Hello please help, our lesson is about Limited Growth in Calculus using differential equations.

Example 7 A new cell phone is introduced. The company estimates they will sell 200 thousand phones. After 1 month they have sold 20 thousand. How many will they have sold after 9 months? In this case there is a maximum amount of phones they expect to sell, so M = 200 thousand. Modeling the sales, y: in thousands of phones, we can write the differential equation y' = k:(200 - y). Since it was a new phone, y(0) = 0. We also know the sales after one month, y(1) = 20. Solving the differential equation: dy _k(200 - y) dt dy =k dt Separate the variables. 200 - y - In 200 - yl =kt + C Integrate both sides. On the left use the substitution. In 200 y -kt+C Multiply both sides by -1, and exponentiale both sides. 200 - y =Be *t Simplify. y =Ae -* + 200 Subtract 200, divide by -1, and simplify. Using the initial condition y(0) = 0, 0 = Ae *(0) + 200, so 0 = A + 200, giving A = -200. Using the value y(1) = 20: 20 - 200e () + 200 = 180 =0.9 = ek -200 Subtract 200 and divide -200. In (0.9) =In(e *) = -k Take the In of both sides. k = - In(0.9) = 0.105 Divide by -1. As a quick sanity check, this value is positive as we would expect, indicating that the sales are growing over time. We now have the equation for the sales of phones over time: A = -200e-0.105t + 200. Finally, we can evaluate this at t = 9 to find the sales after 9 months: A(9) = -200e -19() + 200 # 122.26 thousand phones