How can you solve for y in a difficult separable equation?

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Homework Help Overview

The discussion revolves around solving a separable differential equation given by dy/dx = (y cos x) / (1+y²). Participants are exploring the challenges of expressing the solution in terms of y as a function of x.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the progress made in solving the equation, noting the presence of ln(y) as a complication. Questions arise regarding the necessity of expressing the solution explicitly as y=f(x). Some suggest the use of the Lambert W function as a theoretical approach, while others express skepticism about its requirement in the context of the problem.

Discussion Status

The discussion is active, with participants sharing insights about the solution process and questioning the need for a specific form of the solution. There is acknowledgment of the Lambert W function as a potential method, but uncertainty remains about its applicability in this context.

Contextual Notes

Participants are navigating the constraints of the problem statement and the implications of expressing the solution in a particular form. There is a sense of ambiguity regarding the expectations for the solution format.

Jamin2112
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Homework Statement



dy/dx = (y cos x) / (1+y2)

Homework Equations



Meh

The Attempt at a Solution



I've made it to this point:

ln(y) + y2/2 = sinx + C.

But we can't figure out to solve for y. It seems impossible with the ln(y) hanging around.
 
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As far as the differential equation goes, you've solved it. Does your problem statement requre you do express the solution as y=f(x) ?
 
Theoretically, you can solve for y using the Lambert W function, defined as the inverse function to x*e^(x). But I doubt they'd make you do that.
 
Char. Limit said:
Theoretically, you can solve for y using the Lambert W function, defined as the inverse function to x*e^(x). But I doubt they'd make you do that.

Is their another way, brah?
 
PAllen said:
As far as the differential equation goes, you've solved it. Does your problem statement requre you do express the solution as y=f(x) ?

PAllen = Paul Allen?
 
jamin2112 said:
pallen = paul allen?

myob
 
Jamin2112 said:
Is their another way, brah?

Not really. I checked wolframalpha, and that's the only way.

log(y^2)+y^2 = 2 sin(x) + 2C

e^{log(y^2)+y^2} = e^{2 sin(x) + 2C}

y^2 e^{y^2} = e^{2 sin(x)+ 2C}

y^2 = W\left( e^{2 sin(x)+2C}\right)

y= \pm \sqrt{W\left( e^{2 sin(x)+2C}\right)}
 

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