week_03_2.4 - dy t y dt t y ) 6 3 ( ) 3 4 ( 2 2 4 3 3 3 y y...

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MATH2860U: Chapter 2 cont… 1 FIRST ORDER DIFFERENTIAL EQUATIONS cont… Exact Equations (Section 2.4 of Zill and Cullen, pg. 62) Recall: So far, we’ve considered two types of 1 st order equations that we learned how to solve. Now, we’ll consider one more class of linear first-order equations which also has a well- defined method of solution. Again, let’s begin with a simple example: 0 1 2 3 2 ) 3 ( 2 2 dy x xy dx y xy Question: What did we do in the above example and why did it work?
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MATH2860U: Chapter 2 cont… 2 Definition and Theorem: Let the functions M(x, y) and N(x, y) be continuous and have continuous first partial derivatives in the rectangular region d y c b x a R , : . Then 0 ) , ( ) , ( dy y x N dx y x M is an exact differential equation in R if and only if x N y M at each point of R . [i.e. it is only under this condition that there exists a function f that satisfies ) , ( y x M x f and ) , ( y x N y f ] Examples: Is the differential equation exact or not? 0 cos ) sin (
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Unformatted text preview: dy t y dt t y ) 6 3 ( ) 3 4 ( 2 2 4 3 3 3 y y y x x y x Summary of Overall Technique: To solve an exact first-order equation: 1. Put the equation in the form ) , ( ) , ( dy y x N dx y x M 2. Start with M x f and integrate with respect to x to find f . f will involve arbitrary function of y 3. Differentiate f from step 2 with respect to y and set this equal to N . Can now solve for arbitrary function of y 4. C y x f ) , ( is the solution. MATH2860U: Chapter 2 cont… 3 Example: ) 3 ( ) 1 6 ( 2 dy y x dx xy Example: Solve x y y x y 2 ) 2 ( subject to 7 ) ( y MATH2860U: Chapter 2 cont… 4 Example: Solve dy y x dx x y 1 2 2 subject to 2 ) 1 ( y Example: ) sin ( ) 3 cos 2 ( 2 3 2 dx dy y y x x y x y x...
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This note was uploaded on 10/29/2010 for the course DEFFERENTI 2080 taught by Professor Kidnan during the Spring '10 term at UOIT.

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week_03_2.4 - dy t y dt t y ) 6 3 ( ) 3 4 ( 2 2 4 3 3 3 y y...

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