# Thread: Innocent looking 1st order DE

1. ## Innocent looking 1st order DE

$\displaystyle 2y e^{y/x} \frac{dy}{{dx}} - 2x - y = 0$

I'm not sure how difficult you guys at this forum might find this. It took me a long long long time to figure out (I am not a math major). Should need at least two substitutions.

(The solution, in implicit form, is $\displaystyle y^2 - x^2 e^{y/x} = C$).

2. Originally Posted by milodino
$\displaystyle 2y e^{y/x} \frac{dy}{{dx}} - 2x - y = 0$

I'm not sure how difficult you guys at this forum might find this. It took me a long long long time to figure out (I am not a math major). Should need at least two substitutions.

(The solution, in implicit form, is $\displaystyle y^2 - x^2 e^{y/x} = C$).
If you divide your ODE by x then

$\displaystyle 2\frac{y}{x} \,e^{\frac{y}{x}} \,\frac{dy}{{dx}} - 2 - \frac{y}{x} = 0$

which suggests the substitution $\displaystyle u = \frac{y}{x}$ and the ODE separates. Not really sure how you got your answer. An ODE that looks a lot like your and gives rise to your solution is

$\displaystyle \left( 2y \,e^{-\frac{y}{x}} -x\right) y' - 2x + y = 0$

3. u = y/x is one of the substitutions, but it doesn't actually neatly separate after that.

$\displaystyle x \frac{du}{dx} + u = \frac{2+u}{2u e^u + 1}$

Admittedly it does reach a pretty nice form eventually but it looked pretty nasty here.