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# Chapter 2: Second-Order Differential Equations

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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 : an interval)

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

The graph of Given the initial condition

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

○ 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:

※ 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

○ 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

○ Let If : linear dep., then or Assume

○ 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.

(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

。 Example 2.2: are solutions of : linearly independent

。 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

○ 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

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

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

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

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

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

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

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

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)

i) Solutions : : linearly independent The general solution:

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

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

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:

iii) Let The general sol.:

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

Maclaurin expansions:

。 Eq. (2.5),

Find any two independent solutions Take
The general sol.:

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

Substituting into Eq. (2. 7), i. e. , --------(2
Substituting into Eq. (2.7), i.e., (2.8) Steps: (1) Solve (2) Substitute (3) Obtain

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

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

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

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

。 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

ii) Nonhomogeneous solution Let

iii) The general solution:

(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

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

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

。 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

Try Substituting into (A),

○ 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

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

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

。 Example 2.25: The general solution: where homogeneous solutions

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