Answer to Question #127128 in Differential Equations for Abdul

Question

Solve the initial value problem

(1+e^x)y' = e^(x-y) y(0) =1

Answer 1

(e^y)'=e^x/(e^x+1).

e^y=ln(e^x+1)+c.

if y(0)=1 then c=e-ln(2).

y=ln(ln(e^x+1)+e-ln(2)).