chapter2.page18

chapter2.page18 - x = f ( t,x ) is to draw a segment of...

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18 1. EXPLICITLY SOLVABLE FIRST ORDER DIFFERENTIAL EQUATIONS In Example ?? , c = 1 is a sink, c = - 2 , 3 are source. From phase diagram and the sign of derivative of x 00 = df ( x ) dt = df ( x ) dx dx dt = f ( x ) df ( x ) dx , we can get an very good picture about the behavior of the solutions. Example 5.2 . For x 0 = f ( x ) = ( x - 1)( x + 2)( x + 3) , we have f ( x ) df ( x ) dx = ( x - 1)( x +2)( x +3)(3 x 2 - 4 x - 5) , which has the following sign distribution, (a) d 2 x ( t ) dt 2 > 0 on intervals, ( - 2 , 2 3 - 1 3 19) , (1 , 2 3 + 1 3 19) , and (3 , ) . (b) d 2 x ( t ) dt 2 < 0 on intervals, ( , - 2) , ( 2 3 - 1 3 19 , 1) , and ( 2 3 + 1 3 19 , 3) . With this information and the phase diagram we have the following picture of typical solutions for x 0 = ( x - 1)( x + 2)( x - 3) , Figure 11. Typical solutions for x 0 = ( x - 1)( x + 2)( x + 3) 5.2. Vector field diagram. We know that the derivative x 0 ( t 0 ) of x ( t ) at t 0 is the slope of the tangent line to the graph of x ( t ) at the point ( t 0 ,x ( t 0 )) , and the equation of the tangent line is y - x ( t 0 ) = x 0 ( t 0 )( t - t 0 ) . The idea of vector field for an ODE
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Unformatted text preview: x = f ( t,x ) is to draw a segment of tangent line to many chosen points ( t ,x ) with slope x ( t ) = f ( t ,x ) . When we draw enough such line segments, patterns shall emerge and we shall have a clear idea(with some training) of the behavior of solutions to the equation. The following diagram shows the vector eld of x = sin ( t ) . Clearly we see the wave pattern of typical solution x ( t ) =-cos ( t ) + C. Here two diagrams are shown, the right one is provided for novel eyes, which has solutions curves superimposed on the vector eld. For complicated vector eld only trained eyes will see the subtle patterns...
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This note was uploaded on 10/04/2011 for the course MAP 4231 taught by Professor Dr.han during the Spring '11 term at UNF.

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