Question: Consider the differential equation dy/dx = 2x - y.

(a) Sketch the slope field for this differential equation on the grid provided.

(b) Find the particular solution to the differential equation that passes through the point (0, 1).

(c) Find the particular solution to the differential equation that passes through the point (1, 2).

(d) Determine whether the particular solutions found in parts (b) and (c) intersect, and if so, find their point of intersection.

Answer:

(a) To sketch the slope field for the given differential equation dy/dx = 2x - y, we evaluate the slope 2x - y at various points in the xy-plane. The slope at a given point (x, y) will be the value of 2x - y at that point. We use this information to draw short line segments with the corresponding slope at each point.

(b) To find the particular solution passing through the point (0, 1), we solve the differential equation with the initial condition y(0) = 1.

We can separate variables and then integrate to solve for y: dy = (2x - y)dx dy + y = 2xdx (ye^x)dx + y = 2xdx Integrating both sides: ∫(ye^x)dx + ∫y dy = ∫2x dx y ∙ e^x + 1/2 y^2 = x^2 + C, where C is the constant of integration Using the given initial condition y(0) = 1: 1 ∙ e^0 + 1/2 (1)^2 = 0^2 + C 1 + 1/2 = C C = 3/2 So, the particular solution passing through (0, 1) is y ∙ e^x + 1/2 y^2 = x^2 + 3/2.

(c) To find the particular solution passing through the point (1, 2), we use the same method as in part (b) but this time with the initial condition y(1) = 2. Solving the differential equation with this new initial condition, we get: y ∙ e^x + 1/2 y^2 = x^2 + 3/2. Plugging in the new initial condition (1, 2), we find: 2 ∙ e^1 + 1/2 (2)^2 = 1^2 + 3/2 2e + 2 = 1 + 3/2 2e + 1/2 = 5/2 2e = 4/2 e = 2 Therefore, the particular solution passing through the point (1, 2) is y ∙ e^x + 1/2 y^2 = x^2 + 3/2.

(d) To determine whether the particular solutions from parts (b) and (c) intersect, we solve the system of equations to find their point of intersection. We have the two particular solutions: y ∙ e^x + 1/2 y^2 = x^2 + 3/2 and y ∙ 2^x + 1/2 y^2 = x^2 + 3/2.

To find their point of intersection, we solve this system of equations. Subtracting the second equation from the first, we get: y ∙ e^x - y ∙ 2^x = 0 y (e^x - 2^x) = 0 This equation gives us the possible points of intersection. Solving for y, we get: y = 0 or e^x = 2^x The first solution implies that the point of intersection is (0, 0) thus not valid. For the second case, taking natural log of both sides, we get: x = ln(2)x x = 0, 0 is not a valid point of intersection.

Therefore, the particular solutions from parts (b) and (c) do not intersect.