Solve: `(dy)/(dx)+y=1`

Explanation

To solve the differential equation \(\frac{dy}{dx} + y = 1\), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = 1 - y \]

### Step 2: Taking the Reciprocal
Next, we take the reciprocal of both sides:
\[ \frac{dx}{dy} = \frac{1}{1 - y} \]

### Step 3: Integrating Both Sides
Now we will integrate both sides. The left side integrates to \(x\) and the right side requires integration:
\[ \int dx = \int \frac{1}{1 - y} dy \]
This gives us:
\[ x = -\log|1 - y| + C \]
where \(C\) is the constant of integration.

### Step 4: Rearranging the Equation
To express \(y\) in terms of \(x\), we can rearrange the equation:
\[ \log|1 - y| = -x + C \]

### Step 5: Exponentiating Both Sides
Exponentiating both sides to eliminate the logarithm:
\[ |1 - y| = e^{-x + C} = e^C e^{-x} \]
Let \(k = e^C\), then we have:
\[ |1 - y| = k e^{-x} \]

### Step 6: Solving for \(y\)
Now we can solve for \(y\):
\[ 1 - y = \pm k e^{-x} \]
Thus, we have two cases:
1. \(1 - y = k e^{-x}\) which gives \(y = 1 - k e^{-x}\)
2. \(1 - y = -k e^{-x}\) which gives \(y = 1 + k e^{-x}\)

### Final Solution
The general solution of the differential equation is:
\[ y = 1 - k e^{-x} \quad \text{or} \quad y = 1 + k e^{-x} \]
where \(k\) is a constant.

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