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Verify that the indicated function y = p(x) is an explicit solution of the given first-order differential equation. 2y = y³ cos(x); y = (1-sin(x))-1/2 When y (1 sin(x))-1/2, = 2y' Thus, in terms of x, y³ cos(x) = -1/2 Since the left and right hand sides of the differential equation are equal when (1 - sin(x))¯¹ is substituted for y, y = (1 - sin(x))-¹/2 is a solution. Proceed as in Example 6, by considering y simply as a function and give its domain. O {x|x * 2nm} # 0{x|x=+2nm} 0{x|x*^#} 2 O{x|x = 2n} 0 {x|xx = +291²}

Verify that the indicated function y = p(x) is an explicit solution of the given first-order differential equation. 2y = y³ cos(x); y = (1-sin(x))-1/2 When y (1 sin(x))-1/2, = 2y' Thus, in terms of x, y³ cos(x) = -1/2 Since the left and right hand sides of the differential equation are equal when (1 - sin(x))¯¹ is substituted for y, y = (1 - sin(x))-¹/2 is a solution. Proceed as in Example 6, by considering y simply as a function and give its domain. O {x|x * 2nm} # 0{x|x=+2nm} 0{x|x*^#} 2 O{x|x = 2n} 0 {x|xx = +291²}

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Verify that the indicated function y = p(x) is an explicit solution of the given first-order differential equation. 2y = y³ cos(x); y = (1-sin(x))-1/2 When y (1 sin(x))-1/2, = 2y' Thus, in terms of x, y³ cos(x) = Since the left and right hand sides of the differential equation are equal when (1 - sin(x))¯¹ -1/2 is substituted for y, y = (1 - sin(x))-¹/2 is a solution. Proceed as in Example 6, by considering y simply as a function and give its domain. O {x|x * 2nm} # 0 {x|x=+2nx} o{x|x*} 2 O{x|x = 2n} ढ 0{x|x */ +27= } Then by considering p as a solution of the differential equation, give at least one interval I of definition. (27πT, DO) 0 (7,00) π 5x 2 2 [27π, 00) (2,57) O
Transcribed Image Text:Verify that the indicated function y = p(x) is an explicit solution of the given first-order differential equation. 2y = y³ cos(x); y = (1-sin(x))-1/2 When y (1 sin(x))-1/2, = 2y' Thus, in terms of x, y³ cos(x) = Since the left and right hand sides of the differential equation are equal when (1 - sin(x))¯¹ -1/2 is substituted for y, y = (1 - sin(x))-¹/2 is a solution. Proceed as in Example 6, by considering y simply as a function and give its domain. O {x|x * 2nm} # 0 {x|x=+2nx} o{x|x*} 2 O{x|x = 2n} ढ 0{x|x */ +27= } Then by considering p as a solution of the differential equation, give at least one interval I of definition. (27πT, DO) 0 (7,00) π 5x 2 2 [27π, 00) (2,57) O
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