Solving Differential Equations of the Form y'+Cy^2=D

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Discussion Overview

The discussion centers around solving the differential equation of the form y' + Cy² = D, where C and D are constants. Participants explore various methods for solving this equation, including separation of variables and potential connections to Riccati equations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses curiosity about solving the equation by hand, noting that they found separation of variables to be time-consuming.
  • Another participant argues that linear equations are easier to solve and suggests that the equation can be integrated directly.
  • A third participant references the Riccati differential equation, indicating that it can be solved using a specific method outlined in an external source.
  • One participant challenges the use of the Riccati equation, asserting that the equation is separable and can be integrated directly, providing a breakdown of the integration process based on the signs of C and D.
  • Another participant proposes an alternative form of the equation, suggesting that it can be rewritten as dy/(D - Cy²) = dx, which leads to a different solution method.
  • There is a reiteration of the desire to find a solution method that does not rely on separation of variables.

Areas of Agreement / Disagreement

Participants express differing opinions on the best method to solve the differential equation, with some advocating for separation of variables and others suggesting alternative approaches, including the Riccati equation. The discussion remains unresolved regarding the most efficient method.

Contextual Notes

Participants note that the method of integration may depend on the signs of constants C and D, which introduces additional complexity to the discussion.

gamesguru
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I'm curious about the differential equation which takes the general form of,
[tex]y'+Cy^2=D[/tex].
Where C and D are constants. According to mathematica, the answer is:
[tex]\sqrt{\frac{D}{C}} \tanh{(x \sqrt{CD})}[/tex].
But I'd like to know how this is done by hand, I was able to do this with separation of variables and finding x, then solving for y, but it was a large waste of time and I'm curious if there is a general way to solve this just as linear ones can be solved.
 
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no that is why linear equations are nice they are much easier to solve and can always (at least in principle) be solved. It is a simple integration.
 
I don't know why you would use the Riccatti equation. This is a simple, separable, first order differential equation. It can be integrated directly. I don't know exactly what gamesguru did or why he considers it a "large waste of time" but it is not difficult. It separates into
[tex]\frac{dy}{Cy^2+ D}= dx[/itex]<br /> multiplying on both sides by D, we get<br /> [tex]\frac{dy}{\frac{C}{D}y^2+ 1}= Ddx[/tex]<br /> Now, what we do depends upon the signs of C and D. If C and D have opposite signs so that C/D is negative, write the equation as <br /> [tex]\frac{dy}{1- \left|\frac{C}{D}}\right|y^2}= Ddx[/tex]<br /> and let [itex]u= \sqrt{|C/D|}[/itex]. Then the equation becomes<br /> [tex]\sqrt{|C/D|}\frac{du}{u^2+ 1}= Ddt[/tex]<br /> and the solution is exactly what gamesguru said.<br /> <br /> If C and D have the same sign, let u= [itex]\sqrt{C/D}y[/itex] so that the equation becomes<br /> [tex]\sqrt{\frac{D}{C}}\frac{du}{u^2+ 1}= Ddt[/tex]<br /> and integrating gives the same thing with "tan" rather than "tanh".[/tex]
 
Shouldn't it be:
[tex] \frac{dy}{D-Cy^2}= dx[/tex]
?
The answer changes, but the method is the same.
 
HallsofIvy said:
I don't know why you would use the Riccatti equation.

Yes, but he wanted to know if there was a way to do it without separation of variables.
 

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