# Thread: Having Trouble With ODE Step

1. ## Having Trouble With ODE Step

Hey. Looking through my notes here and I'm having trouble with a jump they've made.

$\displaystyle e^X+(Ye^X-X)\frac{dX}{dY}=0$

This can be written as:

$\displaystyle \frac{dY}{dX}+Y=Xe^{-X}$

I've been sitting trying to see how the top line can be written as the bottom line but I seem to be stumped.

Any help would be greatly appreciated .

2. $\displaystyle e^{x}+(ye^{x}-x)\frac{dy}{dx}=0$

Mult by $\displaystyle \frac{dx}{dy}$ to get

$\displaystyle e^{x}\frac{dx}{dy}+(ye^{x}-x)=0$

Divide by $\displaystyle e^{x}$ to get

$\displaystyle \frac{dx}{dy}+y-xe^{-x}=0$

3. Ahh, It's seems very simple now. Thanks man .

4. Originally Posted by Macca567
Hey. Looking through my notes here and I'm having trouble with a jump they've made.

$\displaystyle e^X+(Ye^X-X)\frac{dX}{dY}=0$

This can be written as:

$\displaystyle \frac{dY}{dX}+Y=Xe^{-X}$

I've been sitting trying to see how the top line can be written as the bottom line but I seem to be stumped.

Any help would be greatly appreciated .
It is easy

$\displaystyle e^x + \left(ye^x - x\right)\frac{dx}{dy} = 0$

$\displaystyle e^x\frac{dy}{dx} + e^xy = x$

$\displaystyle \left(e^xy\right)^{'} = x$

$\displaystyle e^xy = \frac{x^2}{2} + C$

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

5. Originally Posted by Ment
It is easy

$\displaystyle e^x + \left(ye^x - x\right)\frac{dx}{dy} = 0$

$\displaystyle e^x\frac{dy}{dx} + e^xy = x$
The question was how to get to this line! You can't say "It is easy" and them just write it down!

$\displaystyle \left(e^xy\right)^{'} = x$

$\displaystyle e^xy = \frac{x^2}{2} + C$

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