The solution of `(dy)/(dx)+1 = e^(x+y) ` is

Options

A. ` e^(-(x+y))+x+C=0`
B. ` e^((-x+y)) - x + C = 0 `
C. ` e^(x+y) + x+C = 0`
D. ` e^(x+y)- x + C = 0 `

Explanation

To solve the differential equation \(\frac{dy}{dx} + 1 = e^{x+y}\), we will follow a systematic approach.

### Step 1: Rewrite the equation
We start with the given equation:
\[ \frac{dy}{dx} + 1 = e^{x+y} \]
Subtract 1 from both sides:
\[ \frac{dy}{dx} = e^{x+y} - 1 \]

### Step 2: Substitute \(t = x + y\)
Let \(t = x + y\). Then, differentiating both sides with respect to \(x\) gives:
\[ \frac{dt}{dx} = 1 + \frac{dy}{dx} \]
Substituting \(\frac{dy}{dx}\) from the previous step, we have:
\[ \frac{dt}{dx} = 1 + (e^{x+y} - 1) = e^{t} \]

### Step 3: Rearrange the equation
Now we can rearrange the equation:
\[ \frac{dt}{dx} = e^{t} \]
This can be rewritten as:
\[ \frac{dt}{e^{t}} = dx \]

### Step 4: Integrate both sides
Now we integrate both sides:
\[ \int \frac{dt}{e^{t}} = \int dx \]
The left side integrates to:
\[ -\int e^{-t} dt = -e^{-t} + C_1 \]
The right side integrates to:
\[ x + C_2 \]
Combining these results gives:
\[ -e^{-t} = x + C \]
where \(C = C_2 - C_1\).

### Step 5: Substitute back for \(t\)
Recall that \(t = x + y\), so we substitute back:
\[ -e^{-(x+y)} = x + C \]

### Step 6: Rearranging the equation
To express the solution clearly, we can rearrange it:
\[ e^{-(x+y)} = -x - C \]
or
\[ e^{x+y} = \frac{1}{-x - C} \]

### Final Solution
Thus, the solution of the differential equation \(\frac{dy}{dx} + 1 = e^{x+y}\) is:
\[ e^{x+y} = \frac{1}{-x - C} \]