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Match the differential equation with its direction field (labeled I-IV). Give reasons for your answer.$ y' = 2 - y $
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Calculus 2 / BC
Chapter 9
Differential Equations
Section 2
Direction Fields and Euler's Method
Oregon State University
Idaho State University
Boston College
Lectures
13:37
A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. An ordinary differential equation (ODE) is a differential equation containing one or more derivatives of a function and their rates of change with respect to the function itself; it can be used to model a wide variety of phenomena. Differential equations can be used to describe many phenomena in physics, including sound, heat, electrostatics, electrodynamics, fluid dynamics, elasticity, quantum mechanics, and general relativity.
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Match the differential equ…
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3-6 Match the differential…
You'Re getting a differential equation and 4 slope fields of differential equations were asked to find a slope field which corresponds to this differential equation. The equation is given by prime equals to minus y, and the slope fields are 123 and 4 determine, which is the slope of equation, we'll find some equilibrium solutions for this equation. So we know that y equals c is an equilibrium solution. If you only, we have that y prime, which is equal to 0, is equal to 3. Minus y is equal to 2, minus c or c equals 2 point. So the only equilibrium solution is equal 2. Therefore, we should expect a horizontal asymptote at y equals 2. We see that slope 1 and slope our 3 and this horizontal asymptote slope fields 2 and 4, so we integrate 1 of 3 determine which of these is the slope's plug in the initial that is so supposition. 0 is equal to 0 and the solution of this initial value is going to have slope x, equals 0 of 2 minus 0, which is equal to 2 and then see that in graph 1 y pine 0, the slope at the origin is going to be 0. So it cannot be graph 1 and you see that so field 3. The slope at the origin is indeed a positive number, so you can see that slope. Field 3 is out.
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