Calculus of a Single Variable, Chapter 6, 6.2, Section 6.2, Problem 4

Apply direct integration both sides: intN(y) dy= int M(x) dx to solve for the general solution of a differential equation.
For the given first order ODE: (dy)/(dx)=6-y it can be rearrange by cross-multiplication into:
(dy)/(6-y)=dx
Apply direct integration on both sides: int(dy)/(6-y)=int dx
For the left side, we consider u-substitution by letting:
u=6-y then du = -dy or -du=dy
The integral becomes: int(dy)/(6-y)=int(-du)/(u)
Applying basic integration formula for logarithm:
int(-du)/(u)= -ln|u|
Plug-in u = 6-y on -ln|u| , we get:
int(dy)/(6-y)=-ln|6-y|
For the right side, we apply the basic integration: int dx= x+C

Combing the results from both sides, we get the general solution of the differential equation as:
-ln|6-y|= x+C
y =6-e^((-x-C))
or
y = 6-Ce^(-x)