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Using the initial condition at (0, 1.5), sketch a potential solution to the differential equation\\ $\frac{dy}{dx} = \frac{y}{25}(8 - y)$.\\ This problem must be manually graded, so no matter what you draw, the problem will probably not be\\ given full credit until your instructor enters an actual grade.\\ Consider the solution satisfying $y(0) = 1.5$.\\ $\lim_{x \to \infty} y(x) = $

          Using the initial condition at (0, 1.5), sketch a potential solution to the differential equation\\
$\frac{dy}{dx} = \frac{y}{25}(8 - y)$.\\
This problem must be manually graded, so no matter what you draw, the problem will probably not be\\
given full credit until your instructor enters an actual grade.\\
Consider the solution satisfying $y(0) = 1.5$.\\
$\lim_{x \to \infty} y(x) = $
        
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Using the initial condition at (0, 1.5), sketch a potential solution to the differential equation

(dy)/(dx) = (y)/(25)(8 - y).

This problem must be manually graded, so no matter what you draw, the problem will probably not be

given full credit until your instructor enters an actual grade.

Consider the solution satisfying y(0) = 1.5.

limx →∞ y(x) =

Added by Jose Francisco H.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Using the initial condition at (0,1.5), sketch a potential solution to the differential equation (dy)/(dx)=(y)/(25)*(8-y). This problem must be manually graded, so no matter what you draw, the problem will probably not be given full credit until your instructor enters an actual grade. -2+ Draw: Freehand Draw Eraser Consider the solution satisfying y(0)=1.5. lim_(x->infty )y(x)= Using the initial condition at (0, 1.5), sketch a potential solution to the differential equation dy y 8-y) dx 25 This problem must be manually graded, so no matter what you draw, the problem will probably not be given full credit until your instructor enters an actual grade. 15 14 13 12 1 / 1 11 10 10 12 1314 15 1617 18 1920 Clear AllDraw:Freehand DrawEraser Consider the solution satisfying y(0) = 1.5. lim yx Submit Question
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Transcript

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00:01 Hi, in the given problem we have y prime is equal to y times y minus 1 times y minus.
00:07 So the equilibrium y prime is equal to zero at equilibrium at equilibrium.
00:21 So this means y times y minus 1 times y minus 2 is equal to zero.
00:27 So this gives y is equal to zero y is equal to one y is equal to two.
00:34 So at y is equal to zero unstable equation unstable equation at y is equal to one this is stable equilibrium and y is equal to two unstable equilibrium unstable equilibrium...
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