Solving a Differential Equation: y'/(1+y'^2) = 2y^2 + C

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Homework Help Overview

The discussion revolves around solving a differential equation involving the derivative of a function and its relation to a quadratic expression in terms of the function itself. The equation presented is of a form that suggests potential methods for separation of variables.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the possibility of separating variables by rearranging the equation to express dy/dx as a function of y. There are attempts to manipulate the equation into a more manageable form for integration.

Discussion Status

Some participants have suggested methods for simplifying the equation and integrating, while others have raised concerns about the complexity introduced by certain substitutions. The discussion is ongoing, with various approaches being considered without a clear consensus on the best path forward.

Contextual Notes

There is mention of the difficulty in integrating the transformed equation, particularly due to the presence of a second power of y in the square root, which complicates the substitution process. The original poster's reference to a book suggests a context of academic study, possibly with specific constraints or guidelines in mind.

psid
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Homework Statement



How to solve the following DE:
[tex]\frac{1}{\sqrt{1+(dy/dx)^{2}}}=\frac{2y^{2}}{2}+C[/tex]?
 
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I suppose solving it for dy/dx might enable you to do a separation of variables...

I.e. (since you are posting this in advanced physics): write
dy/dx = f(y)
for some function f only depending on y; then integrate
dx = dy / f(y)
and invert to find y(x).

Granted, it's probably easier said than done, but you can give it a try.
 
It is indeed separable. I get it into the following form, but don't know how to integrate
[tex]dx=\sqrt{\frac{((2/\gamma)y^{2}+C)^{2}}{1-((2/\gamma)y^{2}+C)^{2}}}dy[/tex]
 
psid said:
It is indeed separable. I get it into the following form, but don't know how to integrate
[tex]dx=\sqrt{\frac{((2/\gamma)y^{2}+C)^{2}}{1-((2/\gamma)y^{2}+C)^{2}}}dy[/tex]

This is an elegant problem.

Superb.

First: Let's try to make the equation a bit less horrendous.

Take [tex]\sqrt{1-((2/\gamma)y^{2}+C)^{2}} = t[/tex]

Proceed with that. Simplify it well and then take

[tex]t= sin\theta[/tex]

Simplify it and then use De moivre's theorem.
 
May i know the name of the book.
 
But the problem with this substitution is that there is a second power of y in the square root. Thus there will be a term including y for the expression for dt...
 

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