2006a_x2b_sols - Spring 2006 Math 152 Exam 2B Solutions c...

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Spring 2006 Math 152 Exam 2B: Solutions Mon, 27/Mar c 2006, Art Belmonte 1. (e) The arc length of the curve r ( t ) = [ x ( t ), y ( t ) ] = t 2 , t 2 + t , 0 t 1 , is represented by the following integral. L = b a r 0 ( t ) dt = 1 0 k [2 t , 2 t + 1] k dt = 1 0 4 t 2 + 4 t 2 + 4 t + 1 dt = 1 0 1 + 4 t + 8 t 2 dt 2. (d) The differential equation dy dx = xy 2 + x 2 y = xy ( y + x ) is not separable. Other choices are separable, as follows. (b) Rewrite dy dx = sin x cos y as sec y dy = sin x dx . (c) Rewrite dy dx = xy + x 2 y = y x + x 2 as 1 y dy = x + x 2 dx . (e) Rewrite dy dx = e x + y = e x e y as e - y dy = e x dx . (a) Repeated factoring gives dy dx = 1 + x + y + xy dy dx = ( 1 + x ) + y ( 1 + x ) dy dx = ( 1 + x ) ( 1 + y ) , whence 1 1 + y dy = ( 1 + x ) dx . 3. (c) The linear differential equation y 0 + ( 2 sin 2 x ) y = cos 4 x is already in standard linear form (SLF). Accordingly, an integrating factor is μ = exp ( 2 sin 2 x dx ) = e - cos 2 x . 4. (e) Let y = y ( t ) be the amount of salt in the tank at time t . The classical balance law gives dy dt = rate in - rate out dy dt = 0 . 1 kg L × 10 L min - y kg 100 L × 10 L min dy dt = 1 - y 10 kg min . Since the tank initially contains pure water, we have y ( 0 ) = 0 kg of salt in the tank at the start. Therefore, dy dt = 1 -
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