Math250_Chapter2

# Math250_Chapter2 - Chapter 2 First Order Different...

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Chapter 2: First Order Different Equations Yanxiang Zhao Department of Mathematics, The Pennsylvaina State University January 23, 2011

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Outline 1 First Order Linear Equations: Integrating Factors Equations we are able to solve General linear equations 2 1st Order Nonlinear Equations I: Separable ODE 3 1st Order Nonlinear Equations II: Non-separable ODE Exact Equations Inexact equations 4 Existence and Uniqueness of the Solution of 1st ODE 5 Applications Autonomous Equations Mixture Problems
Type I: d d t f ( t ) = g ( t ) or f ( t ) = g ( t ) , g ( t ) is known Example 1: Solve d d t f ( t ) = sin t . f ( t ) = sin t d t = cos t + C . Example 2: Solve d d t f ( t ) = e π t . f ( t ) = e π t d t = 1 π e π t + C . Example 3: Solve d d t f ( t ) = g ( t ) , g ( t ) is known. f ( t ) = g ( t ) d t = t t 0 g ( s ) d s .

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Type II: d d t μ ( t ) · f ( t ) = g ( t ) or μ ( t ) · f ( t ) = g ( t ) , μ ( t )& g ( t ) are known Example 1: Solve d d t t 2 f ( t ) = cos t . t 2 · f ( t ) = cos t d t = sin t + C f ( t ) = 1 t 2 ( sin t + C ) . Example 2: Solve d d t μ ( t ) · f ( t ) = g ( t ) . μ ( t ) · f ( t ) = g ( t ) d t f ( t ) = 1 μ ( t ) g ( t ) d t
Type III: μ ( t ) d d t f ( t ) + f ( t ) d d t μ ( t ) = g ( t ) , μ ( t )& g ( t ) are known Notice that μ ( t ) d d t f ( t ) + f ( t ) d d t μ ( t ) = d d t μ ( t ) · f ( t ) Type II = Type III Example 1: Solve t 4 f ( t ) + 4 t 3 f ( t ) = e t . t 4 f ( t ) + 4 t 3 f ( t ) = e t t 4 f ( t ) = e t t 4 f ( t ) = e t + C f ( t ) = 1 t 4 ( e t + C ) .

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Type IV: d d t f ( t ) = g · f ( t ) , g is a constant. Example 1: Solve d d t f ( t ) = f ( t ) . d f ( t ) d t = f ( t ) d f ( t ) f ( t ) = d t d f ( t ) f ( t ) = d t ln f ( t ) = t + C f ( t ) = e t + C = C · e t . Example 2: Solve d d t f ( t ) = g · f ( t ) . d f ( t ) d t = g · f ( t ) d f ( t ) f ( t ) = g d t d f ( t ) f ( t ) = g d t ln f ( t ) = g t + C f ( t ) = e g t + C = C · e g t .
Type V: d d t f ( t ) = g ( t ) · f ( t ) , g ( t ) is a given function. d f ( t ) d t = g ( t ) · f ( t ) d f ( t ) f ( t ) = g ( t ) d t d f ( t ) f ( t ) = g ( t ) d t ln f ( t ) = g ( t ) d t f ( t ) = e g ( t ) d t .

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General Form: d f d t + p ( t ) f = q ( t ) , p & q are given functions. p ( t ) = 0 Type I. q ( t ) = 0 Type V. If neither p ( t ) nor q ( t ) is zero, how can we solve the equation? Compare μ ( t ) d d t f ( t ) + d d t μ ( t ) f ( t ) = g ( t ) d d t f ( t ) + p ( t ) f ( t ) = q ( t ) μ ( t ) d d t f ( t ) + μ ( t ) p ( t ) f ( t ) = μ ( t ) q ( t )
Method of Integrating Factors To change the general form into solvable Type III, we set d d t μ ( t ) = p ( t ) μ ( t ) ( Type V ) μ ( t ) = e p ( t ) d t Then general form can be solved, d f d t + p ( t ) f = q ( t ) μ ( t ) d f d t + μ ( t ) p ( t ) f = μ ( t ) q ( t ) μ ( t ) d f d t + d d t μ ( t ) f = μ ( t ) q ( t ) μ ( t ) f ( t ) = μ ( t ) q ( t ) f ( t ) = 1 μ ( t ) μ ( t ) q ( t ) d t = 1 μ ( t ) t t 0 μ ( s ) q ( s ) d s

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Method of Integrating Factors: Examples d y d t + 1 t y = t 3 ( t > 0 ) . d y d x + y tan x = sin ( 2 x )
Separable ODE A separable ODE is the one which can be rewritten as M ( x ) d x = N ( y ) d y , or M ( x ) d x + N ( y ) d y = 0 . d y d x = x 2 1 y 2 , y + y 2 sin x = 0 , xy = ( 1 y 2 ) 1 / 2 are separable. y = 1 + xy , d y d x = sin ( xy ) are non-separable.

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Solving Separable ODE: Separation of Variables M ( x ) d x + N ( y ) d y = 0 M ( x ) d x + N ( y ) d y = 0 M ( x ) d x = N ( y ) d y M ( x ) d x = N ( y ) d y .
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