342_pdfsam_math 54 differential equation solutions odd

# 342_pdfsam_math 54 differential equation solutions odd -...

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Chapter 5 x 0 4 = x 5 , x 0 5 = x 6 , x 0 6 = x 2 x 3 . 13. With the notation used in (1) on page 264 of the text, f ( x, y )=4 4 y, g ( x, y )= 4 x, and the phase plane equation (see equation (2) on page 265 of the text) can be written as dy dx = g ( x, y ) f ( x, y ) = 4 x 4 4 y = x y 1 . This equation is separable. Separating variables yields ( y 1) dy = xdx Z ( y 1) dy = Z xdx ( y 1) 2 + C = x 2 or x 2 ( y 1) 2 = C ,where C is an arbitrary constant. We Fnd the critical points by solving the system f ( x, y )=4 4 y =0 , g ( x, y )= 4 x =0 y =1 , x =0 . So, (0 , 1) is the unique critical point. ±or y> 1, dx dt =4(1 y ) < 0 , which implies that trajectories ﬂow to the left. Similarly, for
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Unformatted text preview: y < 1, trajectories ﬂow to the right. Comparing the phase plane diagram with those given on ±igure 5.12 on page 270 of the text, we conclude that the critical point (0 , 1) is a saddle (unstable) point. 15. Some integral curves and the direction Feld for the given system are shown in ±igure 5-B. Comparing this picture with ±igure 5.12 on page 270 of the text, we conclude that the origin is an asymptotically stable spiral point. 338...
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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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