# Thread: limit to infinity question

1. ## limit to infinity question

$\displaystyle y'=\frac{e^{x-y}}{y-1}$
when x goes to minus infinity its solution y goes to zero.
find y(ln2)?

i know that y is some sort of exponent of negative power variable or a fraction because for both of them infinity turns the whole thing into zero.
$\displaystyle (y-1)dy=e^{x-y}dx$

$\displaystyle (y-1)dy-e^{x-y}dx=0$
there is no way i solve it by integration coefficient method because we have both x and y in the power of the exponent and no defertiation we change that

2. $\displaystyle y'=\frac{e^{x-y}}{y-1}$

$\displaystyle \frac{dy}{dx}=\frac{e^x}{e^y(y-1)}$

It is seperable
$\displaystyle e^y(y-1) \; dy = e^x \; dx$

3. thanks