Program Explanation Given information: The differential equation is d x d t = k x ( x − M ) − h and the values are H = 1 2 ( M + M 2 + 4 h / k ) > 0 and K = 1 2 ( M − M 2 + 4 h / k ) < 0 . Explanation: The given differential equation is d x d t = k x ( x − M ) − h . The critical points of the function are those pints where the derivative of the function would be 0. So put the given differential equation equals to 0 as, 0 = k x ( x − M ) − h It can be observed that that the above equation is a quadratic equation as 0 = k x 2 − k x M − h . Therefore, use the quadratic rule to find the two critical points. d x d t = k x ( x − M ) − h Therefore, the critical points are x 1 and x 2 . x 1 , x 2 = 1 2 ( M ± M 2 + 4 h k ) It can be observe that the value of the obtained critical points x 1 , x 2 is same of the value of H and K

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

5th Edition

ISBN: 9780321816252

Author: C. Henry Edwards, David E. Penney, David Calvis

Publisher: PEARSON

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Representation of Polynomial

It is necessary for the polynomial expression that each expression consists of two parts:

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Chapter 2.2, Problem 28P

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Program Description: Purpose ofproblem is to show that the differential equation dxdt=kx(x−M)−h can be rewritten as dxdt=k(x−H)(x−K) where H=12(M+M2+4h/k)>0 and K=12(M−M2+4h/k)<0 .

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Chapter 2.2, Problem 27P

Chapter 2.2, Problem 29P

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Chapter 2 Solutions

Differential Equations: Computing and Modeling (5th Edition), Edwards, Penney & Calvis

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Program Description: Purpose ofproblem is to show that the differential equation d x d t = k x ( x − M ) − h can be rewritten as d x d t = k ( x − H ) ( x − K ) where H = 1 2 ( M + M 2 + 4 h / k ) > 0 and K = 1 2 ( M − M 2 + 4 h / k ) < 0 .

Solution Summary: The author explains that the differential equation can be rewritten as dx. The critical points are those pints where the derivative of the function would be 0.

Concept explainers

Program Description: Purpose ofproblem is to show that the differential equation dxdt=kx(x−M)−h can be rewritten as dxdt=k(x−H)(x−K) where H=12(M+M2+4h/k)>0 and K=12(M−M2+4h/k)<0 .

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Chapter 2 Solutions

Program Description: Purpose ofproblem is to show that the differential equation d x d t = k x ( x − M ) − h can be rewritten as d x d t = k ( x − H ) ( x − K ) where H = 1 2 ( M + M 2 + 4 h / k ) &gt; 0 and K = 1 2 ( M − M 2 + 4 h / k ) &lt; 0 .

Solution Summary: The author explains that the differential equation can be rewritten as dx. The critical points are those pints where the derivative of the function would be 0.

Concept explainers

Program Description: Purpose ofproblem is to show that the differential equation dxdt=kx(x−M)−h can be rewritten as dxdt=k(x−H)(x−K) where H=12(M+M2+4h/k)>0 and K=12(M−M2+4h/k)<0 .

Want to see the full answer?

Chapter 2 Solutions

Program Description: Purpose ofproblem is to show that the differential equation d x d t = k x ( x − M ) − h can be rewritten as d x d t = k ( x − H ) ( x − K ) where H = 1 2 ( M + M 2 + 4 h / k ) > 0 and K = 1 2 ( M − M 2 + 4 h / k ) < 0 .