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Question:

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

Solution:

The given differential equation is:

$\frac{d y}{d x}+y=1$

$\Rightarrow d y+y d x=d x$

 

$\Rightarrow d y=(1-y) d x$\

Separating the variables, we get:

$\Rightarrow \frac{d y}{1-y}=d x$

Now, integrating both sides, we get:

$\int \frac{d y}{1-y}=\int d x$

$\Rightarrow \log (1-y)=x+\log \mathrm{C}$

$\Rightarrow-\log \mathrm{C}-\log (1-y)=x$

$\Rightarrow \log \mathrm{C}(1-y)=-x$

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

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

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

$\Rightarrow y=1+A e^{-x}\left(\right.$ where $\left.A=-\frac{1}{\mathrm{C}}\right)$

This is the required general solution of the given differential equation.

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