302_pdfsam_math 54 differential equation solutions odd

302_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 Case 1 : y>x and y< 1 . (This region is above the line y = x but below the line y =1.) In this case, y x> 0 but y 1 < 0. Thus, dx/dt =( y x )( y 1) < 0. Hence, x ( t ) will be decreasing here. Therefore, the flow along the trajectories will be from right to left and so the movement is clockwise. Case 2 : y>x and y> 1 . (This region is above the lines y = x and y = 1.) In this case, we see that y x> 0and y 1 > 0. Hence, dx/dt =( y x )( y 1) > 0. Thus, x ( t ) will be increasing and the flow along the trajectories in this region will still be clockwise. Case 3 : y<x and y< 1 . (This region is below the lines y = x and y = 1.) In this case, y x< 0and y 1 < 0. Thus, dx/dt > 0andso x ( t ) is increasing. Thus, the movement is from left to right and so the flow along the trajectories will be counterclockwise. Case 4 : y<x and y> 1 . (This region is below the line y = x but above the line y =1.) In this case, y x< 0and y 1 > 0. Thus, dx/dt < 0andso x ( t ) will be decreasing here. Therefore, the flow is from right to left and, thus, counterclockwise here also.
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Unformatted text preview: Therefore, above the line y = x the ow is clockwise and below that line the ow is counter-clockwise. See Figure B.31 in the answers of the text. 15. From Denition 1 on page 266 of the text, we must solve the system of equations given by 2 x + y + 3 = 0 , 3 x 2 y 4 = 0 . By eliminating y in the rst equation we obtain x + 2 = 0 and by eliminating x in the rst equation we obtain y + 1 = 0 . Thus, we observe that x = 2 and y = 1 will satisfy both equations. Therefore ( 2 , 1) is a critical point. From Figure B.32 in the answers of the text we see that all solutions passing near the point ( 2 , 1) do not stay close to it therefore the critical point ( 2 , 1) is unstable. 298...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.

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