## Calculus: Early Transcendentals 8th Edition

Answer in an implicit form : $y = e^y -2x -sinx + C$
$(e^y -1)y' = 2 + cos x$ can be rewritten as: $(e^y -1) dy/dx = 2 + cos x$ Let's separate y on one side and x on the other side: $(e^y -1) dy= (2 + cos x) dx$ Now let's integrate both parts: $\int(e^y -1) dy= \int (2 + cos x) dx$ $e^y - y + C= 2x + sin x$ $y = e^y -2x -sinx + C$ (Answer is in implicit form.)