The general solution of the differential equation x `(dy)/(dx) + y = 0` is

Options

A. `xy = C `
B. `x = Cy`
C. `x + y = C
D. `x^(2) + y^(2) = C `

Answer

To solve the differential equation \( x \frac{dy}{dx} + y = 0 \), we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ x \frac{dy}{dx} + y = 0 \]
We can rearrange this to isolate \( \frac{dy}{dx} \):
\[ x \frac{dy}{dx} = -y \]
Now, divide both sides by \( x \):
\[ \frac{dy}{dx} = -\frac{y}{x} \]

### Step 2: Separate the variables
Next, we separate the variables \( y \) and \( x \):
\[ \frac{dy}{y} = -\frac{dx}{x} \]

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

### Step 4: Simplify the equation
We can rewrite the equation by using properties of logarithms:
\[ \ln |y| + \ln |x| = C \]
This can be simplified to:
\[ \ln |xy| = C \]

### Step 5: Exponentiate both sides
To remove the logarithm, we exponentiate both sides:
\[ |xy| = e^C \]
Let \( k = e^C \) (where \( k \) is a new constant), so we have:
\[ xy = k \]

### Step 6: Write the general solution
Thus, the general solution of the differential equation is:
\[ xy = C \]
where \( C \) is an arbitrary constant.