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(a) What can you say about a solution of the equation y' = −(1/3)y² just by looking at the differential equation? The function y must be decreasing (or equal to 0) on any interval on which it is defined. The function y must be increasing (or equal to 0) on any interval on which it is defined. O The function y must be strictly decreasing on any interval on which it is defined. The function y must be strictly increasing on any interval on which it is defined. O The function y must be equal to 0 on any interval on which it is defined. (b) Verify that all members of the family y = 3/(x + C) are solutions of the equation in part (a). 3 3 y = X + C LHS = y'= - y = → 3 (x + c)² (x + c)² == (d) Find a solution of the initial-value problem. y' = -(1/3)y² y(0) = 0.5 3 X+ C 1)² = -√²/1/1² = (c) Can you think of a solution of the differential equation y' = −(1/3)y² that is not a member of the family in part (b)? O Every solution of y'= -(1/3)y² is a member of the family in part (b). O y = x is a solution of y' = −(1/3)y² that is not a member of the family in part (b). y = 0 is a solution of y'= -(1/3)y² that is not a member of the family in part (b). O y = 3 is a solution of y' = −(1/3)y² that is not a member of the family in part (b). O y = ³x is a solution of y' = -(1/3)y² that is not a member of the family in part (b). = RHS

(a) What can you say about a solution of the equation y' = −(1/3)y² just by looking at the differential equation? The function y must be decreasing (or equal to 0) on any interval on which it is defined. The function y must be increasing (or equal to 0) on any interval on which it is defined. O The function y must be strictly decreasing on any interval on which it is defined. The function y must be strictly increasing on any interval on which it is defined. O The function y must be equal to 0 on any interval on which it is defined. (b) Verify that all members of the family y = 3/(x + C) are solutions of the equation in part (a). 3 3 y = X + C LHS = y'= - y = → 3 (x + c)² (x + c)² == (d) Find a solution of the initial-value problem. y' = -(1/3)y² y(0) = 0.5 3 X+ C 1)² = -√²/1/1² = (c) Can you think of a solution of the differential equation y' = −(1/3)y² that is not a member of the family in part (b)? O Every solution of y'= -(1/3)y² is a member of the family in part (b). O y = x is a solution of y' = −(1/3)y² that is not a member of the family in part (b). y = 0 is a solution of y'= -(1/3)y² that is not a member of the family in part (b). O y = 3 is a solution of y' = −(1/3)y² that is not a member of the family in part (b). O y = ³x is a solution of y' = -(1/3)y² that is not a member of the family in part (b). = RHS

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter11: Differential Equations
Section11.CR: Chapter 11 Review
Problem 1CR
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(a) What can you say about a solution of the equation y'= -(1/3)y² just by looking at the differential equation? The function y must be decreasing (or equal to 0) on any interval on which it is defined. The function y must be increasing (or equal to 0) on any interval on which it is defined. O The function y must be strictly decreasing on any interval on which it is defined. The function y must be strictly increasing on any interval on which it is defined. O The function y must be equal to 0 on any interval on which it is defined. (b) Verify that all members of the family y = 3/(x + C) are solutions of 3 X + C y = LHS = y'= - y = → 3 (x + c)² == (d) Find a solution of the initial-value problem. y' = -(1/3)y² y(0) = 0.5 3 equation in part (a). (x + C) (c) Can you think of a solution of the differential equation y' = −(1/3)y² that is not a member of the family in part (b)? Every solution of y'= -(1/3)y² is a member of the family in part (b). O y = x is a solution of y'= -(1/3)y² that is not a member of the family in part (b). y = 0 is a solution of y'= -(1/3)y² that is not a member of the family in part (b). O y = 3 is a solution of y'= -(1/3)y² that is not a member of the family in part (b). e3x O y = ³x is a solution of y'= -(1/3)y² that is not a member of the family in part (b). e X + C ² = -1/² = = RHS
Transcribed Image Text:(a) What can you say about a solution of the equation y'= -(1/3)y² just by looking at the differential equation? The function y must be decreasing (or equal to 0) on any interval on which it is defined. The function y must be increasing (or equal to 0) on any interval on which it is defined. O The function y must be strictly decreasing on any interval on which it is defined. The function y must be strictly increasing on any interval on which it is defined. O The function y must be equal to 0 on any interval on which it is defined. (b) Verify that all members of the family y = 3/(x + C) are solutions of 3 X + C y = LHS = y'= - y = → 3 (x + c)² == (d) Find a solution of the initial-value problem. y' = -(1/3)y² y(0) = 0.5 3 equation in part (a). (x + C) (c) Can you think of a solution of the differential equation y' = −(1/3)y² that is not a member of the family in part (b)? Every solution of y'= -(1/3)y² is a member of the family in part (b). O y = x is a solution of y'= -(1/3)y² that is not a member of the family in part (b). y = 0 is a solution of y'= -(1/3)y² that is not a member of the family in part (b). O y = 3 is a solution of y'= -(1/3)y² that is not a member of the family in part (b). e3x O y = ³x is a solution of y'= -(1/3)y² that is not a member of the family in part (b). e X + C ² = -1/² = = RHS
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