For the differential equations in given question find the general solution.

$\frac{d y}{d x} + y = 1 (y \neq 1) .$

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Solution

Given, $\frac{d y}{d x} + y = 1$
On separating the variables, we get dy+ydx=dx $\Rightarrow$ dy=dx-ydx
$\Rightarrow d y = d x (1 - y) \Rightarrow d x = \frac{d y}{1 - y}$
On integrating both sides, we get $\int \frac{d y}{(1 - y)} = \int d x \Rightarrow - l o g | 1 - y | = x - C \Rightarrow - x + C = l o g | 1 - y | \Rightarrow | 1 - y | = e^{- x + C} \Rightarrow y - 1 = (\pm e^{C}) e^{- x} \Rightarrow y = 1 + A e^{- x}$ (where, $(A = \pm e^{C}$ )
which is the required general solution of the given differential equation.

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