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

Explanation

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

### Step 1: Rewrite the equation
We start by rewriting the given equation in a more manageable form:
\[ \frac{dy}{dx} = \frac{1}{x+y+1} \]

### Step 2: Separate variables
Next, we can separate the variables \(x\) and \(y\):
\[ dy = \frac{1}{x+y+1} dx \]

### Step 3: Change the form
To make it easier to solve, we can express this in terms of \(dx\) and \(dy\):
\[ dx = (x+y+1) dy \]

### Step 4: Rearranging
Rearranging gives us:
\[ dx - x dy = (y + 1) dy \]

### Step 5: Identify the linear form
This is in the form of a linear differential equation:
\[ \frac{dx}{dy} - x = y + 1 \]

### Step 6: Finding the integrating factor
The standard form of a linear differential equation is:
\[ \frac{dx}{dy} + P(y)x = Q(y) \]
where \(P(y) = -1\) and \(Q(y) = y + 1\). The integrating factor \(I(y)\) is given by:
\[ I(y) = e^{\int P(y) dy} = e^{-\int 1 dy} = e^{-y} \]

### Step 7: 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 8: Left-hand side as a derivative
The left-hand side can be rewritten as the derivative of a product:
\[ \frac{d}{dy}(e^{-y} x) = e^{-y}(y + 1) \]

### Step 9: Integrate both sides
Now we 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 10: Solve the integral on the right
To solve the integral on the right, we can use integration by parts. Let \(u = y + 1\) and \(dv = e^{-y} dy\). Then \(du = dy\) and \(v = -e^{-y}\):
\[ \int e^{-y}(y + 1) dy = -(y + 1)e^{-y} + \int e^{-y} dy = -(y + 1)e^{-y} - e^{-y} + C \]
Thus,
\[ e^{-y} x = -(y + 2)e^{-y} + C \]

### Step 11: Solve for \(x\)
Multiply through by \(e^{y}\):
\[ x = -(y + 2) + Ce^{y} \]
or
\[ x + y + 2 = Ce^{y} \]

### Final Answer
The solution to the differential equation is:
\[ x + y + 2 = Ce^{y} \]