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# 16bf10lec16 - Math 16b Thomas Scanlon Autumn 2010 Thomas...

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Unformatted text preview: Math 16b Thomas Scanlon Autumn 2010 Thomas Scanlon () Math 16b Autumn 2010 1 / 38 Autonomous di erential equations A di erential equation of the form y = g ( y ) where g is a function of a single variable is called an autonomous ( rst-order, ordinary) di erential equation. Scholium In principle, we can solve autonomous equations using the method separation of variables as e G ( y ( T ))- e G ( y ( )) = Z T y ( t ) dt g ( y ( t )) = Z T dt = T where e G ( x ) = 1 g ( x ) . Thomas Scanlon () Math 16b Autumn 2010 2 / 38 Relating the ( x , z )-graph of z = g ( x ) and the ( t , y )-graph of y = y ( t ) From the equation y = g ( y ) we can relate properties of the graph of z = g ( x ) to the shape of the solution curves to the di erential equation. Thomas Scanlon () Math 16b Autumn 2010 3 / 38 Example Consider y = g ( y ) = y 2- 2 y Figure: z = g ( y ) = y 2- 2 y Figure: y = g ( y ) Thomas Scanlon () Math 16b Autumn 2010 4 / 38 Constant solutions If g ( c ) = and y = g ( y ) and y ( a ) = c for some number a , then y ≡ a . Thomas Scanlon () Math 16b Autumn 2010 5 / 38 Sign of g and direction of y If g ( c ) > and y ( a ) = c then y is increasing at a . Likewise, if g ( c ) < 0 and y ( a ) = c , then y is decreasing at a . Thomas Scanlon () Math 16b Autumn 2010 6 / 38 In ection points If c is a relative extremum of g and y ( a ) = c , then because y ( a ) = g ( y ( a )) = g ( c ) there is an in ection point on the graph of y = y ( t ) at ( a , c ) . Figure: z = g ( y ) = y 2- 2 y Figure: y = g ( y ) Thomas Scanlon () Math 16b Autumn 2010 7 / 38 Direction between zeros If g ( y ) is a continuous function (which we shall always assume) and a < b are two consecutive zeros, then the sign of g is constant between a and b . Consequently, if y = g ( y ) and a < y ( ) < b , then y is always increasing if g ( y ( )) > 0 and is always decreasing if g ( y ( )) < 0. Thomas Scanlon () Math 16b Autumn 2010 8 / 38 Returning to the example of y = g ( y ) = y 2- 2 y The zeros of g ( y ) = y 2- 2 y = y ( y- 2 ) are y = 0 and y = 2. Hence, the constant solutions are y ≡ 0 and y ≡ 2. Di erentiating, we nd g ( y ) = 2 y- 2 and g 00 ( y ) = 2 so that the only relative extremum of g is a minimum at y = 1. Hence, any solution of y = g ( y ) will have an in ection point at ( a , 1 ) if y ( a ) = 1. g ( y ) > 0 for y < 0 and y > 2 while g ( y ) < 0 for 0 < y < 2. Hence, solutions with y ( ) < 0 or y ( ) > 2 will be increasing while those with < y ( ) < 2 will be decreasing. Thomas Scanlon () Math 16b Autumn 2010 9 / 38 Example: y = g ( y ) = y 3- 3 y 2 + y + 1 Consider y = g ( y ) = y 3- 3 y 2 + y + 1 Figure: z = g ( y ) = y 3- 3 y 2 + y + 1 Figure: y = g ( y ) Thomas Scanlon () Math 16b Autumn 2010 10 / 38 Example: y = g ( y ) = y 2 ( e y- 2 ) Consider...
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