$\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