MATH 2300 Practice-Exam 2 (Spring 2007)

# MATH 2300 Practice-Exam 2 (Spring 2007) - MATH...

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MATH 2300 (Calculus 2), Spring 2007 — 2nd Test Practice Exam — SOLUTIONS 1 MATH 2300 (Calculus 2) Spring 2007 2nd midterm PRACTICE exam — SOLUTIONS 1. [** pts] Solve d y d x + 1 x y = cos x by the method of integrating factors. SOLUTION From the general form of the equation d y d x + p ( x ) y = q ( x ) we obtain p ( x ) = 1 x and q ( x ) = cos x . Therefore μ ( x ) = exp p ( x ) d x = exp 1 x d x = exp (ln x ) = x and d y d x + 1 x y = cos x = d d x ( x y ( x )) = x cos x = d d x ( x y ( x )) d x = x cos x d x = x y ( x ) = C + cos x + x sin x = y ( x ) = C x + cos x x sin x 2. [** pts] Solve d y d x = 3 x 2 2 y + sin y by the method of separation of variables. Note: Leave the solution in implicit form . Do not attempt to solve the equation for y ! SOLUTION d y d x = 3 x 2 2 y + sin y = (2 y + sin y ) d y = 3 x 2 d x = (2 y + sin y ) d y = 3 x 2 d x = y 2 - cos y = x 3 + C 3. [** pts] A tank contains 10 Kg of salt dissolved in 1000 L of water. Fresh water enters into the tank at a rate of 2 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. How much slat remains in the tank after two hours? SOLUTION Let S be the amount of salt in the tank measured in Kilograms, then d S d t = [rate salt] i n - [rate salt] o ut , S (0) = 10 Kg with [rate salt] i n = [rate brine] i n × [salt concentration in brine] [rate salt] o ut = [rate fluid] i n × [salt concentration fluid] Therefore [rate salt] i n = 2 L min × 0 Kg L = 0 Kg min [rate salt] o ut = 2 L min × S Kg 1000 L = S 500 Kg min

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MATH 2300 (Calculus 2), Spring 2007 — 2nd Test Practice Exam — SOLUTIONS 2 The di ff erential equation that models the problem is then d S d t = - S 500 , S (0) = 10 with [ S ] = Kg and [ t ] = min.
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