SecondOrder Differential Equations 2 1 Preliminary Concepts Secondorder

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. Theories of Solution ○ Consider y contains two parameters c and d 2

The graph of. for different values of and Given the initial condition The graph of 3

Given another initial condition The graph of ◎ The initial value problem (I. V. P): 4

Existence and Uniqueness of 1 st order equation Theorem 1 • Suppose P(x) and Q(x) are continuous on the interval (a, b) that contains the point x 0. Then the initial value problem: • y + P(x)y = Q(x), y(x 0)=y 0 for any given y 0 has a unique solution on (a, b). 5 5

Existence and Uniqueness of 2 nd order equation • Theorem 2. Let p(x), q(x) and g(x) be continuous on an interval (a, b), and x 0 (a, b). Then the I. V. P. 6 6

※ Linear independent solutions Two solutions are linearly independent if 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 ○ Wronskian test whether two solutions of a homogeneous differential equation are linearly independent Define: Wronskian of solutions is the 2 by 2 determinant 8

○ Let If are linear dep. , then or Assume 9

○ Theorem 2. 3: 1) Either 2) or are linearly independent iff Proof (2): (i) if are linearly independent, then if then , are linearly dependent. 10

(ii) Conversely, if , then independent. if linear dep. then are linearly : 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

Fundamental solution set • A pair of solutions [y 1, y 2] of L[y] = 0, on (a, b) where L[y] = y +py +qy is called a fundamental solution set, if W[y 1, y 2](x 0) 0 for some x 0 (a, b). A simple example: Consider • L[y] = y +9 y. It is easily checked that y 1 = cos 3 x and y 2 = sin 3 x are solutions of L[y] = 0. Since the corresponding Wronskian W[y 1, y 2](x) = 3 0 , thus {cos 3 x, sin 3 x} forms a fundamental solution set to the homogenenous eq: y + 9 y = 0. We see that any linear combination c 1 y 1 + c 2 y 2 also satisfies L[y] = 0. This is known as a general solution 14

Homogeneous 2 nd order o. d. e with constant coefficients The general formula for such equation is To solve this equation we assume the solution in the form of exponential function: If then and the equation will change into after dividing by the eλx we obtain We obtained a quadratic characteristic equation. The roots are 15

There exist three types of solutions according to the discriminant D 1) If D>0, the roots λ 1, λ 2 are real and distinct 2) If D=0, the roots are real and identical λ 12 =λ 3) If D<0, the roots are complex conjugate λ 1, λ 2 where α and ω are real and imaginary parts of the root 16

EXAMPLE 1 Solve auxiliary function 17

EXAMPLE 2 Solve auxiliary function 18

EXAMPLE 3 Solve auxiliary function 19

Non-linear 2 nd O. D. E. - Equations where the independent variables does not occur explicitly • They are solved by differentiation followed by the p substitution. • When the p substitution is made in this case, the second derivative of y is replaced as Let 20

Solve Let and therefore Separating the variables 21

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 22

Substituting into (2. 9), ------(2. 11) Solve (2. 10) and (2. 11) for . Likewise, 23

。 Example 2. 15: ------(A) i) General homogeneous solution : Let. Substitute into (A) The characteristic equation: Complex solutions: Real solutions: : independent 24

ii) Nonhomogeneous solution Let 25

iii) The general solution: 26

(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 27

。 Example 2. 19: ---(A) It’s derivatives can be multiples of or Try Compute Substituting into (A), 28

: linearly independent and The homogeneous solutions: The general solution: 29

。 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 30

Try Substituting into (A), 31

○ 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 32

○ Guess Let from : a given polynomial , : polynomials with unknown coefficients Guessed 33

2. 6. 3. Superposition Let be a solution of is a solution of (A) 34

。 Example 2. 25: The general solution: where homogeneous solutions 35