The solution of the differential equation `(dy)/(dx)=(y^(2)+xlnx)/(2xy)` is (where, c is the constant of integration)

Options

A. `2x^(2)=y(lnx)^(2)+2cy`
B. `2y^(2)=x(lnx)^(2)+2cx`
C. `x^(2)=y(lnm)^(2)+c`
D. `2y^(2)=(x)/(y)(lnx)^(2)+cx`

Answer

To solve the differential equation \(\frac{dy}{dx} = \frac{y^2 + x \ln x}{2xy}\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given differential equation:
\[ \frac{dy}{dx} = \frac{y^2 + x \ln x}{2xy} \]
Multiply both sides by \(2xy\):
\[ 2xy \frac{dy}{dx} = y^2 + x \ln x \]

### Step 2: Rearranging the equation
Now, we can rearrange the equation:
\[ 2y \frac{dy}{dx} = \frac{y^2}{x} + \ln x \]
Next, divide both sides by \(x\):
\[ \frac{2y}{x} \frac{dy}{dx} = \frac{y^2}{x^2} + \frac{\ln x}{x} \]

### Step 3: Substituting \(y^2\)
Let \(T = y^2\). Then, differentiate \(T\) with respect to \(x\):
\[ \frac{dT}{dx} = 2y \frac{dy}{dx} \]
Substituting this into our equation gives:
\[ \frac{dT}{dx} = \frac{T}{x} + \ln x \]

### Step 4: Forming a linear differential equation
Now we have a linear first-order differential equation:
\[ \frac{dT}{dx} - \frac{T}{x} = \ln x \]
Here, \(P = -\frac{1}{x}\) and \(Q = \ln x\).

### Step 5: Finding the integrating factor
The integrating factor \(I\) is given by:
\[ I = e^{\int P \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln x} = \frac{1}{x} \]

### Step 6: Multiplying through by the integrating factor
Multiply the entire differential equation by the integrating factor:
\[ \frac{1}{x} \frac{dT}{dx} - \frac{T}{x^2} = \frac{\ln x}{x} \]
This simplifies to:
\[ \frac{d}{dx}\left(\frac{T}{x}\right) = \frac{\ln x}{x} \]

### Step 7: Integrating both sides
Integrate both sides:
\[ \int \frac{d}{dx}\left(\frac{T}{x}\right) \, dx = \int \frac{\ln x}{x} \, dx \]
The left side simplifies to \(\frac{T}{x}\). The right side can be solved using integration by parts, letting \(u = \ln x\) and \(dv = \frac{1}{x}dx\):
\[ \int \ln x \, d(\ln x) = \frac{(\ln x)^2}{2} + C \]
Thus, we have:
\[ \frac{T}{x} = \frac{(\ln x)^2}{2} + C \]

### Step 8: Solving for \(T\)
Multiplying through by \(x\):
\[ T = x \left(\frac{(\ln x)^2}{2} + C\right) \]

### Step 9: Substituting back for \(y\)
Recall that \(T = y^2\):
\[ y^2 = x \left(\frac{(\ln x)^2}{2} + C\right) \]

### Final Solution
Thus, the solution of the differential equation is:
\[ 2y^2 = x(\ln x)^2 + 2Cx \]