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For some reason, I cannot see a clever way to solve this (I know the way of doing it like in Wolframalapha) but I am pretty sure there is a double angle identity to crack this puzzle. Could someone hint a bit to get this puzzle onwards?

Firstly, I thought to use some rules such as $(x+y)^{2}=x^{2}+2xy+y^{2}$ or $(x-y)^{2}=x^{2}-2xy+y^{2}$ but I think some trigonometric substitution could solve this problem.

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up vote 2 down vote accepted

If we write the equation as,

$$\frac{{dy}}{{dx}} = \frac{{2xy}}{{2{x^2} + {y^2}}}$$

and then divide through $x^2$ we will get:

$$\frac{{dy}}{{dx}} = \frac{{2\dfrac{y}{x}}}{{2 + {{\left( {\dfrac{y}{x}} \right)}^2}}}$$

This suggests that we simplify the previous equation in terms of $$f\left( v \right) = \frac{{2v}}{{2 + {v^2}}}$$

So putting

$$\eqalign{ & \frac{y}{x} = v \cr & y = vx \cr & y' = v'x + v \cr} $$

We get

$$\frac{{dv}}{{dx}}x + v = \frac{{2v}}{{2 + {v^2}}}$$


$$\eqalign{ & \frac{{dv}}{{dx}}x = - \frac{{{v^3}}}{{2 + {v^2}}} \cr & \frac{{dx}}{x} = - \frac{{2 + {v^2}}}{{{v^3}}}dv \cr & \frac{{dx}}{x} = \left( { - \frac{2}{{{v^3}}} - \frac{1}{v}} \right)dv \cr} $$

Upon integration we have:

$$\log x + C = \frac{1}{{{v^2}}} - \log v$$

Let's substitute back

$$\eqalign{ & \log x + C = \frac{{{x^2}}}{{{y^2}}} - \log \frac{y}{x} \cr & \log x + C = \frac{{{x^2}}}{{{y^2}}} - \log y + \log x \cr & \log y = \frac{{{x^2}}}{{{y^2}}} - C \cr & y = {C_1}\exp \left( {{x^2}{y^{ - 2}}} \right) \cr} $$

You can find $y$ in terms of $x$, but I don't think the inverse is possible, at least with everyday functions.

$$y\sqrt {\log y + C} = x$$

Ok, using the Lambert W we have

$${y^2}\left( {\log y + C} \right) = {x^2}$$

Use the exponential:

$${e^{{y^2}}}y{e^C} = {e^{{x^2}}}$$

Square and multiply by two

$$2{y^2}{e^{2{y^2}}}{e^{2C}} = 2{e^{2{x^2}}}$$

Use the Lambert W

$$2{y^2} = W\left( {\frac{{2{e^{2{x^2}}}}}{{{e^{2C}}}}} \right)$$

$$y = \sqrt {\frac{1}{2}W\left( {\frac{{2{e^{2{x^2}}}}}{{{e^{2C}}}}} \right)} $$

Another aproach would be

$$\eqalign{ & \log y + C = \frac{{{x^2}}}{{{y^2}}} \cr & y{e^C} = {e^{\frac{{{x^2}}}{{{y^2}}}}} \cr & {y^2}{e^{2C}} = {e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr & 2\frac{{{x^2}}}{{{y^2}}}{y^2}{e^{2C}} = 2\frac{{{x^2}}}{{{y^2}}}{e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr & 2{x^2}{e^{2C}} = 2\frac{{{x^2}}}{{{y^2}}}{e^{2\frac{{{x^2}}}{{{y^2}}}}} \cr & W\left( {2{x^2}{e^{2C}}} \right) = 2\frac{{{x^2}}}{{{y^2}}} \cr & {y^2} = \frac{{2{x^2}}}{{W\left( {2{x^2}{e^{2C}}} \right)}} \cr & y = \frac{{\sqrt 2 x}}{{\sqrt{W\left( {2{x^2}{e^{2C}}} \right)}}} \cr} $$

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...any idea what is this function here? We had once the Lambert function, does this have also some name? – hhh Feb 26 '12 at 22:01
@hhh I added the use of the Lambert W. – Pedro Tamaroff Feb 26 '12 at 22:22

Hint: it's an homogeneous differential equation.

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Rewrite equation into form :


Substitute :

$$z =\frac{y}{x} \Rightarrow y'=xz'+z$$

Therefore :

$$xz'+z=\frac{2z}{2+z^2} \Rightarrow xz'=\frac{-z^3}{2+z^2} \Rightarrow \int \frac {2+z^2}{z^3} \,dz= -\int \frac {dx}{x} $$

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