What is the general solution for a differential equation involving e^x and e^-y?

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

Thanks
Matt
 
intenzxboi said:
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
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
[tex]y e^y dy= \frac{1+ e^{-2x}}{e^x} dx[/tex]
just integrate both sides.

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

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