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311_pdfsam_math 54 differential equation solutions odd

# 311_pdfsam_math 54 differential equation solutions odd - (1...

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Exercises 5.5 (d) For the system found in (a), f ( y, v ) = v, g ( y, v ) = 0 , if | y | < 1 , v = 0 , y + sign( y ) , if | y | ≥ 1 , v = 0 , y sign( v ) , if v = 0 . Since f ( y, v ) = 0 v = 0 and g ( y, 0) = 0 , if | y | < 1 , y + sign( y ) , if | y | ≥ 1 , we consider two cases. If y < 1, then g ( y, 0) 0. This means that any point of the interval 1 < y < 1 is a critical point. If | y | ≥ 1, then g ( y, 0) = y + sign( y ) which is 0 if y = ± 1. Thus the critical point set is the segment v = 0, 1 y 1. (e) According to (c), the mass released at (7 . 5 , 0) goes in the lower half plane from right to left along a semicircle centered at (1 , 0). The radius of this semicircle is 7 . 5 1 = 6 . 5, and its other end is (1 6 . 5 , 0) = ( 5 . 5 , 0). From this point, the mass goes from left to right in the upper half plane along the semicircle centered at ( 1 , 0) and of the radius 1 ( 5 . 5) = 4 . 5, and comes to the point ( 1 + 4 . 5 , 0) = (3 . 5 , 0). Then the mass again goes from right to left in the lower half plane along the semicircle centered at
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Unformatted text preview: (1 , 0) and of the radius 3 . 5 1 = 2 . 5, and comes to the point (1 2 . 5 , 0) = ( 1 . 5 , 0). From this point, the mass goes in the upper half plane from left to right along the semicircle centered at ( 1 , 0) and of the radius 1 ( 1 . 5) = 0 . 5, and comes to the point ( 1 + 0 . 5 , 0) = ( . 5 , 0). Here it comes to rest because | . 5 | &amp;lt; 1, and there is not a lower semicircle starting at this point. See the colored curve in Figure B.42 of the text. EXERCISES 5.5: Coupled Mass-Spring Systems, page 284 1. For the mass m 1 there is only one force acting on it; that is the force due to the spring with constant k 1 . This equals k 1 ( x y ). Hence, we get m 1 x = k 1 ( x y ) . 307...
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