Explanation Proof: Given: The differential equation is P ′ ( t ) = r P ( 1 − P K ) with its initial point P ( 0 ) = P 0 . Calculation: The differential equation is P ′ ( t ) = r P ( 1 − P K ) . Rewrite the above equation as follows, P ′ ( t ) P ( 1 − P K ) = r P ′ ( t ) P ( K − P K ) = r K P ′ ( t ) P ( K − P ) = r Integrate on both sides of the above expression with respect to t , ∫ K P ′ ( t ) P ( K − P ) d t = ∫ r d t ∫ ( P ′ ( t ) P ( t ) + P ′ ( t ) K − P ( t ) ) d t = ∫ r d t ∫ P ′ ( t ) P ( t ) d t + ∫ P ′ ( t ) K − P ( t ) d t = ∫ r d t ln ( P ( t ) ) − ln ( K − P ( t ) ) = r t + C On Further simplification, ln ( P ( t ) ) ( K − P ( t ) ) = r t + C Where C is constant. Take exponential on both sides of the above solution as follows, e ln ( P ( t ) ) ( K − P ( t ) ) = e r t + C ( P ( t ) ) ( K − P ( t ) ) = C 1 e r t Reciprocate the above expression

Calculus: Single Variable, Early Transcendentals and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition) (Briggs, Cochran, Gillett & Schulz, Calculus Series)

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ISBN: 9780134996714

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

Publisher: PEARSON

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Chapter 9.5, Problem 33E

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To show: The given differential equation has a solution P(t)=KP0(K−P0)e−rt+P0 .

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Chapter 9.5, Problem 32E

Chapter 9.5, Problem 34E

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Chapter 9 Solutions

Calculus: Single Variable, Early Transcendentals and MyLab Math with Pearson eText -- Title-Specific Access Card Package (3rd Edition) (Briggs, Cochran, Gillett & Schulz, Calculus Series)

Ch. 9.1 - What are the orders of the equations in Example 2?...Ch. 9.1 - Prob. 2QC Ch. 9.1 - Prob. 3QC Ch. 9.1 - Prob. 4QC Ch. 9.1 - In Example 7, if the height function were given by...Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - The solution to the initial value problem y(t) = 2...

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The given differential equation has a solution P ( t ) = K P 0 ( K − P 0 ) e − r t + P 0 .

Solution Summary: The author explains that the given differential equation has a solution P(t)=KP_0

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To show: The given differential equation has a solution P(t)=KP0(K−P0)e−rt+P0 .

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