Chapter 2 SecondOrder Differential Equations 2 1 Preliminary

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

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

The graph of Given the initial condition The graph of 3

Given another initial condition The graph of ◎ The initial value problem: ○ Theorem 2. 1: : continuous on I, has a unique solution 4

2. 2. 1. Homogeous Equation ○ Theorem 2. 2: : solutions of Eq. (2. 2) solution of Eq. (2. 2) : real numbers Proof: 5

※ 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 6

○ 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

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

○ Theorem 2. 3: 1) Either 2) or : linearly independent iff Proof (2): (i) (if : linear indep. (P), then (Q) then if : linear dep. ( ( Q) , P) ) : linear dep. 9

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

。 Example 2. 2: are solutions of : linearly independent 11

。 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

◎ 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

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

2. 2. 2. Nonhomogeneous Equation ○ Theorem 2. 5: : linearly independent homogeneous solutions of : a nonhomogeneous solution of any solution has the form 15

Proof: Given , solutions : a homogenous solution of : linearly independent homogenous solutions (Theorem 2. 4) 16

○ 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

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

For symlicity, let c = 1, : independent solutions 。 Example 2. 4: : a solution Let 19

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

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

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

。 Example 2. 6: Let , Then Substituting into (A), The characteristic equation: The general solution: 23

ii) By the reduction of order method, Let Substituting into (2. 4) 24

Choose : linearly independent The general sol. : 。 Example 2. 7: Characteristic eq. : The repeated root: The general solution: 25

iii) Let The general sol. : 26

。 Example 2. 8: Characteristic equation: Roots: The general solution: ○ Find the real-valued general solution 。 Euler’s formula: 27

Maclaurin expansions: 28

。 Eq. (2. 5), 29

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

2. 5. Euler’s Equation , A , B : constants -----(2. 7) Transform (2. 7) to a constant coefficient equation by letting 31

Substituting into Eq. (2. 7), i. e. , ----(2. 8) Steps: (1) Solve (2) Substitute (3) Obtain 32

。 Example 2. 11: ------(A) -------(B) (i) Let Substituting into (A) Characteristic equation: Roots: General solution: 33

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

2. 6. Nonhomogeneous Linear Equation ------(2. 9) The general solution: ◎ Two methods for finding (1) Variation of parameters -- Replace with homogeneous solution in the general Let Assume ------(2. 10) Compute 35

Substituting into (2. 9), ------(2. 11) Solve (2. 10) and (2. 11) for . Likewise, 36

。 Example 2. 15: ------(A) i) General homogeneous solution : Let. Substitute into (A) The characteristic equation: Complex solutions: Real solutions: : independent 37

ii) Nonhomogeneous solution Let 38

iii) The general solution: 39

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

。 Example 2. 19: ---(A) It’s derivatives can be multiples of or Try Compute Substituting into (A), 41

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

。 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

Try Substituting into (A), 44

○ 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

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

2. 6. 3. Superposition Let be a solution of is a solution of (A) 47

。 Example 2. 25: The general solution: where homogeneous solutions 48