For differential equation `2xlnx. (dy)/(dx) + 2y = 3 lnx`. Let y(x) is solution and y(1) = 0 then y(x) =

Options

A. `3/2 (lnx)^2`
B. `3lnx + x`
C. `3/4 lnx`
D. `3/2 lnx - x`

Answer

To solve the differential equation \( 2x \ln x \frac{dy}{dx} + 2y = 3 \ln x \) with the initial condition \( y(1) = 0 \), we will follow these steps:

### Step 1: Rearranging the Differential Equation
We start with the given equation:
\[ 2x \ln x \frac{dy}{dx} + 2y = 3 \ln x \]
We can rearrange this to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} + \frac{2y}{2x \ln x} = \frac{3 \ln x}{2x \ln x} \]
This simplifies to:
\[ \frac{dy}{dx} + \frac{y}{x \ln x} = \frac{3}{2x} \]

### Step 2: Identifying \( p(x) \) and \( q(x) \)
Now we identify \( p(x) \) and \( q(x) \) from the standard form \( \frac{dy}{dx} + p(x)y = q(x) \):
- \( p(x) = \frac{1}{x \ln x} \)
- \( q(x) = \frac{3}{2x} \)

### Step 3: Finding the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int \frac{1}{x \ln x} \, dx} \]
Let \( u = \ln x \), then \( du = \frac{1}{x}dx \):
\[ \int \frac{1}{x \ln x} \, dx = \int \frac{1}{u} \, du = \ln |\ln x| + C \]
Thus, the integrating factor becomes:
\[ \mu(x) = e^{\ln |\ln x|} = \ln x \]

### Step 4: Multiplying the Equation by the Integrating Factor
Now we multiply the entire differential equation by \( \ln x \):
\[ \ln x \frac{dy}{dx} + \frac{y}{x} = \frac{3 \ln x}{2x} \]

### Step 5: Left Side as a Derivative
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx}(y \ln x) = \frac{3 \ln x}{2x} \]

### Step 6: Integrating Both Sides
Integrate both sides:
\[ \int \frac{d}{dx}(y \ln x) \, dx = \int \frac{3 \ln x}{2x} \, dx \]
The left side simplifies to:
\[ y \ln x \]
For the right side, we can use integration by substitution, letting \( u = \ln x \):
\[ \int \frac{3 \ln x}{2x} \, dx = \frac{3}{2} \int u \, du = \frac{3}{4} u^2 + C = \frac{3}{4} (\ln x)^2 + C \]
Thus, we have:
\[ y \ln x = \frac{3}{4} (\ln x)^2 + C \]

### Step 7: Solving for \( y \)
Now, we isolate \( y \):
\[ y = \frac{3}{4} \frac{(\ln x)^2}{\ln x} + \frac{C}{\ln x} = \frac{3}{4} \ln x + \frac{C}{\ln x} \]

### Step 8: Applying the Initial Condition
We apply the initial condition \( y(1) = 0 \):
\[ 0 = \frac{3}{4} \ln 1 + \frac{C}{\ln 1} \]
Since \( \ln 1 = 0 \), we need to take the limit as \( x \to 1 \):
\[ 0 = 0 + C \cdot \text{undefined} \]
This implies \( C = 0 \).

### Final Solution
Thus, the particular solution is:
\[ y = \frac{3}{4} \ln x \]