Show that each of the following differential equations is homogeneous and solve each of them :
`(3xy+y^(2))dx=(x^(2)+xy)dy`

Answer

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

### Step 1: Rewrite the Differential Equation
We start by rewriting the given differential equation in the standard form:
\[ (3xy + y^2)dx - (x^2 + xy)dy = 0 \]

### Step 2: Check for Homogeneity
A differential equation is homogeneous if both sides can be expressed as functions of \( \frac{y}{x} \). We will check if the equation is homogeneous by dividing each term by \( x^2 \):
\[ \frac{3xy}{x^2} + \frac{y^2}{x^2} = \frac{3y}{x} + \frac{y^2}{x^2} \]
\[ \frac{x^2}{x^2} + \frac{xy}{x^2} = 1 + \frac{y}{x} \]
Both sides can be expressed in terms of \( \frac{y}{x} \), confirming that the equation is homogeneous.

### Step 3: Substitute \( y = vx \)
Let \( y = vx \), where \( v \) is a function of \( x \). Then, \( dy = vdx + xdv \).

### Step 4: Substitute into the Equation
Substituting \( y = vx \) and \( dy \) into the differential equation:
\[ (3x(vx) + (vx)^2)dx = (x^2 + x(vx))(vdx + xdv) \]
This simplifies to:
\[ (3vx^2 + v^2x^2)dx = (x^2 + vx^2)(vdx + xdv) \]
Now, we can factor out \( x^2 \):
\[ (3v + v^2)dx = (1 + v)(vdx + xdv) \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ (3v + v^2)dx = (1 + v)vdx + (1 + v)xdv \]
This leads to:
\[ (3v + v^2 - (1 + v)v)dx = (1 + v)xdv \]
Simplifying the left-hand side:
\[ (3v + v^2 - v - v^2)dx = (1 + v)xdv \]
\[ (2v)dx = (1 + v)xdv \]

### Step 6: Separate Variables
Now, we separate the variables:
\[ \frac{dx}{x} = \frac{(1 + v)}{2v} dv \]

### Step 7: Integrate Both Sides
Integrating both sides:
\[ \int \frac{dx}{x} = \int \frac{(1 + v)}{2v} dv \]
The left side integrates to:
\[ \ln |x| = \frac{1}{2} \int \left( \frac{1}{v} + 1 \right) dv \]
The right side integrates to:
\[ \ln |x| = \frac{1}{2} (\ln |v| + v) + C \]

### Step 8: Solve for \( y \)
Substituting back \( v = \frac{y}{x} \):
\[ \ln |x| = \frac{1}{2} \left( \ln \left| \frac{y}{x} \right| + \frac{y}{x} \right) + C \]
Exponentiating both sides gives:
\[ |x| = e^C \cdot \left( \frac{y}{x} \right)^{1/2} \cdot e^{y/x/2} \]
This can be rearranged to find the relationship between \( x \) and \( y \).

### Final Solution
The final solution can be expressed in terms of \( y \) and \( x \) as:
\[ \ln |y| - \ln |x| = 2\ln |x| + \frac{y}{x} + C \]