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In order to solve with we first write and Setting to get sin 2xdx + cos 3ydy = 0 y(π/2) = π/3 cos 3ydy - sin 2xdx 1 1 sin3y=cos2x+c cos 2x + c. and x=π/2 we find that 1 3 = . It follows that 1 cos 2x + cos² (x). - 2 1 sin 3y=-=-cos 2

In order to solve with we first write and Setting to get sin 2xdx + cos 3ydy = 0 y(π/2) = π/3 cos 3ydy - sin 2xdx 1 1 sin3y=cos2x+c cos 2x + c. and x=π/2 we find that 1 3 = . It follows that 1 cos 2x + cos² (x). - 2 1 sin 3y=-=-cos 2

Calculus: Early Transcendentals
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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In order to solve sin 2xdx + cos 3ydy = 0 with y(T/2) = 1/3 we first write cos 3ydy sin 2xdx and to get 1 sin 3y = cos 27 1 cos 2x + c. 3 Setting and x=t/2 we find that . It follows that sin 3y = 2com 1 1 cos 2x + 1 cos (2). 2 = COS We can solve for . Indeed,
Transcribed Image Text:In order to solve sin 2xdx + cos 3ydy = 0 with y(T/2) = 1/3 we first write cos 3ydy sin 2xdx and to get 1 sin 3y = cos 27 1 cos 2x + c. 3 Setting and x=t/2 we find that . It follows that sin 3y = 2com 1 1 cos 2x + 1 cos (2). 2 = COS We can solve for . Indeed,
1 y = A -arcsin(3 cos? æ) 3 where We can also solve this equation by using another method. We let M sin 2x and N = cos 3y. It follows from M, = N, that the equation is Then we integrate with respect to x to get 1 cos 2x + C(y). 2
Transcribed Image Text:1 y = A -arcsin(3 cos? æ) 3 where We can also solve this equation by using another method. We let M sin 2x and N = cos 3y. It follows from M, = N, that the equation is Then we integrate with respect to x to get 1 cos 2x + C(y). 2
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