Form differential equation for `y=A log x+B`

Options

A. `xy_(2)+y_(1)=0`
B. `xy_(1)=+y_(2)=0`
C. `y_(1)+y_(2)=x`
D. `y_(2)=x+y_(1)`

Answer

To form the differential equation for \( y = A \log x + B \), we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)

Given:
\[ y = A \log x + B \]

We differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(A \log x + B) \]

Since \( B \) is a constant, its derivative is 0. Thus:
\[ \frac{dy}{dx} = A \cdot \frac{1}{x} = \frac{A}{x} \]

### Step 2: Express \( A \) in terms of \( y \) and \( \frac{dy}{dx} \)

From the equation \( \frac{dy}{dx} = \frac{A}{x} \), we can express \( A \) as:
\[ A = x \frac{dy}{dx} \]

### Step 3: Substitute \( A \) back into the equation for \( y \)

Now substituting \( A \) in the original equation:
\[ y = A \log x + B \implies y = \left(x \frac{dy}{dx}\right) \log x + B \]

### Step 4: Differentiate again to find \( \frac{d^2y}{dx^2} \)

Now, we differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{A}{x}\right) = \frac{d}{dx}\left(\frac{x \frac{dy}{dx}}{x}\right) \]

Using the product rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(x \frac{dy}{dx}) \cdot \frac{1}{x} - \frac{y_1}{x^2} \]

### Step 5: Form the differential equation

From the previous steps, we know:
- \( A = x \frac{dy}{dx} \)
- Substitute this into the equation to eliminate \( A \) and \( B \).

Using the relationship \( xy_2 + y_1 = 0 \) (where \( y_1 = \frac{dy}{dx} \) and \( y_2 = \frac{d^2y}{dx^2} \)), we arrive at the differential equation:
\[ xy \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0 \]

### Final Differential Equation

Thus, the required differential equation is:
\[ xy \frac{d^2y}{dx^2} + \frac{dy}{dx} = 0 \]

---