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: Rearranging the Equation
First, we can multiply both sides by \(2xy\) to eliminate the fraction:

\[ 2xy \frac{dy}{dx} = y^2 + x \ln x \]

### Step 2: Isolate Terms
Next, we can rearrange the equation to isolate the terms involving \(y\):

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

### Step 3: Substitution
Now, let's make a substitution. Let \(t = y^2\). Then, differentiating gives us:

\[ 2y \frac{dy}{dx} = \frac{dt}{dx} \]

Substituting this into our equation gives:

\[ \frac{dt}{dx} = \frac{t}{x} + \ln x \]

### Step 4: Rearranging to Linear Form
Rearranging the equation gives us:

\[ \frac{dt}{dx} - \frac{t}{x} = \ln x \]

This is now in the standard form of a linear differential equation:

\[ \frac{dt}{dx} + P(x)t = Q(x) \]

where \(P(x) = -\frac{1}{x}\) and \(Q(x) = \ln x\).

### Step 5: Finding the Integrating Factor
The integrating factor \(I(x)\) is given by:

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

### Step 6: Multiply the Equation by the Integrating Factor
Now, 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: Integrate Both Sides
Integrating both sides with respect to \(x\):

\[ \int \frac{d}{dx}\left(\frac{t}{x}\right) \, dx = \int \frac{\ln x}{x} \, dx \]

The left side integrates to:

\[ \frac{t}{x} = \int \frac{\ln x}{x} \, dx \]

The right side can be solved using integration by parts. Let \(u = \ln x\) and \(dv = \frac{1}{x}dx\):

\[ du = \frac{1}{x}dx, \quad v = x \]

Thus,

\[ \int \frac{\ln x}{x} \, dx = \frac{(\ln x)^2}{2} + C \]

### Step 8: Substitute Back
Now substituting back, we have:

\[ \frac{y^2}{x} = \frac{(\ln x)^2}{2} + C \]

Multiplying through by \(x\):

\[ y^2 = x \left(\frac{(\ln x)^2}{2} + C\right) \]

### Step 9: Final Form
Thus, the solution of the differential equation is:

\[ y^2 = \frac{x (\ln x)^2}{2} + Cx \]