# Find the general solution $\large\frac{dy}{dx}$$+y=1(y\neq1) ## 1 Answer Toolbox: • \int\large\frac{ 1}{(y - 1)}$$= \log(y - 1)$
Step 1:
Given : $\large\frac{dy}{dx}$$+ y = 1 \large\frac{dy}{dx}$$= 1-y$
seperating the variables we get
$dy(y - 1)= - dx$
Step 2:
Integrating on both sides
Using the information from the tool box
$\log(y-1) = -x - \log A$
$\log_eA(y-1) = -x$
Writing this in the exponential form
$Ae^{-x} = y - 1.$
This is the required solution.