Question

Solve :

$\frac{dy}{dx}+y=1(y\ne 1)$

Answer 1

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

$dy+ydx=dx$

$dy=(1-y)dx$

Separating the variables,

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

Integrating both sides,

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

$log(1-y)=x+logC$

$logC(1-y)=-x$

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

$1-y=\frac{1}{C}{e}^{-x}$

$y=1-\frac{1}{C}{e}^{-x}$

$y=1+A{e}^{-x}$ $(A=\frac{-1}{C})$