Explanation Given: A differential equation d B d t + 0.1 B − 10 = 0 ; B ( 2 ) = 3 . Formula Used: The general solution to the differential equation d y d t = k ( y − A ) is given as y = A + C e k t , where C is any constant Calculation: The differential equation is given as d B d t + 0.1 B − 10 = 0 d B d t = − 0.1 B + 10 d B d t = − 0.1 ( B − 100 ) Compare the above differential equation with d y d t = k ( y − A ) , General solution is given as B = − 100 + C e − 0

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ISBN: 9781119399254

Author: Hughes-Hallett

Publisher: WILEY

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Chapter 9.5, Problem 8P

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Find the particular solution for the differential equation dBdt+0.1B−10=0; B(2)=3.

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Chapter 9.5, Problem 7P

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Find the particular solution for the differential equation d B d t + 0.1 B − 10 = 0 ; B ( 2 ) = 3 .

Solution Summary: The author compares the general solution to the differential equation dyt=k(y-A) with B=-100+Ce

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Find the particular solution for the differential equation dBdt+0.1B−10=0; B(2)=3.

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Chapter 9 Solutions