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: Rearranging the equation
We start with the given equation:
\[ (x+y+1)\frac{dy}{dx}=1 \]
We can rearrange this to isolate \(\frac{dx}{dy}\):
\[ \frac{dx}{dy} = \frac{1}{x+y+1} \]

### Step 2: Rearranging further
Next, we can rearrange the equation to express it in a linear form:
\[ \frac{dx}{dy} - x = \frac{1}{y+1} \]
This is now a linear first-order differential equation in the form \(\frac{dx}{dy} + P(y)x = Q(y)\), where \(P(y) = -1\) and \(Q(y) = \frac{1}{y+1}\).

### Step 3: Finding the integrating factor
To solve this linear differential equation, we need to find the integrating factor \(IF\):
\[ IF = e^{\int P(y) \, dy} = e^{\int -1 \, dy} = e^{-y} \]

### Step 4: Multiplying through by the integrating factor
Now we multiply the entire equation by the integrating factor:
\[ e^{-y}\frac{dx}{dy} - e^{-y}x = e^{-y}\frac{1}{y+1} \]

### Step 5: Recognizing the left-hand side as a derivative
The left-hand side can be recognized as the derivative of a product:
\[ \frac{d}{dy}(e^{-y}x) = e^{-y}\frac{1}{y+1} \]

### Step 6: Integrating both sides
Now we integrate both sides with respect to \(y\):
\[ \int \frac{d}{dy}(e^{-y}x) \, dy = \int e^{-y}\frac{1}{y+1} \, dy \]
The left side simplifies to:
\[ e^{-y}x = \int e^{-y}\frac{1}{y+1} \, dy + C \]
where \(C\) is the constant of integration.

### Step 7: Solving the integral on the right side
To solve the integral \(\int e^{-y}\frac{1}{y+1} \, dy\), we can use integration by parts or look it up in a table of integrals. The result is:
\[ \int e^{-y}\frac{1}{y+1} \, dy = -e^{-y}(y+1) + C \]
Thus, we have:
\[ e^{-y}x = -e^{-y}(y+1) + C \]

### Step 8: Isolating \(x\)
Now, we can isolate \(x\):
\[ x = -(y+1) + Ce^{y} \]

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