First Order, Non-Linear DE - Not Seperable

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    First order Non-linear
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Discussion Overview

The discussion revolves around solving a first-order, non-linear differential equation presented in a practice test. Participants explore various methods to approach the equation, including considerations of separability, exactness, and potential transformations.

Discussion Character

  • Homework-related, Technical explanation, Debate/contested

Main Points Raised

  • One participant expresses difficulty in solving the equation cos(x) + y² + (2yx - 1)y' = 0, noting that it cannot be separated and does not fit the criteria for exact equations or Bernoulli's method.
  • Another participant suggests that the equation appears to be exact and implies that the solution may not be straightforward, recommending the use of the functions P and Q for finding a potential function.
  • A later reply acknowledges the suggestion regarding exactness, indicating a shift in understanding.
  • Another participant reformulates the equation, presenting it in a different form: cos(x) + (xy² - y)' = 0, which may suggest an alternative approach to the problem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the method to solve the equation, with differing views on its exactness and the appropriate approach to take.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the equation's properties, as well as the potential need for additional mathematical steps to clarify the solution process.

Berkshire
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Hello, I'm studying for a test and this is a question on a practice test...

cos(x)+y^2+(2yx-1)y'=0


I can't separate the variables (it's not homogeneous, either), this isn't exact and bernoulli won't work...

dy/dx=-cos(x)/(2yx-1)-y^2/(2yx-1)

I changed the equation so it would look like this but I can't simplify it any more than that and I can't just take the integral of it here...If anyone could give me some help with this problem it would be much appreciated. Thanks!
 
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Looks like an exact equation, if it's on your test you should know how to solve it. That said, it doesn't look like the solution is anything obvious so you'll have to go through the usual P = df/dx and Q = df/dy
 
Oh you're right, thanks!
 
Hi !

cos(x)+y²+2yxy'-y'=0
cos(x)+(x y² -y)'=0
 

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