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Let y(x) be the solution of the initial value problem y' = 1 + + 60-y/x х , y(1)=0 Then y(4) 4 ln(k) with k equal to = Enter an integer value in the box provided. Your Answer: Answer Which of the following is a solution of the initial value problem - √(x+2) y² dx + x (y¯¹ − 1) dy = 0 \y(1) = 1 -2 −1 У +Y X -2 1 - 2 In(x) = 12 12 - ½ ½ y² ² + y² ¹ + x + 2 ln(x) = 3/4 2y-3 — -2 3 y -4 - y¯¹ -1 + x + 2 ln(x) = 0 12 + x + 2 ln(x) = }}} 12 -y²+ y¹ + + 1/1 x² + 2 x = 52 329

Let y(x) be the solution of the initial value problem y' = 1 + + 60-y/x х , y(1)=0 Then y(4) 4 ln(k) with k equal to = Enter an integer value in the box provided. Your Answer: Answer Which of the following is a solution of the initial value problem - √(x+2) y² dx + x (y¯¹ − 1) dy = 0 \y(1) = 1 -2 −1 У +Y X -2 1 - 2 In(x) = 12 12 - ½ ½ y² ² + y² ¹ + x + 2 ln(x) = 3/4 2y-3 — -2 3 y -4 - y¯¹ -1 + x + 2 ln(x) = 0 12 + x + 2 ln(x) = }}} 12 -y²+ y¹ + + 1/1 x² + 2 x = 52 329

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter2: Graphical And Tabular Analysis
Section2.1: Tables And Trends
Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...
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Let y(x) be the solution of the initial value problem y' = 1 + + 60-y/x х , y(1)=0 Then y(4) 4 ln(k) with k equal to = Enter an integer value in the box provided. Your Answer: Answer
Transcribed Image Text:Let y(x) be the solution of the initial value problem y' = 1 + + 60-y/x х , y(1)=0 Then y(4) 4 ln(k) with k equal to = Enter an integer value in the box provided. Your Answer: Answer
Which of the following is a solution of the initial value problem - √(x+2) y² dx + x (y¯¹ − 1) dy = 0 \y(1) = 1 -2 −1 У +Y X -2 1 - 2 In(x) = 12 12 - ½ ½ y² ² + y² ¹ + x + 2 ln(x) = 3/4 2y-3 — -2 3 y -4 - y¯¹ -1 + x + 2 ln(x) = 0 12 + x + 2 ln(x) = }}} 12 -y²+ y¹ + + 1/1 x² + 2 x = 52 329
Transcribed Image Text:Which of the following is a solution of the initial value problem - √(x+2) y² dx + x (y¯¹ − 1) dy = 0 \y(1) = 1 -2 −1 У +Y X -2 1 - 2 In(x) = 12 12 - ½ ½ y² ² + y² ¹ + x + 2 ln(x) = 3/4 2y-3 — -2 3 y -4 - y¯¹ -1 + x + 2 ln(x) = 0 12 + x + 2 ln(x) = }}} 12 -y²+ y¹ + + 1/1 x² + 2 x = 52 329
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