Question:

Consider the slope field for the differential equation dy/dx = x^2 - 4x.

(a) On the given slope field, identify the point (0, 1.5) and sketch the solution curve that passes through this point.

(b) Determine the particular solution to the given differential equation that satisfies the initial condition y(0) = 2.

(c) Find the x-coordinate of any horizontal tangent lines to the particular solution from part (b).

Answer:

(a) To sketch the solution curve that passes through the point (0, 1.5), we can use the slope field. The slope at each point on the field indicates the derivative of the solution curve at that point.

Looking at the slope field, we can see that when dy/dx = x^2 - 4x, at the point (0, 1.5), the slope is positive and relatively steep. Hence, the solution curve passing through this point will have a steep positive slope.

(b) To determine the particular solution that satisfies the initial condition y(0) = 2, we need to integrate the given differential equation and solve for the unknown constant.

Given: dy/dx = x^2 - 4x

We integrate both sides with respect to x:

∫ dy = ∫ (x^2 - 4x) dx

Integrating, we get:

y = (1/3)x^3 - 2x^2 + C

To find the particular solution, we substitute the initial condition y(0) = 2:

2 = (1/3)(0)^3 - 2(0)^2 + C

2 = 0 + 0 + C C = 2

Therefore, the particular solution to the given differential equation that satisfies the initial condition y(0) = 2 is:

y = (1/3)x^3 - 2x^2 + 2

(c) To find the x-coordinate of any horizontal tangent lines to the particular solution from part (b), we need to find the values of x for which dy/dx = 0.

Given: dy/dx = x^2 - 4x

Setting dy/dx to zero and solving for x:

0 = x^2 - 4x x^2 - 4x = 0 x(x - 4) = 0

So, either x = 0 or x - 4 = 0.

Therefore, the x-coordinate of the horizontal tangent lines to the particular solution is x = 0 or x = 4.