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Calculation of differential equation $\displaystyle \frac{dy}{dx} = \frac{1}{x^2+y^2}$

I have tried like this way put $y=vx$ and $\displaystyle \frac{dy}{dx} = v+x.\frac{dv}{dx}$

so $\displaystyle v+ x.\frac{dv}{dx} = \frac{1}{x^2.(1+v^2)}$

Now after that how can i proceed

Thanks

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1 Answer 1

Without loss of generality, we can switch $x$ and $y$, and write $$y'=x^2+y^2.$$ Plugging this into wolfram alpha, we see this is an example of the Riccatis' equations:

http://www.wolframalpha.com/input/?i=y%27%3Dx%5E2%2By%5E2

Wikipedia has an article on this, where they give instructions on how to convert it to a second order linear differential equation.

http://en.wikipedia.org/wiki/Riccati_equation#Reduction_to_a_second_order_linear_equation

This results in an equation $$u''+x^2u=0.$$ I take it this problem was not given in a typical course in differential equations?

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$x'$ not $y'$ in the second line. ;-) –  Babak S. Mar 24 '13 at 2:13

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