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UT M 408D - HW04-solutions

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tapia jat4858 HW04 clark 52990 This print out should have 20 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 1 y 10 0 points 2 Use the direction field of the differential equation y y 5 to sketch a solution curve that passes through the point 0 1 4 2 2 x 2 x 2 x 1 y 2 2 2 x 2 y 2 2 5 2 y correct 2 2 2 2 x 2 y 2 2 6 2 3 y 2 2 2 2 2 x Explanation The vector field for y y 5 with a sample curve through our initial condition y 0 1 looks like the following tapia jat4858 HW04 clark 52990 Explanation Looking at the graph of y we can see that y 2 0 and y 2 5 4 Using this we can see the following choices cannot have our solution y 2 2 2 x y x y sin x y x y cos 2x y x x sin 2y 2 002 2 In each of these options y 2 6 0 Now we can notice that y 4 0 and y 4 0 This rules out 10 0 points Consider the following graph of y x y x y sin 2x y x y sin 4x y 6 In the first one y 4 0 while in the second y 4 0 Hence we are left with only 4 y y sin 2x Now we will check this answer by generating the direction field for y y sin 2x 2 2 2 y x 6 Choose the equation whose solution is graphed and satisfies the initial condition y 0 2 4 1 y y sin 2x 2 2 y y sin 2x correct 3 y x sin 2y 4 y y sin 4x 5 y y cos 2x 6 y y sin x 2 2 x As we can see from the above plot our graph of y is indeed a solution to the differential equation y y sin 2x tapia jat4858 HW04 clark 52990 3 003 10 0 points Consider the graph 6 6 y y 4 2 4 x 2 x 4 2 2 2 Using the above graph of y x choose the equation whose solution is graphed and satisfies the initial condition y 0 4 1 y y 3 x2 2 y y x 3 y y 2 x 4 y y x 1 5 y y 1 correct 6 y y 4 5 7 y y 2 x2 Explanation Let us begin by approximating the tangent line about y 0 4 2 2 2 From the above graph we can see the tangent line about y 0 has a slope of about 3 This tells us that y 0 3 Now only three of our answer choices match this property Those are y1 x y x 1 y2 x y x x 1 y3 x y x 4 5 Now we ll look similarly at x 4 Notice that y 4 0 and y 4 1 If we use this value in our remaining 3 answer choices we get y1 4 y 4 1 1 1 0 y2 4 y 4 4 1 1 4 1 4 y3 4 y 4 4 5 1 4 5 3 5 Now we can see only y1 x can be the derivative of our function To show that it works tapia jat4858 HW04 clark 52990 let us now look at a direction field of y1 6 4 2 x 4 2 table to display our values i 0 1 2 3 y 4 x 1 0 1 4 1 8 2 2 y 5 000 4 600 4 443 4 590 y 1 000 0 392 0 368 0 875 Notice in this table y is computed in each row using the values for y and y in the previous row while y in each row is determined using y and x in the current row Hence 2 y 2 2 y3 4 59 2 From this we can see that the given graph is indeed a solution to the differential equation y y 1 004 10 0 points Use Euler s method with step size 0 4 to estimate y 2 2 where y x is the solution of dy the initial value problem 2x2 y 9x2 dx y 1 5 1 y 2 2 4 942 2 y 2 2 3 353 3 y 2 2 2 343 4 y 2 2 1 934 5 y 2 2 4 59 correct Explanation In Euler s Method we predict the value of yi using yi 1 yi 1 and dx with the following equation yi yi 1 yi 1 dx Notice that y x y 9x2 2x2 y and y 1 5 Hence in this problem y0 5 y0 y 1 5 1 and dx 0 4 In order to visualize this better let us use the following 005 10 0 points Use Euler s method with step size 0 1 to compute the approximate y values y1 y2 and y3 of the solution of the initial value problem y 1 6x 3y y 1 2 1 y1 2 200 y2 1 800 y3 2 061 2 y1 2 100 y2 1 770 y3 2 041 3 y1 2 100 y2 2 370 y3 2 588 4 y1 2 100 y2 2 230 y3 2 381 correct 5 y1 2 200 y2 1 880 y3 2 364 Explanation In Euler s Method we predict the value of yi using yi 1 yi 1 and dx with the following equation yi yi 1 yi 1 dx Recall that y x y 1 6x 3y and y 1 2 Hence in this problem y0 2 y0 y 1 2 1 and dx 0 1 In order to visualize this better let us use the following table to display our values i 0 1 2 3 x 1 0 1 1 1 2 1 3 y 2 000 2 100 2 230 2 381 y 1 000 1 300 1 510 1 657 tapia jat4858 HW04 clark 52990 Notice in this table y is computed in each row using the values for y and y in the previous row while y in each row is determined using y and x in the current row Hence 5 Hence y0 x 24 3 ln x 1 4 in which case y1 2 100 y2 2 230 y3 2 381 006 y0 e 24 3 1 4 131 4 10 0 points 007 If y0 satisfies the equations 4 3 dy 0 dx x y3 y 1 2 If y y0 x is the solution of the differential equation for x y 0 find the value of y0 e p 1 y0 e 51 3 1 4 2 y0 e 13 10 0 points 25 x2 dy 2xy 0 dx which satisfies y 5 9 find the value of y0 0 correct 3 y0 e 191 4 1 y0 0 9e10 4 y0 e 101 4 2 y0 0 9e 10 3 y0 0 …


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