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

Options

A. ` x + y + 2 = Ce ^(y)`
B. `x+y +4=c c bgy`
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: Rewrite the equation
We start with the given equation:
\[ (x + y + 2) dy = dx \]
We can rearrange this to express \(dy\) in terms of \(dx\):
\[ dy = \frac{dx}{x + y + 2} \]

### Step 2: Express \(\frac{dy}{dx}\)
From the rearranged equation, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{x + y + 2} \]

### Step 3: Introduce a substitution
Let \(v = x + y + 2\). Then, differentiating \(v\) with respect to \(x\) gives:
\[ \frac{dv}{dx} = 1 + \frac{dy}{dx} \]
Substituting \(\frac{dy}{dx}\) into this equation:
\[ \frac{dv}{dx} = 1 + \frac{1}{v} \]

### Step 4: Rearrange the equation
Now, we can rearrange the equation:
\[ \frac{dv}{dx} - 1 = \frac{1}{v} \]
This simplifies to:
\[ \frac{dv}{dx} = 1 + \frac{1}{v} \]

### Step 5: Separate variables
Now, we can separate the variables:
\[ \frac{v}{v + 1} dv = dx \]

### Step 6: Integrate both sides
Integrating both sides:
\[ \int \frac{v}{v + 1} dv = \int dx \]
We can split the left-hand side:
\[ \int \left(1 - \frac{1}{v + 1}\right) dv = \int dx \]
This gives:
\[ v - \log|v + 1| = x + C \]

### Step 7: Substitute back for \(v\)
Now, substitute back \(v = x + y + 2\):
\[ (x + y + 2) - \log|x + y + 3| = x + C \]

### Step 8: Simplify the equation
Simplifying gives:
\[ y + 2 - \log|x + y + 3| = C \]
This can be rearranged to:
\[ \log|x + y + 3| = y + 2 - C \]

### Step 9: Exponentiate both sides
Exponentiating both sides:
\[ x + y + 3 = e^{y + 2 - C} \]
Let \(k = e^{-C}\), we have:
\[ x + y + 3 = k e^{y + 2} \]

### Final Solution
Thus, the general solution of the differential equation is:
\[ x + y + 3 = k e^{y + 2} \]