlinear: e.g., 2.2. Theory of Solution ○ Consider y contains two parameters c and d 2

3 The graph of Given the initial condition The graph of 3

4 Given another initial condition The graph of ◎ The initial value problem: ○ Theorem 2.1: : continuous on I, has a unique solution 4

5 2.2.1.Homogeous Equation ○ Theorem 2.2: : solutions of Eq. (2.2) solution of Eq. (2.2) : real numbers Proof: 5

6 6 Two solutions are linearly independent. Their linear combination provides an infinity of new solutions ○ Definition 2.1: f, g : linearly dependent If s.t. or ; otherwise f, g : linearly independent In other words, f and g are linearly dependent only if for ※

7 ○ Wronskian test -- Test whether two solutions of a homogeneous differential equation are linearly independent Define: Wronskian of solutions to be the 2 by 2 determinant 7

8 ○ Let If : linear dep., then or Assume 8

9 ○ Theorem 2.3: 1) Either or 2) : linearly independent iff Proof (2): (i) (if : linear indep. (P), then (Q) if ( Q), then : linear dep. ( P) ) : linear dep. 9

10 (ii) (if (P), then : linear indep. (Q) if : linear dep. ( Q), then ( P)) : linear dep., ※ Test at just one point of I to determine linear dependency of the solutions 10

11 11 。 Example 2.2: are solutions of : linearly independent

12 。 Example 2.3: Solve by a power series method The Wronskian of at nonzero x would be difficult to evaluate, but at x = 0 are linearly independent 12

13 ◎ Find all solutions ○ Definition 2.2: 1. : linearly independent : fundamental set of solutions 2. : general solution : constant ○ Theorem 2.4: : linearly independent solutions on I Any solution is a linear combination of 13

14 Proof: Let be a solution. Show s.t. Let and Then, is the unique solution on I of the initial value problem 14

15 Nonhomogeneous Equation ○ Theorem 2.5: : linearly independent homogeneous solutions of : a nonhomogeneous solution of any solution has the form 15

16 Proof: Given, solutions : a homogenous solution of : linearly independent homogenous solutions (Theorem 2.4) 16

17 ○ Steps: 1. Find the general homogeneous solutions of 2. Find any nonhomogeneous solution of 3. The general solution of is 2.3. Reduction of Order -- A method for finding the second independent homogeneous solution when given the first one 17

18 ○ Let Substituting into ( : a homogeneous solution ) Let (separable) 18

19 For symlicity, let c = 1, : independent solutions 19 。 Example 2.4: : a solution Let

20 Substituting into (A), For simplicity, take c = 1, d = 0 : independent The general solution: 20

21 2.4. Constant Coefficient Homogeneous A, B : numbers (2.4 ) The derivative of is a constant (i.e., ) multiple of Constant multiples of derivatives of y, which has form, must sum to 0 for (2,4) ○ Let Substituting into (2,4), (characteristic equation) 21

22 i) Solutions : : linearly independent The general solution: 22

23 。 Example 2.6: Let, Then Substituting into (A), The characteristic equation: The general solution: 23

24 ii) By the reduction of order method, Let Substituting into (2.4) 24

25 Choose : linearly independent The general sol.: 。 Example 2.7: Characteristic eq. : The repeated root: The general solution: 25

26 iii) Let The general sol.: 26

27 。 Example 2.8: Characteristic equation: Roots: The general solution: ○ Find the real-valued general solution 。 Euler’s formula: 27

28 Maclaurin expansions: 28

29 。 Eq. (2.5), 29

30 Find any two independent solutions Take 30 The general sol.:

31 2.5. Euler’s Equation, A, B : constants -----(2.7) Transform (2.7) to a constant coefficient equation by letting 31

32 Substituting into Eq. (2.7), i.e., (2.8) Steps: (1) Solve (2) Substitute (3) Obtain 32

33 。 Example 2.11: (A) (B) (i) Let Substituting into (A) Characteristic equation: Roots: General solution: 33

34 34 ○ Solutions of constant coefficient linear equation have the forms: Solutions of Euler’s equation have the forms:

35 2.6. Nonhomogeneous Linear Equation (2.9) The general solution: ◎ Two methods for finding (1) Variation of parameters -- Replace with in the general homogeneous solution Let Assume------(2.10) Compute 35

36 Substituting into (2.9), (2.11) Solve (2.10) and (2.11) for. Likewise, 36

37 。 Example 2.15: (A) i) General homogeneous solution : Let. Substitute into (A) The characteristic equation: Complex solutions: Real solutions: :independent 37

38 ii) Nonhomogeneous solution Let 38

39 iii) The general solution: 39

40 (2) Undetermined coefficients Apply to A, B: constants Guess the form of from that of R e.g. : a polynomial Try a polynomial for : an exponential for Try an exponential for 40

41 。 Example 2.19: ---(A) It’s derivatives can be multiples of or Try Compute Substituting into (A), 41

42 : linearly independent and The homogeneous solutions: The general solution: 42

43 。 Example 2.20: (A), try Substituting into (A), * This is because the guessed contains a homogeneous solution Strategy: If a homogeneous solution appears in any term of, multiply this term by x. If the modified term still occurs in a homogeneous solution, multiply by x again 43

44 Try Substituting into (A), 44

45 ○ Steps of undetermined coefficients: (1) Find homogeneous solutions (2) From R(x), guess the form of If a homogeneous solution appears in any term of, multiply this term by x. If the modified term still occurs in a homogeneous solution, multiply by x again (3) Substitute the resultant into and solve for its coefficients 45

46 ○ Guess from Let : a given polynomial, : polynomials with unknown coefficients 46 Guessed

47 Superposition Let be a solution of is a solution of (A) 47

48 。 Example 2.25: The general solution: wherehomogeneous solutions 48


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