AP Calculus AB Exam Question - Slope Fields

Consider the differential equation given by:

dy/dx = x - y

1. Sketch the slope field for the above differential equation over the interval [-2, 2] for both x and y values.

2. Determine the solution to the differential equation that passes through the point (0, 1). Provide your answer in the form of a function.

Answer:

1. To sketch the slope field, we will select several representative points in the x-y plane within the given interval and evaluate the slope at each point using the given differential equation. The slope at each point will be represented by a short line segment or an arrow.

The slope at any point (x, y) is given by the equation dy/dx = x - y, where x denotes the x-coordinate and y denotes the y-coordinate of the point.

We can choose evenly spaced x and y values within the interval [-2, 2]. Let's select x values of -2, -1, 0, 1, and 2, and y values of -2, -1, 0, 1, and 2.

Using the selected values, we can now calculate the slope at each of these points using the given differential equation:

(x, y) = (-2, -2):
dy/dx = (-2) - (-2) = 0

(x, y) = (-2, -1):
dy/dx = (-2) - (-1) = -1

(x, y) = (-2, 0):
dy/dx = (-2) - (0) = -2

(x, y) = (-2, 1):
dy/dx = (-2) - (1) = -3

(x, y) = (-2, 2):
dy/dx = (-2) - (2) = -4

(x, y) = (-1, -2):
dy/dx = (-1) - (-2) = 1

(x, y) = (-1, -1):
dy/dx = (-1) - (-1) = 0

(x, y) = (-1, 0):
dy/dx = (-1) - (0) = -1

(x, y) = (-1, 1):
dy/dx = (-1) - (1) = -2

(x, y) = (-1, 2):
dy/dx = (-1) - (2) = -3

... and so on for the remaining points.

We repeat this process for all selected points to obtain the slope at each point. Finally, we sketch line segments or arrows with these slopes at their respective points. The resulting slope field should resemble the following:

  |     ===    ===    ===    ===    ===    ===    ===     |
  |   -----  -----  -----  -----  -----  -----  -----   |
  |  |  ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| |||  |  
  |  | -||| ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| |||- |
  |  |  ||| -||| ||| ||| ||| ||| ||| ||| ||| ||| ||| |||  |
  |  |  ||| ||| -||| ||| ||| ||| ||| ||| ||| ||| ||| |||  |
  |  |  ||| ||| ||| -||| ||| ||| ||| ||| ||| ||| ||| |||  |
  |  |  ||| ||| ||| ||| -||| ||| ||| ||| ||| ||| ||| |||  |
  |  |  ||| ||| ||| ||| ||| -||| ||| ||| ||| ||| ||| |||  |
  |  |  ||| ||| ||| ||| ||| ||| -||| ||| ||| ||| ||| |||  |
  |  |  ||| ||| ||| ||| ||| ||| ||| -||| ||| ||| ||| |||  |
  |  |  ||| ||| ||| ||| ||| ||| ||| ||| -||| ||| ||| |||  |
  |  |  ||| ||| ||| ||| ||| ||| ||| ||| ||| -||| ||| |||  |
  |  |  ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| -||| |||  |
  |  |  ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| -|||  |
  |  |  ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| |||  |  
  |  | -||| ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| |||- |
  |  |  ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| ||| |||  | 
  |   -----  -----  -----  -----  -----  -----  -----   |
  |     ===    ===    ===    ===    ===    ===    ===     |
  ---------------------------------------------------------
          -2    -1     0      1      2     x-axis

2. To determine the solution to the differential equation that passes through the point (0, 1), we can use separation of variables.

Starting with the given differential equation:

dy/dx = x - y

Rearranging, we have:

dy = (x - y) dx

Now, we can separate the variables and integrate both sides:

∫(1/y - 1) dy = ∫x dx

Integrating, we get:

ln|y - 1| - y = (1/2)x^2 + C

Next, we'll substitute the initial condition (0, 1) to find the value of the constant C:

ln|1 - 1| - 1 = (1/2)(0)^2 + C
ln(0) - 1 = 0 + C

Since the natural logarithm of 0 is undefined, we can conclude that C = -1.

Substituting C back into the equation, we get:

ln|y - 1| - y = (1/2)x^2 - 1

Simplifying and exponentiating both sides, we obtain:

|y - 1|e^-y = e^[(1/2)x^2 - 1]

Taking the positive and negative cases of the absolute value, we have:

y - 1 = ±e^[(1/2)x^2 - 1]e^[-y]

Finally, solving for y:

y = 1 ± e^[(1/2)x^2 - 1]e^[-y]

Therefore, the solution to the differential equation that passes through the point (0, 1) is:

y = 1 ± e^[(1/2)x^2 - 1]e^[-y]