Consider the differential equation dy/dx = 1/3x(y-2)^2. (a) a slope field for the given differential equation is shown below. sketch the solution curve that passes through the point (0, 2), and sketch the solution curve that passes through the point (1, 0). (b) let y = f(x) be the particular solution to the given differential equation with initial condition f(1) = 0. write an equation for the line tangent to the graph of y = f(x) at x = 1. use your equation to approximate f(0.7). (c) find the particular solution y = f(x) to the given differential equation with initial condition f(1) = 0.(this was on the no-calculator section of the recently-released ap calculus ab 2018 exam so i appreciate it if you tried to limit calculator usage)

A) Judging by the slope field, it would appear the solution passing through (0, 2) would just be the line . This holds up for the given ODE, since if , then both sides of the ODE reduce to 0.

Since we can surmise that is an equilibrium for this ODE (that is, the derivative along this line is 0 everywhere), and the slope at (1, 0) is positive, we know that as , the function will converge to 2. In other words, as gets larger, the slope field suggests that the solution curve through (1, 0) will start to plateau and steadily approach . On the other hand, as , the slope field tells us that the curve would rapidly diverge off to . (When you actually draw the solution, you would end up with something resembling the plot of .)

b) The tangent line to at , given that , takes the form




When , we have , so



and so the tangent line to at is



Using the tangent line as an approximation, we would find



c) The ODE is separable, so we can write



Integrating both sides gives us



Given that , we get



so the particular solution is



You're asked to find the solution in the form , so you should solve for . You would end up with