Solve the following differential equations :
`(dy)/(dx)=(x+1)/(2-y)(y != 2)`

Answer

To solve the differential equation

\[ \frac{dy}{dx} = \frac{x + 1}{2 - y} \quad (y \neq 2) \]

we will follow these steps:

### Step 1: Separate the variables
We can rearrange the equation to separate the variables \(y\) and \(x\):

\[ (2 - y) dy = (x + 1) dx \]

### Step 2: Integrate both sides
Now, we will integrate both sides. The left side becomes:

\[ \int (2 - y) dy = \int 2 dy - \int y dy = 2y - \frac{y^2}{2} \]

The right side becomes:

\[ \int (x + 1) dx = \int x dx + \int 1 dx = \frac{x^2}{2} + x \]

### Step 3: Combine the results
After integrating, we have:

\[ 2y - \frac{y^2}{2} = \frac{x^2}{2} + x + C \]

where \(C\) is the constant of integration.

### Step 4: Rearrange the equation
To make it more presentable, we can multiply through by 2 to eliminate the fraction:

\[ 4y - y^2 = x^2 + 2x + 2C \]

### Step 5: Final form
Rearranging gives us:

\[ y^2 - 4y + x^2 + 2x + 2C = 0 \]

We can let \(2C = C'\) (where \(C'\) is a new constant) to simplify our final answer:

\[ y^2 - 4y + x^2 + 2x + C' = 0 \]

### Final Answer:
Thus, the solution to the differential equation is:

\[ y^2 - 4y + x^2 + 2x + C = 0 \]