Math 2214 Solution Test 1D Spring 2015

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1 Math 2214 Solution Test 1D Spring 2015 Problem 1: A 600 gallon open top tank initially holds 300 gallons of fresh water. At t = 0, a brine solution containing 3 lbs of salt per gallon is poured into the tank at a rate of 4 gallons per minute while a well stirred solution leaves the tank at a rate of 2 gallons per minute. a) Set up the differential equation that models this scenario. (do not solve) b) At what time will overflow occur? Solution: (8pts) a) Set up the differential equation that models this scenario. (do not solve) 12, (0) t Solution: (4pts) c) At what time will overflow occur? Volume = = 600, so t = 150 min. Problem 2: time t is given to be Using Newton s law of cooling we have the temperature of a given object at any T t ( ) e F a) What was the initial temperature of the object? b) What was the limiting temperature for the object? Solution: (8pts) a) What was the initial temperature of the object? At t = 0, T(0) = e 0 = = 60 b) What was the limiting temperature for the object? Limit t ( e ) I would like to comment here that is not number value and you should not try to use this notation in your calculations. Also, when the limit is zero then you should write the zero in the appropriate place.

2 Problem 3: Use Euler s method (not Euler s formula) to evaluate y(4) when you are given the y following non-linear differential equation. y, with the initial value y(2) = 1 and h = 1. 2 t Discuss if you think the estimate would be close. Explain why or why not. Solution: (16pts) the method involves finding a point, slope and the equation of the tangent line. Then evaluate the point of interest. y 1 Part 1) We are given the point (2, ) and the slope is y 2, and the equation is t 4 y 1 (1 4)( t 2) to simplify to y ( t 4) (1 2) and y(3) = 54 Part 2) We now have the point (3, 54) and the slope is 5 4 5, (3) 36 y and the equation is 2 y 5 4 (5 36)( t 3) to simplify to y (5 t 36) 5 6 and y(4) = 2518 In general the step value between the t values is probably too large to be a close estimation Problem 4: A body of weight 160 lbs is dropped with an initial velocity of 4 ftsec. The resistance constant due to air is given to be k = 10. Using the given differential equation, find the equation for velocity at any time t. mv mg kv Solution: (16pts) Since the weight is given to be 160 lbs we know that mg = m(32) = 160 so m = 5. Using mv mg kv and inserting appropriate values we have 5v v 32 2v 2 t so [ ve ] dt 32e dt v v 2v 32 find the integrating factor to be u = e ve 16e C v 16 Ce using v(0) = 4 (since velocity is negative) 4 16 C and C = 12, so v e

3 Problem 5: Given the differential equation y 1 cos t e Solutions: (10pts) Consider the given solution: y 1 e ye cost e cost cost [ ye ] [ e 1] cos t cos t cost y p t y g t ( ) ( ) with a particular solution, find p(t) and g(t). Show all work and explain reasoning. 1 Take the derivative on both sides to get the following: cost cost cos t y e sin( t) e y sin( t) e 0 [ y sin( t) y sin( t)] e Clearly the integrating factor is u(t) = cost e, p(t) = -sin(t) and g(t) = sin(t) Problem 6: Given the following direction field, find the possible autonomous differential equation associated with it and clearly justify your answer. 1) y ( y1)( y 2) 2) y ( y 2)( y1) 3) y (1 y)( y 2) Solution (10 pts) We first notice the equilibrium isoclines are found at y = 1 and y = -2. By investigating the equations above we notice that when we set y = 0 we have the following results: 1) 0 = (y +1)(y + 2) and so y = -2 and y = - 1 which matches up with the equilibrium isocline y = -2 but not y = 1.

4 2) 0 = (-y 2)(y - 1) and so y = -2 and y = 1 which matches up with both equilibrium isoclines y = 1 and y = -2. 3) 0 = - (1- y)(y + 2) and so y = -2 and y = 1 which matches up with both equilibrium isoclines y = 1 and y = -2. Since 2) and 3) are each a possibility, we will check other isoclines in the direction field to see if slopes match. We will test the slope when y = 0. So in 2) y = ( - 0 2)(0-1) = 2 which is a positive slope that corresponds with the graph. Now check 3) so y = - (1-0)(0 + 2) = -2 which is negative and dose not match. We should test 2) for one more isocline. Let y = -4 to get y = (-(-4)-2)(-4 1) = - 10 which is a negative slope. Therefore 2) is the DE that generates this direction field. Problem 7: Demonstrate how to use the Existence and Uniqueness Theorem to find the largest rectangular region that guarantees a unique solution for the non-linear differential equation y ( t 3) sec( y), with initial value y( 4) = 1. Sketch the region. Solution:(12pts) sec( y) f ( t, y) y t 3 f sec( y) tan( y) y t3 notice 3 < t < and 2 y 2 for the initial value y( 4) =1 To sketch the region you need to plot the initial value point and enclose it in the rectangular region that is bounded by the intervals given above

5 Problem 8: differential equation In a certain agency a computer virus is spreading at a rate modeled by the k where the initially there were 20 computers infected. In a day a 2 total of 100 were infected. Use the method separation of variables to solve for the number computers infected at any given time t. Solution: (16pts) 2 2 (0) = kt C and with initail values this gives us C = kt and with value t = 1day and = 100, k = so k d k dt d 2 dt kdt 1 1 t 20 25

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