Second-Order Differential Equations: Preliminary Concepts and Theory of Solutions

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

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

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

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

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

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

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

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.

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

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

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

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

15. 2.2.2. Nonhomogeneous Equation ○ Theorem 2.5: : linearly independent homogeneous solutions of : a nonhomogeneous solution of any solution has the form

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

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

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

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

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

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)

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

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

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

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

26. iii) Let The general sol.:

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

28. Maclaurin expansions:

29. 。 Eq. (2.5),

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

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

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

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

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

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

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

38. ii) Nonhomogeneous solution Let

39. iii) The general solution:

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

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

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

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

44. Try Substituting into (A),

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

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

47. 2.6.3. Superposition Let be a solution of is a solution of (A)

48. 。 Example 2.25: The general solution: where homogeneous solutions