ch02_6 - Ch 2.6: Exact Equations & Integrating Factors...

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Ch 2.6: Consider a first order ODE of the form Suppose there is a function ψ such that and such that ( x,y ) = c defines y = φ ( x ) implicitly. Then and hence the original ODE becomes Thus ( x,y ) = c defines a solution implicitly. In this case, the ODE is said to be exact . 0 ) , ( ) , ( = + y y x N y x M ) , ( ) , ( ), , ( ) , ( y x N y x y x M y x y x = = [ ] ) ( , ) , ( ) , ( x x dx d dx dy y x y y x N y x M = + = + [ ] 0 ) ( , = x x dx d
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Suppose an ODE can be written in the form where the functions M , N , M y and N x are all continuous in the rectangular region R : ( x , y ) ( α , β ) x ( γ , δ ). Then Eq. (1) is an exact differential equation iff That is, there exists a function ψ satisfying the conditions iff M and N satisfy Equation (2). ) 1 ( 0 ) , ( ) , ( = + y y x N y x M ) 2 ( ) , ( ), , ( ) , ( R y x y x N y x M x y 2200 = ) 3 ( ) , ( ) , ( ), , ( ) , ( y x N y x y x M y x y x = = Theorem 2.6.1
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Example 1: Exact Equation (1 of 4) Consider the following differential equation. Then and hence From Theorem 2.6.1, Thus 0 ) 4 ( ) 4 ( 4 4 = - + + - + - = y y x y x y x y x dx dy y x y x N y x y x M - = + = 4 ) , ( , 4 ) , ( exact is ODE ) , ( 4 ) , ( = = y x N y x M x y y x y x y x y x y x - = + = 4 ) , ( , 4 ) , ( ψ ( 29 ) ( 4 2 1 4 ) , ( ) , ( 2 y C xy x dx y x dx y x y x x + + = + = =
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Example 1: Solution (2 of 4) We have and It follows that Thus By Theorem 2.6.1, the solution is given implicitly by y x y x y x y x y x - = + = 4 ) , ( , 4 ) , ( ψ
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ch02_6 - Ch 2.6: Exact Equations & Integrating Factors...

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