The solution of the differential equation `2xy"dy"/"dx"=x^2+3y^2` is

Options

A. `x^3+y^2=px^2`
B. `x^2/2+y^2/2=y^2+p`
C. `x^2/2+y^3/x=y^2`
D. `x^2+y^2=px^3`

Answer

To solve the differential equation \( 2xy \frac{dy}{dx} = x^2 + 3y^2 \), we will follow these steps:

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

### Step 2: Substituting \( y = bx \)
To simplify the equation, we can use the substitution \( y = bx \), where \( b \) is a function of \( x \). Then, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = b + x \frac{db}{dx} \]
Substituting this into our rearranged equation gives:
\[ b + x \frac{db}{dx} = \frac{x^2 + 3(bx)^2}{2x(bx)} \]

### Step 3: Simplifying the Equation
Substituting \( y = bx \) into the right-hand side:
\[ b + x \frac{db}{dx} = \frac{x^2 + 3b^2x^2}{2bx^2} \]
This simplifies to:
\[ b + x \frac{db}{dx} = \frac{1 + 3b^2}{2b} \]

### Step 4: Isolating \(\frac{db}{dx}\)
Rearranging gives:
\[ x \frac{db}{dx} = \frac{1 + 3b^2}{2b} - b \]
This can be further simplified to:
\[ x \frac{db}{dx} = \frac{1 + 3b^2 - 2b^2}{2b} = \frac{1 + b^2}{2b} \]

### Step 5: Separating Variables
Now we can separate the variables:
\[ 2b \, db = \frac{1 + b^2}{x} \, dx \]

### Step 6: Integrating Both Sides
Integrate both sides:
\[ \int 2b \, db = \int \frac{1 + b^2}{x} \, dx \]
The left side integrates to:
\[ b^2 \]
The right side integrates to:
\[ \ln |x| + C \]
Thus, we have:
\[ b^2 = \ln |x| + C \]

### Step 7: Back Substituting \( b \)
Recalling that \( b = \frac{y}{x} \), we substitute back:
\[ \left(\frac{y}{x}\right)^2 = \ln |x| + C \]
Multiplying through by \( x^2 \) gives:
\[ y^2 = x^2 (\ln |x| + C) \]

### Final Solution
The solution to the differential equation is:
\[ y^2 = x^2 (\ln |x| + C) \]