# Math Help - Help solving y'=e^(x-y)

1. ## Help solving y'=e^(x-y)

Prompt: y'=e^(x-y); y(0)=1

I understand what to do for the particular solution but am having trouble finding the general equation.

2. ## Re: Help solving y'=e^(x-y)

Originally Posted by Lamont
Prompt: y'=e^(x-y); y(0)=1

I understand what to do for the particular solution but am having trouble finding the general equation.
The equation is seperable!

$\frac{dy}{dx}=e^{x}e^{-y} \iff e^{y}dy=e^{x}dx \iff e^{y}=e^{x}+C$

Now just plug in your point to get

$e=e^{0}+C \implies C=e-1$

This gives

$e^y=e^x+e-1 \implies y=\ln(e^x+e-1)$

Thanks!