Explanation Given: The equation one is d x 1 d t = a − b x 1 − β x 1 x 2 + γ x 3 . The equation two is d x 2 d t = β x 1 x 2 − ( b + α + v ) x 2 . The equation third is d x 3 d t = v x 2 − ( β + γ ) x 3 . Approach: First calculate the non trivial equilibrium point by equating each equation with zero. The value of x 3 must be greater than zero to be positive value. Calculation: Equate the first differential equation with zero. d x 1 d t = a − b x 1 − β x 1 x 2 + γ x 3 0 = a − b x 1 − β x 1 x 2 + γ x 3 b x 1 + β x 1 x 2 = a + γ x 3 ... ( I ) Equate the second differential equation with zero. d x 2 d t = β x 1 x 2 − ( b + α + v ) x 2 0 = β x 1 x 2 − ( b + α + v ) x 2 0 = β x 1 − ( b + α + v ) x 1 = ( b + α + v ) β ... ( II ) Equate the third differential equation with zero. d x 3 d t = v x 2 − ( β + γ ) x 3 0 = v x 2 − ( β + γ ) x 3 ( β + γ ) x 3 v = x 2

Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)

2nd Edition

ISBN: 9780137553457

Author: Raymond Greenwell, Nathan Ritchey

Publisher: PEARSON+

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Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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The verification for the ab>b+α+vβ.

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The verification for the a b > b + α + v β .

Solution Summary: The author calculates the non trivial equilibrium point by equating each equation with zero. The value of x_3 must be greater than zero to be positive.

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To find:

The verification for the ab>b+α+vβ.

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Chapter 11 Solutions

The verification for the a b &gt; b + α + v β .

Solution Summary: The author calculates the non trivial equilibrium point by equating each equation with zero. The value of x_3 must be greater than zero to be positive.

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To find: The verification for the ab>b+α+vβ.

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Chapter 11 Solutions

The verification for the a b > b + α + v β .

To find:

The verification for the ab>b+α+vβ.