Question

For the differential equation find the general solution of
$\frac{dy}{dx}+y=1(y\ne 1)$

Answer 1

Given differential equation is

$\frac{dy}{dx}+y=1$

$dy=(1-y)dx$

$\frac{dy}{y-1}=-dx$

Integrating both sides
We get,

$\int \frac{dy}{y-1}=\int -dx$

$\log (y-1)=-x+c$

$y-1=e^{(-x+c)}$

$y=e^{-x}\times e^c+1$

Putting $e^c=K$

$y=K{e}^{-x}+1$

Final Answer:
Hence, the required general solution is

$y=K{e}^{-x}+1$