Solve the differential equation: `(x+y)(dy)/(dx)=1`

Explanation

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

### Step 1: Rewrite the Equation
We start by rewriting the given differential equation:
\[ \frac{dy}{dx} = \frac{1}{x+y} \]

### Step 2: Find \(\frac{dx}{dy}\)
Next, we take the reciprocal of both sides to express \(\frac{dx}{dy}\):
\[ \frac{dx}{dy} = x + y \]

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

### Step 4: Find the Integrating Factor
To solve this linear differential equation, we need to find the integrating factor \(I(y)\):
\[ I(y) = e^{\int P(y) \, dy} = e^{\int -1 \, dy} = e^{-y} \]

### Step 5: Multiply by the Integrating Factor
We multiply the entire differential equation by the integrating factor:
\[ e^{-y} \frac{dx}{dy} - e^{-y} x = e^{-y} y \]

### Step 6: Recognize the Left Side as a Derivative
The left side can be recognized as the derivative of a product:
\[ \frac{d}{dy}(e^{-y} x) = e^{-y} y \]

### Step 7: Integrate Both Sides
Now we integrate both sides with respect to \(y\):
\[ \int \frac{d}{dy}(e^{-y} x) \, dy = \int e^{-y} y \, dy \]
The left side simplifies to:
\[ e^{-y} x = \int e^{-y} y \, dy \]
For the right side, we will use integration by parts. Let \(u = y\) and \(dv = e^{-y} dy\):
- Then \(du = dy\) and \(v = -e^{-y}\).

Using integration by parts:
\[ \int y e^{-y} \, dy = -y e^{-y} - \int -e^{-y} \, dy = -y e^{-y} + e^{-y} + C \]
Thus,
\[ \int e^{-y} y \, dy = -y e^{-y} + e^{-y} + C \]

### Step 8: Substitute Back
Substituting back into our equation gives:
\[ e^{-y} x = -y e^{-y} + e^{-y} + C \]

### Step 9: Solve for \(x\)
Now, we multiply through by \(e^{y}\) to solve for \(x\):
\[ x = -y + 1 + C e^{y} \]

### Final Solution
Thus, the solution to the differential equation is:
\[ x = -y + 1 + Ce^{y} \]