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slope field for the autonomous differential equation dy/dx =1−y Include in your diagram any equilibrium values of the differential equation, and sketch the solution curve for the initial condition y(1)=0. That is, the solution curve that passes through the point (1,0).

User Singhsumit
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Final answer:

The task involves constructing a slope field and sketching a solution curve for the differential equation dy/dx = 1 - y. The equilibrium value is y = 1, and the solution curve for y(1) = 0 begins at the point (1, 0) and tends to flatten as y approaches 1.

Step-by-step explanation:

The question pertains to creating a slope field for the autonomous differential equation dy/dx = 1 - y. In this context, an autonomous differential equation is one where the rate of change of the variable y depends solely on y itself, and not on x. To understand the behavior of this equation, we consider the equilibrium values, which are values where dy/dx = 0. Setting the right-hand side of the equation to zero gives us 1 - y = 0, so the equilibrium value is y = 1. This means that whenever y is 1, the slope of the tangent to the solution curve at that point is zero, indicating a horizontal line.

To sketch the slope field, we would plot short line segments at various points (x, y) in the xy-plane, where the slope of each segment corresponds to 1 - y at that point. The slope field helps visualize the behavior of potential solution curves for the differential equation. For the given initial condition y(1) = 0, we can draw a solution curve that starts at the point (1, 0) and follows the directional cues provided by the slope field. As y approaches the equilibrium value of 1, the solution curve flattens out since the slopes of the tangent lines are approaching zero.

User Franiis
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