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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd18aae14-5c46-499f-ad8f-4ffb02ea16f6%2Fca39db0a-13ba-424c-b1a9-3bd8377f6491%2Flnkvre8b_processed.png&w=3840&q=75)
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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