# Differential equiations

1. Sep 5, 2009

### intenzxboi

the general solution for ye^x dy/dx = [e^-y ] + [ e^ -2x-y ]is:

so i tried to see if this was a homogenous equation but it is not.

next i tried to simplify and got:

y e^y dy= (1+e^-2x) dx / e^x

2. Sep 5, 2009

### CFDFEAGURU

Could you please re-post this using the homework layout? Also, you are missing the equation that describes the general solution.

Thanks
Matt

3. Sep 6, 2009

### HallsofIvy

Staff Emeritus
It is very difficult to understand what you mean. Do you mean that you are asked to find the general solution? And is "e^-2x- y" supposed to be "e^(-2x-y)" or "e^(-2x) - y"? Guessing at what you mean:
Once you have
$$y e^y dy= \frac{1+ e^{-2x}}{e^x} dx$$
just integrate both sides.

Use integration by parts, letting u= y and $dv= e^ydy$ on the left.
You can rewrite the right side as e-x+ e-3x which should be easy to integrate.