dy A possible solution to the differential equation t dt (2) + 2y = 4t + 1 would be: y = # + 4t 1 3 2 y = 2t² +t +t=1 °y = ? +1+t 2 y = 4t + t

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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dy
A possible solution to the differential equation t
dt
+ 2y = 4t + 1
would be:
y = # +
4t
3
1
y = 2t2 +t + t
y = ? +1+t
2
y = 4t + 4
t
Transcribed Image Text:dy A possible solution to the differential equation t dt + 2y = 4t + 1 would be: y = # + 4t 3 1 y = 2t2 +t + t y = ? +1+t 2 y = 4t + 4 t
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