3 dx – dy = o; y(o) = 6 x² dx + y² dy = o; y(9) = -1 x dx + 2y dy = 0; y(1) = 2 4 cos 2u du – e5v dv = o; v(o) = -6 e'dx + (xe' – 1)dy = o; y(-5) = 6

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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Show that the equation is exact, and obtain its general solution. Also, find the particular solution
corresponding to the given initial condition.

Exercises 3
Show that the equation is exact, and obtain its general solution. Also, find the particular solution
corresponding to the given initial condition.
3 dx – dy = o; y(0) = 6
x² dx + y² dy = 0; y(9) = -1
x dx + 2y dy = 0; y(1) = 2
4 cos 2u du – e*5v dv = o; v(o) = -6
%3!
e'dx + (xe' – 1)dy = o; y(-5) = 6
Transcribed Image Text:Exercises 3 Show that the equation is exact, and obtain its general solution. Also, find the particular solution corresponding to the given initial condition. 3 dx – dy = o; y(0) = 6 x² dx + y² dy = 0; y(9) = -1 x dx + 2y dy = 0; y(1) = 2 4 cos 2u du – e*5v dv = o; v(o) = -6 %3! e'dx + (xe' – 1)dy = o; y(-5) = 6
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