MHB Differential equation y/(x^2+y^2)

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The discussion revolves around the differential equation y/(x^2+y^2) and seeks clarification on its correctness. Participants emphasize the importance of posting progress when asking for help to avoid redundant suggestions. There is a suggestion that the equation might actually be y^2/(x^2+y^2), which would simplify the problem due to its homogeneity. A transformation is proposed to further analyze the equation, indicating a method to approach the solution. The conversation highlights the need for clear communication and foundational understanding in solving differential equations.
Nikolas7
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Can you advice the changes in this diff equation:

$\d{y}{x}=\dfrac{y}{x^2+y^2}$
 
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Hi Nikolas7 and happy new year, :)

We ask that a new question is posted in a new thread rather than tagged at the end of an existing thread (http://mathhelpboards.com/rules/).
That's why I have moved your post to a new thread.

We also ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?
 
Nikolas7 said:
Can you advice the changes in this diff equation:

$\d{y}{x}=\dfrac{y}{x^2+y^2}$

I am wondering if this is the correct DE. Are you sure it's not $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y^2}{x^2 + y^2} \end{align*}$?
 
Prove It said:
I am wondering if this is the correct DE. Are you sure it's not $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y^2}{x^2 + y^2} \end{align*}$?
This would be wonderful if that was true b/c the equation is homogenous. With the right side numerator of power one, the problem is a little more difficult!
 
Make the transformation

$$x = - \dfrac{r s'}{s},\;\;\; y = r$$

where $$s = s(r)$$ and see where that takes you.
 

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