Integrating factor of the differential equation `(x.logx)(dy)/(dx)+y=2logx` is

Options

A. ` e^x`
B. `log x`
C. `log (log x)`
D. `x`

Answer

To find the integrating factor of the given differential equation

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

we will follow these steps:

### Step 1: Rewrite the Equation in Standard Form
First, we need to rewrite the differential equation in the standard linear form:

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

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

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

This simplifies to:

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

### Step 2: Identify \(P(x)\)
From the equation, we can identify \(P(x)\):

\[ P(x) = \frac{1}{x \log x}. \]

### Step 3: Find the Integrating Factor
The integrating factor \(I(x)\) is given by:

\[ I(x) = e^{\int P(x) \, dx} = e^{\int \frac{1}{x \log x} \, dx}. \]

### Step 4: Solve the Integral
To solve the integral \(\int \frac{1}{x \log x} \, dx\), we can use the substitution method. Let:

\[ t = \log x \implies dt = \frac{1}{x} \, dx \implies dx = x \, dt = e^t \, dt. \]

Substituting into the integral gives:

\[ \int \frac{1}{x \log x} \, dx = \int \frac{1}{e^t t} e^t \, dt = \int \frac{1}{t} \, dt = \log |t| + C = \log |\log x| + C. \]

### Step 5: Calculate the Integrating Factor
Now substituting back, we have:

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

Since \(\log x\) is positive for \(x > 1\), we can simplify this to:

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

### Conclusion
The integrating factor of the given differential equation is:

\[ \log x. \]