Math 117 Practice Midterm 3 Please work out each of the given problems.  Credit will be based on the steps that you show towards the final solution.  Show your work. Problem 1  Solve the differential Equation         xy' - 2x  =  x2     y(1)  =  5 Solution Solve for y':         xy' = 2x + x2          y'  =  2 + x Now integrate to get                             x2         y  =  2x +             + C                             2 Problem 2 Determine if                             1         y  =  cos t -       e-t cos t                             5 is a solution to the differential equation         y''' - y'' + y' - y  =  e-t sin t   Solution We take three derivative of y:                               1         y'  =  -sin t +       (e-t cos t + e-t sin t)                               5                                 1         y''  =  -cos t +       (-e-t cos t - e-t sin t - e-t sin t + e-t cos t )                                 5                            1         =  -cos t -        (2e-t sin t )                            5                                2         y'''  =  sin t +       (e-t sin t - e-t cos t)                                5 Now plug the these derivatives into the differential equation:                      2                                               1         sin t +       (e-t sin t - e-t cos t) + cos t +        (2e-t sin t)                     5                                                5                       1                                                 1         - sin t +       (e-t cos t + e-t sin t) - cos t +        e-t cos t                       5                                                 5             2e-t sin t - 2e-t cos t  + 2e-t sin t + e-t cos t + e-t sin t + e-t cos t       =                                                                                                                                                                   5               5e-t sin t       =                      =  e-t sin t                   5   Problem 3  The rate of growth in customers for a small business is proportional to the product of the number of customers who eat at the restaurant divided by the total number of years that the restaurant has been in business.  After one year in business the restaurant averages 20 customers per day, and after two years in business the restaurant averages 25 customers per day.  Approximately how many customers can the restaurant expect to have per day after three years in business?   Solution We write the differential equation         dx                 kx                     =                         dt                   t Separating the variables gives         dx                 kdt                     =                           x                    t Now integrate both sides to get         ln x  =  k ln t  +   C When t  =  1, x  =  20 so that         ln 20  =  k ln 1  +  C  =  C When t  =  2, x  =  25 so that         ln 25  =  k ln 2  +  ln 20         k ln 2  =  ln 25 - ln 20  =  ln 1.25 so that                   ln 1.25         k  =                   =  0.322                     ln 2 We have          ln x  =  0.322 ln t  + ln 20 Plugging in 3 for t gives         ln x  =  0.322 ln 3  + ln 20  =  3.35 Exponentiating both sides gives          x  =  e3.35  =  29 We conclude that there will be 29 customers per day in three years.   Problem 4 Use the sixth degree Taylor polynomial for         f(x)  =  2 cos x to approximate .  Hint:  Use the fact that           Solution We find the first six derivatives of f and evaluate them at 0         f(0)  =  2 cos 0  =  2         f '(0)  =  -2 sin 0  =  0         f ''(0)  =  -2 cos 0  =  -2         f '''(0)  =  2 sin 0  =  0         f (iv)(0)  =  2 cos 0  =  2         f iv)(0)  =  -2 sin 0  =  0         f (vi)(0)  =  -2 cos 0  =  -2 Plugging it into the Taylor polynomial formula gives                                    -2                  2                -2         2 cos x  =  2  +            x2 +             x4   +          x6                                                       2!                 4!                6!                               1                 1         =  2  - x2 +           x4   -           x6                                                12              360 Now plug in p/6 to get         1.73205   Problem 5 Find the Taylor series center at x  =  c for the given function and find its radius of convergence.                         x2            f(x)  =                          (c  =  0)                       1 + x3  Solution We know that           We can write f(x) as                      1          x2                                 1 - (-x3) so                       Problem 6 Use Newton's Method to approximate the intersection of the curves         y  =  sin x        and        y  =  1 - x   Solution We find the root of the difference         f(x)  =  sin x - (1 - x)  =  x - 1 + sin x We find         f '(x)  =  1 + cos x  and find                                      f (x)       N(x)  =        x  -                                                         f '(x)                                       x - 1 + sin x       N(x)  =        x  -                                                                       1 + cos x Now our first guess is 0,          N(0)  =  2        N(2)  =  0.5                 N(0.5)  =  0.51095795        N(0.51095795)  =  0.51097343         N(0.51097343)  =  0.51097343 Hence the root occurs at x  =  0.51097343