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Solve the separable equation y(3 +sin(x²))dy= x cos(x-)da subject to y(0) = -va where a > O is a constant. y=z cos(z)dz subject to y(0) = -Va where a > 0 is a constant. Select one. y(z) = / In(3+sin(r2)) + a - In(3) y(1) = -/In(3 + sin(z²)) + a= In(3) O y(z) = -VIn(3 + sin(z²)) –4+Im(3) a + In(3) y(z) = ±/In(3+ sin(r²))+a- ER)

Solve the separable equation y(3 +sin(x²))dy= x cos(x-)da subject to y(0) = -va where a > O is a constant. y=z cos(z)dz subject to y(0) = -Va where a > 0 is a constant. Select one. y(z) = / In(3+sin(r2)) + a - In(3) y(1) = -/In(3 + sin(z²)) + a= In(3) O y(z) = -VIn(3 + sin(z²)) –4+Im(3) a + In(3) y(z) = ±/In(3+ sin(r²))+a- ER)

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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Solve the separable equation y(3 +sin(x²))dy= x cos(x-)da subject to y(0) = -va where a > O is a constant. y=z cos(z)dz subject to y(0) = -Va where a > 0 is a constant. Select one. y(z) = / In(3+sin(r2)) + a - In(3) y(1) = -/In(3 + sin(z²)) + a= In(3) O y(z) = -VIn(3 + sin(z²)) –4+Im(3) a + In(3) y(z) = ±/In(3+ sin(r²))+a- ER)
Transcribed Image Text:Solve the separable equation y(3 +sin(x²))dy= x cos(x-)da subject to y(0) = -va where a > O is a constant. y=z cos(z)dz subject to y(0) = -Va where a > 0 is a constant. Select one. y(z) = / In(3+sin(r2)) + a - In(3) y(1) = -/In(3 + sin(z²)) + a= In(3) O y(z) = -VIn(3 + sin(z²)) –4+Im(3) a + In(3) y(z) = ±/In(3+ sin(r²))+a- ER)
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