Explanation Given: The differential equation is B ′ ( t ) = 0.005 B − 500 at the condition B ( 0 ) = 50 , 000 . Calculation: The differential equation is B ′ ( t ) = k B + b . Here, k = 0.005 and b = − 500 . d B d t = 0.005 B − 500 d t = d B 0.005 B − 500 Integrate on both sides, ∫ d B 0.005 B − 500 = ∫ d t 1 0.005 ln ( 0.005 B − 500 ) = t + c ln ( 0.005 B − 500 ) = 0.005 t + C Take exponential on both sides, e ln ( 0.005 B − 500 ) = e 0.005 t + C 0.005 B − 500 = C e 0.005 t 0.005 B = 500 + C e 0.005 t B = 500 0.005 + C e 0.005 t That is, B = 100 , 000 + C e 0.005 t . Substitute the initial value B ( 0 ) = 50000 in the equation B = 100000 + C e 0

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ISBN: 9780135243435

Author: Bill Briggs

Publisher: Pearson Education (US)

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Chapter 9.4, Problem 23E

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To solve: The initial value problem of the given differential equation for t≥0 and sketch the solution: determine the first month in which the loan balance is zero.

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Chapter 9.4, Problem 22E

Chapter 9.4, Problem 24E

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The initial value problem of the given differential equation for t ≥ 0 and sketch the solution: determine the first month in which the loan balance is zero.

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To solve: The initial value problem of the given differential equation for t≥0 and sketch the solution: determine the first month in which the loan balance is zero.

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