What is the solution of the equation ln `((dy)/(dx)) + x = 0`

Options

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

Answer

To solve the equation \( \ln \left(\frac{dy}{dx}\right) + x = 0 \), we will follow these steps:

### Step 1: Isolate \( \frac{dy}{dx} \)
We start with the equation:
\[ \ln \left(\frac{dy}{dx}\right) + x = 0 \]
We can isolate \( \ln \left(\frac{dy}{dx}\right) \) by moving \( x \) to the other side:
\[ \ln \left(\frac{dy}{dx}\right) = -x \]

### Step 2: Exponentiate both sides
To eliminate the natural logarithm, we exponentiate both sides:
\[ \frac{dy}{dx} = e^{-x} \]

### Step 3: Integrate both sides
Next, we integrate both sides with respect to \( x \):
\[ \int dy = \int e^{-x} dx \]
The left side integrates to \( y \), and the right side integrates to \( -e^{-x} + C \), where \( C \) is the constant of integration:
\[ y = -e^{-x} + C \]

### Step 4: Rearranging the equation
We can rearrange the equation to express it in a more standard form:
\[ y + e^{-x} = C \]

### Final Solution
Thus, the solution to the differential equation is:
\[ y = -e^{-x} + C \]