Solved on Jan 26, 2024

Sketch direction field and solutions for differential equation dydx=y2x\frac{dy}{dx} = y^2 - x using isoclines for slopes m=1,0,1,2m = -1, 0, 1, 2.

STEP 1

Assumptions
1. We are given the differential equation dydx=y2x\frac{d y}{d x} = y^2 - x.
2. We are to sketch the direction field for this differential equation.
3. We are to sketch isoclines for slopes m=1,0,1,2m = -1, 0, 1, 2.
4. We are to sketch a couple of solutions using the information from these isoclines.
5. The graph will have axes marked from -4 to +4 for both xx and yy.

STEP 2

Understand what an isocline is. An isocline for a differential equation is a curve along which the slope of the solution curves is constant. For the given differential equation, the isoclines are defined by setting dydx=m\frac{d y}{d x} = m, where mm is the constant slope.

STEP 3

Find the isocline for m=1m = -1. We set the differential equation equal to -1 and solve for yy.
y2x=1y^2 - x = -1

STEP 4

Rearrange the equation to express yy in terms of xx.
y2=x1y^2 = x - 1

STEP 5

Solve for yy by taking the square root of both sides. Remember that we will have two solutions, one positive and one negative, since y2y^2 is positive for both y>0y > 0 and y<0y < 0.
y=±x1y = \pm\sqrt{x - 1}

STEP 6

Repeat the process for m=0m = 0. Set the differential equation equal to 0 and solve for yy.
y2x=0y^2 - x = 0

STEP 7

Rearrange the equation to express yy in terms of xx.
y2=xy^2 = x

STEP 8

Solve for yy by taking the square root of both sides.
y=±xy = \pm\sqrt{x}

STEP 9

Repeat the process for m=1m = 1. Set the differential equation equal to 1 and solve for yy.
y2x=1y^2 - x = 1

STEP 10

Rearrange the equation to express yy in terms of xx.
y2=x+1y^2 = x + 1

STEP 11

Solve for yy by taking the square root of both sides.
y=±x+1y = \pm\sqrt{x + 1}

STEP 12

Repeat the process for m=2m = 2. Set the differential equation equal to 2 and solve for yy.
y2x=2y^2 - x = 2

STEP 13

Rearrange the equation to express yy in terms of xx.
y2=x+2y^2 = x + 2

STEP 14

Solve for yy by taking the square root of both sides.
y=±x+2y = \pm\sqrt{x + 2}

STEP 15

Sketch the isoclines on a graph with axes ranging from -4 to +4 for both xx and yy. For each isocline, plot the curve corresponding to the positive square root and the curve corresponding to the negative square root.

STEP 16

On the same graph, sketch the direction field. At various points on the graph, draw short line segments with the slope given by the differential equation dydx=y2x\frac{d y}{d x} = y^2 - x. Use the isoclines as a guide for the slopes of these line segments.

STEP 17

Using the direction field, sketch a couple of solution curves. Start at a point and follow the direction of the field, making sure the curve is tangent to the line segments representing the slope at each point.

STEP 18

Ensure that the solution curves you sketch are consistent with the behavior indicated by the isoclines. For example, near the isocline for m=0m = 0, the solution curve should be nearly horizontal, and near the isocline for m=2m = 2, the solution curve should have a steeper slope.

STEP 19

Review the graph to ensure it accurately represents the direction field, the isoclines, and the solution curves for the given differential equation.
The solution to this problem is the graph itself, which should include the direction field, the isoclines for m=1,0,1,2m = -1, 0, 1, 2, and a couple of solution curves that follow the direction field and pass through these isoclines. Since the graph is a visual representation, it cannot be conveyed through text alone and must be drawn on paper or with the aid of graphing software.

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