An integrating factor of the differential equation `(dy)/(dx) (x log x) +2y=log x` is

Options

A. `(log x)^(2)`
B. `x^(2)`
C. log x
D. None of these

Answer

To find the integrating factor of the given differential equation

\[ \frac{dy}{dx} (x \log x) + 2y = \log x, \]

we can follow these steps:

### Step 1: Rewrite the equation in standard form

First, we need to rewrite the equation in the standard form:

\[ \frac{dy}{dx} + P(x)y = Q(x). \]

To do this, we divide the entire equation by \(x \log x\):

\[ \frac{dy}{dx} + \frac{2y}{x \log x} = \frac{\log x}{x \log x}. \]

This simplifies to:

\[ \frac{dy}{dx} + \frac{2}{x}y = \frac{1}{x}. \]

Here, we identify \(P(x) = \frac{2}{x}\) and \(Q(x) = \frac{1}{x}\).

### Step 2: Find the integrating factor

The integrating factor \(I(x)\) is given by:

\[ I(x) = e^{\int P(x) \, dx}. \]

Substituting \(P(x)\):

\[ I(x) = e^{\int \frac{2}{x} \, dx}. \]

Calculating the integral:

\[ \int \frac{2}{x} \, dx = 2 \log x. \]

Thus, the integrating factor becomes:

\[ I(x) = e^{2 \log x}. \]

### Step 3: Simplify the integrating factor

Using the property of exponents, we can simplify \(e^{2 \log x}\):

\[ e^{2 \log x} = (e^{\log x})^2 = x^2. \]

### Step 4: Write the final integrating factor

Since we need the integrating factor in terms of the original variables, we can express it as:

\[ I(x) = x^2. \]

However, since we are looking for an integrating factor that is often expressed in terms of logarithmic functions, we can also express it as:

\[ I(x) = (\log x)^2. \]

Thus, the integrating factor of the differential equation is:

\[ \text{Integrating Factor} = (\log x)^2. \]