Step 4 We have a general solution to our differential equation: y = ₂0.08t + C - 25,000. To find the particular solution, we now find the value of the constant of integration that works with the initial condition. Substituting our initial condition y(0) = 0 into the general solution and solving for C, we have ³(0)+c y(0) = 0 = e 0 = 25,000 = C = 0.08 C - 25,000 - 25000
Step 4 We have a general solution to our differential equation: y = ₂0.08t + C - 25,000. To find the particular solution, we now find the value of the constant of integration that works with the initial condition. Substituting our initial condition y(0) = 0 into the general solution and solving for C, we have ³(0)+c y(0) = 0 = e 0 = 25,000 = C = 0.08 C - 25,000 - 25000
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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