Explanation The desired differential equation is p ′ ( t ) = r p ( t ) − H for t ≥ 0 Here, p ( t ) is the amount of the resource at time t ≥ 0 , r ≥ 0 is the natural growth rate of the resource and H ≥ 0 is the harvesting rate. An initial condition p ( 0 ) = p 0 is also specified to create an initial value problem. The rate of change p ′ ( t ) has a positive contribution from the natural growth rate and a negative contribution from the harvesting team. For given constants p 0 , r and H , p ( t ) = ( p 0 − H r ) e r t + H r . Differentiating the given equation, d d t p ( t ) = d d t ( p 0 − H r ) e r t + H r p ′ ( t ) = d d t ( p 0 − H r ) e r t + d d t H r p ′ ( t ) = ( p 0 − H r ) r e r t = ( r

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Chapter D1.1, Problem 32E

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To find: The value of p0 for which the amount of the resource is decreasing and constant if r=0.05 , H=500 and the time when the resource vanishes if p0=9000 .

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Chapter D1.1, Problem 31E

Chapter D1.1, Problem 33E

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Chapter D1 Solutions

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Ch. D1.1 - Prob. 1E Ch. D1.1 - Prob. 2E Ch. D1.1 - Prob. 3E Ch. D1.1 - If the general solution of a differential equation...Ch. D1.1 - Does the function y(t) = 2t satisfy the...Ch. D1.1 - Does the function y(t) = 6e3t satisfy the initial...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...Ch. D1.1 - Verifying general solutions Verify that the given...

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The value of p 0 for which the amount of the resource is decreasing and constant if r = 0.05 , H = 500 and the time when the resource vanishes if p 0 = 9000 .

Solution Summary: The author explains the value of p_0 for which the amount of the resource is decreasing and constant.

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To find: The value of p0 for which the amount of the resource is decreasing and constant if r=0.05 , H=500 and the time when the resource vanishes if p0=9000 .

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