Explanation Given: The differential equation is y ′ ( t ) = y ( 3 + y ) ( y − 5 ) . Definition used: “If f ( y ) < 0 on the left of a and f ( y ) > 0 on the right of a , then the equilibrium solution y = a is unstable .” “If f ( y ) > 0 on the left of a and f ( y ) < 0 on the right of a , then the equilibrium solution y = a is stable .” Calculation: Obtain the equilibrium solution by setting y ′ ( t ) = 0 . y ( 3 + y ) ( y − 5 ) = 0 That is, y = 0 , y = − 3 and y = 5 . Therefore, the equilibrium solution is y = 0 , y = − 3 .and y = 5 . The direction field for the equation y ′ ( t ) = y ( 3 + y ) ( y − 5 ) is shown in below Figure 1. From Figure 1, it is clear that the equilibrium solution is unstable at y = − 3 and y = 5 .and stable at the equilibrium point y = 0 . First to check the equilibrium solution y = 0 is stable or unstable. Let y ′ ( t ) = f ( y ) . f ( y ) = y ( 3 + y ) ( y − 5 ) Take any value of y on the left of 0 and obtain f ( y ) . f ( − 1 ) = − 1 ( 3 + ( − 1 ) ) ( ( − 1 ) − 5 ) = − 2 × − 6 = 12 > 0 That is, f ( y ) > 0 when the value of y on the left of 0. Take any value of y on the right of 0 and obtain f ( y ) . f ( 1 ) = 1 ( 3 + 1 ) ( 1 − 5 ) = 4 × − 4 = − 16 < 0 That is, f ( y ) < 0 when the value of y on the right of 0

Single Variable Calculus: Early Transcendentals, Books a la Carte, and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition)

3rd Edition

ISBN: 9780134996103

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

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Differential Equations

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The equilibrium solution of the given equation and check whether it is stable or unstable.

Solution Summary: The author analyzes the equilibrium solution of the given equation and determines whether it is stable or unstable.

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