The integrating factor of the differential equation `3x log_(e) x (dy)/(dx)+y=2 log_(e)x` is given by

Options

A. `(log_(e)x)^(3)`
B. `log_(e) (log_(e)x)`
C. `log_(e)x`
D. `(log_(e)x)^(1//3)`

Answer

To find the integrating factor of the given differential equation \(3x \log_e x \frac{dy}{dx} + y = 2 \log_e x\), we can follow these steps:

### Step 1: Rewrite the Equation
First, we rewrite the equation in the standard form of a first-order linear differential equation:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
To do this, we divide the entire equation by \(3x \log_e x\):
\[ \frac{dy}{dx} + \frac{1}{3x \log_e x} y = \frac{2 \log_e x}{3x \log_e x} \]
This simplifies to:
\[ \frac{dy}{dx} + \frac{1}{3x \log_e x} y = \frac{2}{3x} \]

### Step 2: Identify \(P(x)\) and \(Q(x)\)
From the rewritten equation, we identify:
- \(P(x) = \frac{1}{3x \log_e x}\)
- \(Q(x) = \frac{2}{3x}\)

### Step 3: Find the Integrating Factor
The integrating factor \(I(x)\) is given by:
\[ I(x) = e^{\int P(x) \, dx} \]
Substituting \(P(x)\):
\[ I(x) = e^{\int \frac{1}{3x \log_e x} \, dx} \]

### Step 4: Solve the Integral
Let:
\[ I = \int \frac{1}{3x \log_e x} \, dx \]
To solve this integral, we can use the substitution \(t = \log_e x\), which gives \(dt = \frac{1}{x} \, dx\) or \(dx = x \, dt = e^t \, dt\). Thus:
\[ I = \int \frac{1}{3e^t t} e^t \, dt = \int \frac{1}{3t} \, dt = \frac{1}{3} \log |t| + C = \frac{1}{3} \log |\log_e x| + C \]

### Step 5: Write the Integrating Factor
Now substituting back into the expression for the integrating factor:
\[ I(x) = e^{\frac{1}{3} \log |\log_e x| + C} = e^C \cdot |\log_e x|^{1/3} \]
Since \(e^C\) is a constant, we can denote it as \(K\):
\[ I(x) = K |\log_e x|^{1/3} \]
For the purpose of finding the integrating factor, we can ignore the constant \(K\) and write:
\[ I(x) = |\log_e x|^{1/3} \]

### Final Result
Thus, the integrating factor of the differential equation is:
\[ \boxed{(\log_e x)^{1/3}} \]