Consider the differential equation [x² − (x³ + y³)²/³] dx + y² dy = 0. This is not an exact equation, but we may rearrange the equation as follows. 2 x² dx + y² dy (x³ + y³)2/3 = dx 1 Using the fact that d(x³ + y³) = x² dx + y² dy, / discuss how one could solve the differential equation.
Consider the differential equation [x² − (x³ + y³)²/³] dx + y² dy = 0. This is not an exact equation, but we may rearrange the equation as follows. 2 x² dx + y² dy (x³ + y³)2/3 = dx 1 Using the fact that d(x³ + y³) = x² dx + y² dy, / discuss how one could solve the differential equation.
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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![Consider the differential equation
[x2 – (x3 + y³)2/3] dx + y? dy = 0.
This is not an exact equation, but we may rearrange
the equation as follows.
x² dx + y? dy
dx
(x3 + y3)2/3
1
Using the fact that d(x + y°) = x² dx + y? dy,
3
discuss how one could solve the differential equation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbad2734e-649a-4533-b9cc-3c645832697e%2Fb7cd8beb-9b74-4e34-8ef5-2a320b6d5e23%2F70ygd9w_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the differential equation
[x2 – (x3 + y³)2/3] dx + y? dy = 0.
This is not an exact equation, but we may rearrange
the equation as follows.
x² dx + y? dy
dx
(x3 + y3)2/3
1
Using the fact that d(x + y°) = x² dx + y? dy,
3
discuss how one could solve the differential equation.
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