10Ma2aAnNotes1 - Ma2a (analytical) Fall 2010 PART I: FIRST...

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Ma2a (analytical) Fall 2010 PART I: FIRST ORDER ODEs 1. Introduction 1.1. Fundamental theorem of calculus. The simplest (”trivial”) d.e.: y ( x ) = f ( x ) . Examples: (i) Find all ( maximal ) solutions of the equation y ( x ) = 1 1 + x 2 . Answer: y ( x ) = dx 1 + x 2 = arctan x + C, ( x R ) . (ii) Solve the IVP y ( x ) = 1 1 + x 2 , y (1) = 0 . Answer: y ( x ) = dx 1 + x 2 = arctan x π 4 , ( x R ) . [Set x = 1 in the ”general solution” y ( x ) = arctan x + C .] (iii) Solve the IVP y ( x ) = 1 x , y ( 1) = 0 . Answer: the (maximal) solution is y ( x ) = log( x ) , ( x < 0) . Theorem. Suppose f is a continuous function on some interval I R . Let x 0 I and y 0 R . Then the IVP y = f ( x ) , y ( x 0 ) = y 0 , has ( i ) a unique ( ii ) solution on the whole interval ( iii ) I . The solution is given by the formula ( iv ) y ( x ) = y 0 + x x 0 f ( s ) ds, ( x I ) . Note. Several important concepts: (i) existence of a solution, (ii) uniqueness, (iii) interval/domain of a solution, (iv) solution formula. 1
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2 1.2. First order ODEs. The most general form is F ( x,y,y ) = 0 , where F is a given function of 3 real variables; x is an independent variable; y is the unknown function y = y ( x ); y = dy/dx is the usual ( ordinary ) derivative. Any differentiable function y ( x ) that satisfies the equation (i.e. if we substitute y ( x ) for y , then the equation becomes an identity) is called a solution . Given two numbers x 0 ,y 0 , the corresponding initial value problem (”IVP”) is to find a solution y ( x ) such that y ( x 0 ) = y 0 . Important : we always assume that a solution is defined on some open interval of the real line; the interval can be infinite, e.g. ( −∞ , + ). Example : we don’t consider the function y ( x ) = { log( x ) , x < 0 , log x + C, x > 0 , which is defined on the union of two intervals ( −∞ , 0) (0 , + ), to be a solution of the equation y = 1 /x. Otherwise, the IVP y ( 1) = 0 would have infinitely many solutions. A solution is maximal if it can not be extended to a solution on a larger interval. 1.3. Higher order ODEs. The general form of a 2-nd order ODE is F ( x,y,y ,y ,y ′′ ) = 0 , and solutions are twice differentiable functions (defined on some intervals) satisfying the equation. Solving for the second derivative, we obtain Newton’s equation with 1 degree of freedom (the motion of a particle m = 1 on the real line): ¨ x = f ( t,x, ˙ x ) . [Here we have changed the notation: now time t is an independent variable, the position x = x ( t ) is the unknown function, and dot means time derivative, so ¨ x is acceleration. The RHS is the ”force”; e.g. f = 0 for a free particle, f = kx for a Hooke’s law spring, etc.] Example: the general solution of Eq. ¨ x = 0 is x ( t ) = C 1 + C 2 t, and the IVP x ( t 0 ) = x 0 , ˙ x ( t 0 ) = ˙ x 0 , has a unique solution on R .
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3 The n -th order ODE is F
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This note was uploaded on 11/19/2010 for the course ANTH 122 taught by Professor 323 during the Spring '10 term at Centennial College.

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10Ma2aAnNotes1 - Ma2a (analytical) Fall 2010 PART I: FIRST...

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