Download presentation

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 **

4
**Given another initial condition**

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

5
**○ Theorem 2.2: : solutions of Eq. (2.2)**

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
**○ Definition 2.2: ◎ Find all solutions 1. : linearly independent**

: fundamental set of solutions : 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**

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
**1. Find the general homogeneous solutions**

○ 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, 。 Example 2.4: : a solution Let**

: 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**

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**

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 **

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**

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

33
**Characteristic equation: Roots: General solution:**

。 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**

。 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),**

。 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 **

Guessed

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 **

Similar presentations

OK

Math 3120 Differential Equations with Boundary Value Problems

Math 3120 Differential Equations with Boundary Value Problems

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on electricity crisis in india Ppt on index numbers lecture Ppt on vice president of india Ppt on natural and artificial satellites distance Plant anatomy and physiology ppt on cells Ppt on cleanliness in islam Ppt on extinct species in india Ppt on business model of hul Ppt on indian construction industry Cell surface display ppt on ipad