Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

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

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

Then

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

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

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

share|improve this answer
    
...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.

share|improve this answer

Rewrite equation into form :

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

Substitute :

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

Therefore :

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.