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Match the differential equation with its direction field (labeled I-IV). Give reasons for your answer.$ y' = x(2 - y) $

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$=-1,$ which is true in $1 \mathrm{V}$

Calculus 2 / BC

Chapter 9

Differential Equations

Section 2

Direction Fields and Euler's Method

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Lectures

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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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I asked to find a slope prop which corresponds to this equation. The equation we are given is y prime equals x, 2, minus y and we'll get in slope fields, 12, red and 4. The turn in the correct slope field. Let'S find some equilibrium solutions of y of this equation if they exist, so we know that y is going to be equilibrium solution. It can only we have that 0, which is equal to y prime, is equal to x times 2 minus y, which is equal to x, times, 2 minus c for all x in real numbers and therefore, if x is not equal to 0, which is possible. It follows that 2 minus c must be equal to 0 or that c has to be equal to. So we see that the only thing in the solution is y equals 2 they're gonna be expecting horizontal asymptotes at y equals 2. You see that to fol 1 has this and so field 3. Has this as well that slope, 2 and 4 d? Not our choices, are going to be sub field 1 of 3 determine which of these 2 is the correct slope field in some initial values. So if we have initial value y 0 equals 0, then you know that the slope at the origin 00 going to be 0 times 2 minus 0, which is equal to 0 and therefore expect a horizontal slope at the origin. We see that graph 1 has a horizontal slope at the origin of graph 3 does not graperies. We can conclude that the correct slope field is.

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