Explanation The values of a = 0.005 , m = $ 100 , B 0 = $ 100 . So, the initial condition is s ( 0 ) = 50 . Consider the given differential equation d B d t = a B + m . Simplify the differential equation as follows, d B d t = a B + m ∫ d B a B + m = ∫ d t ln | a B + m | = t + c Take exponential on both sides yields, a B + m = e t + c . Simplify further and substitute the given values as follows. B = e t + c − m a B = C e 0.005 t − 100 0.005 = C e 0.005 t − 20 , 000 Thus, the general solution of the given differential equation is B = C e 0.005 t − 20 , 000 , where C is an arbitrary constant

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Differential Equations

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Chapter 7, Problem 102RE

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To solve: The initial value problem dBdt=aB+m for t≥0, with B(0)=B0 and find out when the balance will reach at $7500.

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Chapter 7, Problem 101RE

Chapter 8.1, Problem 1E

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Calculus: Early Transcendentals (2nd Edition)

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The initial value problem d B d t = a B + m for t ≥ 0 , with B ( 0 ) = B 0 and find out when the balance will reach at $7500.

Solution Summary: The author explains how to solve the initial value problem dBt=aB+m and the general solution of the given differential equation.

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To solve: The initial value problem dBdt=aB+m for t≥0, with B(0)=B0 and find out when the balance will reach at $7500.

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Chapter 7 Solutions