The solution (x+ y+ 2) dy = dx is :

Options

A. ` x + y + 2 = Ce ^(y)`
B. `x+y +4=c c y`
C. `log (x + y+2)=Cy`
D. none of these

Answer

To solve the differential equation \((x + y + 2) dy = dx\), we can follow these steps:

### Step 1: Rearranging the equation
We start by rewriting the equation in a more standard form. We can divide both sides by \(dy\):

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

Now, we can express this as:

\[ \frac{dx}{dy} = x + y + 2 \]

### Step 2: Rearranging to standard linear form
Next, we rearrange this equation to isolate the \(x\) term:

\[ \frac{dx}{dy} - x = y + 2 \]

This is now in the standard linear form:

\[ \frac{dx}{dy} + P(x) = Q(y) \]

where \(P = -1\) and \(Q = y + 2\).

### Step 3: Finding the integrating factor
To solve this linear differential equation, we need to find the integrating factor, which is given by:

\[ e^{\int P \, dy} \]

Substituting \(P = -1\):

\[ \text{Integrating factor} = e^{\int -1 \, dy} = e^{-y} \]

### Step 4: Multiplying through by the integrating factor
We multiply the entire equation by the integrating factor \(e^{-y}\):

\[ e^{-y} \frac{dx}{dy} - e^{-y} x = e^{-y} (y + 2) \]

### Step 5: Recognizing the left-hand side as a derivative
The left-hand side can be recognized as the derivative of a product:

\[ \frac{d}{dy}(e^{-y} x) = e^{-y} (y + 2) \]

### Step 6: Integrating both sides
Now we integrate both sides with respect to \(y\):

\[ \int \frac{d}{dy}(e^{-y} x) \, dy = \int e^{-y} (y + 2) \, dy \]

The left side simplifies to:

\[ e^{-y} x = \int e^{-y} (y + 2) \, dy \]

### Step 7: Solving the integral on the right side
To solve the integral on the right side, we can use integration by parts. Let:

- \(u = y + 2\) and \(dv = e^{-y} dy\)
- Then, \(du = dy\) and \(v = -e^{-y}\)

Using integration by parts:

\[ \int e^{-y} (y + 2) \, dy = -e^{-y}(y + 2) - \int -e^{-y} \, dy \]

The second integral simplifies to:

\[ -e^{-y}(y + 2) + e^{-y} + C \]

### Step 8: Putting it all together
Now substituting back, we have:

\[ e^{-y} x = -e^{-y}(y + 2) + e^{-y} + C \]

Multiplying through by \(e^{y}\):

\[ x = -(y + 2) + 1 + Ce^{y} \]

This simplifies to:

\[ x + y + 3 = Ce^{y} \]

### Final Solution
Thus, the general solution of the differential equation is:

\[ x + y + 3 = Ce^{y} \]