Show that each of the following differential equations is homogeneous and solve each of them :
`(x^(2)+xy)dy+(3xy+y^(2))dx=0`

Answer

To solve the differential equation \((x^{2}+xy)dy+(3xy+y^{2})dx=0\), we will follow these steps:

### Step 1: Show that the differential equation is homogeneous

A differential equation is homogeneous if it can be expressed in the form \(f(\lambda x, \lambda y) = \lambda^n f(x, y)\) for some integer \(n\).

1. Rewrite the equation:
\[ (x^{2} + xy)dy + (3xy + y^{2})dx = 0 \]
This can be rearranged to:
\[ dy = -\frac{3xy + y^{2}}{x^{2} + xy}dx \]

2. Define \(f(x, y) = -\frac{3xy + y^{2}}{x^{2} + xy}\).

3. Now, substitute \(\lambda x\) and \(\lambda y\) into \(f\):
\[ f(\lambda x, \lambda y) = -\frac{3(\lambda x)(\lambda y) + (\lambda y)^{2}}{(\lambda x)^{2} + (\lambda x)(\lambda y)} \]
Simplifying this gives:
\[ = -\frac{3\lambda^2 xy + \lambda^2 y^2}{\lambda^2 x^2 + \lambda^2 xy} = -\lambda^2 \frac{3xy + y^2}{x^2 + xy} \]
This shows that:
\[ f(\lambda x, \lambda y) = \lambda^2 f(x, y) \]
Therefore, the equation is homogeneous of degree 2.

### Step 2: Solve the differential equation

1. Use the substitution \(y = zx\), where \(z = \frac{y}{x}\). Thus, \(dy = zdx + xdz\).

2. Substitute into the equation:
\[ (x^2 + x(zx))(zdx + xdz) + (3x(zx) + (zx)^2)dx = 0 \]
Simplifying gives:
\[ (x^2 + zx^2)(zdx + xdz) + (3zx^2 + z^2x^2)dx = 0 \]
This simplifies to:
\[ (x^2(1 + z))(zdx + xdz) + x^2(3z + z^2)dx = 0 \]

3. Dividing through by \(x^2\) (assuming \(x \neq 0\)):
\[ (1 + z)(z + \frac{dz}{dx}) + (3z + z^2) = 0 \]

4. Rearranging gives:
\[ z + z^2 + z\frac{dz}{dx} + \frac{dz}{dx} + 3z + z^2 = 0 \]
Thus:
\[ (1 + z)\frac{dz}{dx} = -4z \]

5. Rearranging gives:
\[ \frac{dz}{dx} = -\frac{4z}{1 + z} \]

6. Separating variables:
\[ \frac{1 + z}{z}dz = -4dx \]

7. Integrating both sides:
\[ \int \left(\frac{1}{z} + 1\right)dz = -4\int dx \]
This results in:
\[ \ln |z| + z = -4x + C \]

8. Substitute back \(z = \frac{y}{x}\):
\[ \ln \left|\frac{y}{x}\right| + \frac{y}{x} = -4x + C \]

### Final Solution

The final implicit solution of the differential equation is:
\[ \ln |y| - \ln |x| + \frac{y}{x} = -4x + C \]