Integrating factor of linear differential equation ` x(dy)/(dx) + 2y = x^(2) log x ` is

Options

A. `1/x^(2)`
B. `1/x`
C. x
D. `x^(2) `

Answer

To find the integrating factor of the linear differential equation \( x \frac{dy}{dx} + 2y = x^2 \log x \), we can follow these steps:

### Step 1: Rewrite the equation
First, we will divide the entire equation by \( x \) to put it in the standard form of a linear differential equation:

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

### Step 2: Identify \( p(x) \)
In the standard form \( \frac{dy}{dx} + p(x)y = q(x) \), we identify \( p(x) \) as:

\[ p(x) = \frac{2}{x} \]

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by the formula:

\[ \mu(x) = e^{\int p(x) \, dx} \]

Substituting \( p(x) \):

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

### Step 4: Compute the integral
Now we compute the integral:

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

### Step 5: Substitute back to find the integrating factor
Now substituting back into the formula for the integrating factor:

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

Using the property of exponents:

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

### Conclusion
Thus, the integrating factor is:

\[ \mu(x) = x^2 \]

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