If `(dy)/(dx)= (2)/(x+y)`, then `x+y+2=`

Options

A. `ce^(y)`
B. `ce^((y)/(2))`
C. `ce^(-y)`
D. `ce^((y)/(2))`

Answer

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

### Step 1: Rewrite the equation
We start with the given equation:
\[ \frac{dy}{dx} = \frac{2}{x+y} \]
We can rearrange this to:
\[ \frac{dx}{dy} = \frac{x+y}{2} \]

### Step 2: Substitute \(x+y\) with a new variable
Let:
\[ t = x + y \]
Then, differentiating both sides with respect to \(y\), we have:
\[ \frac{dt}{dy} = \frac{dx}{dy} + 1 \]
Thus, we can express \(\frac{dx}{dy}\) as:
\[ \frac{dx}{dy} = \frac{dt}{dy} - 1 \]

### Step 3: Substitute into the equation
Substituting \(\frac{dx}{dy}\) into our rearranged equation gives:
\[ \frac{dt}{dy} - 1 = \frac{t}{2} \]
Rearranging this, we get:
\[ \frac{dt}{dy} = \frac{t}{2} + 1 \]

### Step 4: Separate variables
We can separate the variables:
\[ \frac{dt}{\frac{t}{2} + 1} = dy \]

### Step 5: Integrate both sides
Now we integrate both sides. The left side requires partial fraction decomposition:
\[ \int \frac{dt}{\frac{t}{2} + 1} = \int dy \]
This simplifies to:
\[ 2 \ln\left|\frac{t}{2} + 1\right| = y + C \]

### Step 6: Solve for \(t\)
Exponentiating both sides, we have:
\[ \left|\frac{t}{2} + 1\right| = e^{\frac{y+C}{2}} \]
Let \(C_1 = e^{\frac{C}{2}}\), then:
\[ \frac{t}{2} + 1 = C_1 e^{\frac{y}{2}} \]
Thus:
\[ t = 2C_1 e^{\frac{y}{2}} - 2 \]

### Step 7: Substitute back for \(t\)
Recall that \(t = x + y\):
\[ x + y = 2C_1 e^{\frac{y}{2}} - 2 \]

### Step 8: Express \(x + y + 2\)
Finally, we find:
\[ x + y + 2 = 2C_1 e^{\frac{y}{2}} \]

### Conclusion
Thus, the expression for \(x + y + 2\) is:
\[ x + y + 2 = 2C_1 e^{\frac{y}{2}} \]