Solve the following differential equations.
`(x+y-1) ( dy//dx )=1`

Answer

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

### Step 1: Rearrange the equation
Start by isolating \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{x+y-1} \]

### Step 2: Take the reciprocal
Taking the reciprocal gives us:
\[ \frac{dx}{dy} = x + y - 1 \]

### Step 3: Rearrange the equation
Rearranging the equation, we have:
\[ \frac{dx}{dy} - x = y - 1 \]
This is a linear first-order differential equation in the form \(\frac{dx}{dy} + P(y)x = Q(y)\), where \(P(y) = -1\) and \(Q(y) = y - 1\).

### Step 4: Identify the integrating factor
The integrating factor \(\mu(y)\) is given by:
\[ \mu(y) = e^{\int P(y) \, dy} = e^{\int -1 \, dy} = e^{-y} \]

### Step 5: Multiply through by the integrating factor
Multiply the entire equation by the integrating factor:
\[ e^{-y} \frac{dx}{dy} - e^{-y} x = e^{-y}(y - 1) \]

### Step 6: Rewrite the left side
The left side can be rewritten as:
\[ \frac{d}{dy}(e^{-y} x) = e^{-y}(y - 1) \]

### Step 7: Integrate both sides
Now, integrate both sides with respect to \(y\):
\[ \int \frac{d}{dy}(e^{-y} x) \, dy = \int e^{-y}(y - 1) \, dy \]

The left side simplifies to:
\[ e^{-y} x = \int e^{-y}(y - 1) \, dy \]

### Step 8: Solve the integral on the right side
To solve \(\int e^{-y}(y - 1) \, dy\), we can use integration by parts. Let:
- \(u = y - 1\) and \(dv = e^{-y} dy\)
- Then \(du = dy\) and \(v = -e^{-y}\)

Using integration by parts:
\[ \int e^{-y}(y - 1) \, dy = -(y - 1)e^{-y} - \int -e^{-y} \, dy \]
\[ = -(y - 1)e^{-y} + e^{-y} + C \]
\[ = -ye^{-y} + 2e^{-y} + C \]

### Step 9: Substitute back into the equation
Substituting back gives:
\[ e^{-y} x = -ye^{-y} + 2e^{-y} + C \]

### Step 10: Multiply through by \(e^{y}\)
Multiplying through by \(e^{y}\) to eliminate the exponential:
\[ x = -y + 2 + Ce^{y} \]

### Final Solution
Thus, the general solution of the given differential equation is:
\[ x = -y + 2 + Ce^{y} \]