MHB Differential equation y/(x^2+y^2)

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.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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