Show that each of the following differential equations is homogeneous and solve each of them :
`(dy)/(dx)=(y(y+x))/(x(y-x))`.

Answer

To show that the given differential equation is homogeneous and to solve it, we follow these steps:

### Step 1: Identify the given differential equation
The differential equation given is:
\[ \frac{dy}{dx} = \frac{y(y+x)}{x(y-x)} \]

### Step 2: Check if the function is homogeneous
A function \( f(x, y) \) is homogeneous of degree \( n \) if:
\[ f(\lambda x, \lambda y) = \lambda^n f(x, y) \]
for any non-zero constant \( \lambda \).

Let's define:
\[ f(x, y) = \frac{y(y+x)}{x(y-x)} \]

Now, we will substitute \( x \) with \( \lambda x \) and \( y \) with \( \lambda y \):
\[ f(\lambda x, \lambda y) = \frac{\lambda y (\lambda y + \lambda x)}{\lambda x (\lambda y - \lambda x)} = \frac{\lambda^2 y(y+x)}{\lambda^2 x(y-x)} = \frac{y(y+x)}{x(y-x)} = f(x, y) \]
This shows that \( f(x, y) \) is homogeneous of degree 0.

### Step 3: Substitute \( y = vx \)
To solve the homogeneous differential equation, we can use the substitution \( y = vx \), where \( v \) is a function of \( x \). Then, we have:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]

Substituting \( y = vx \) into the differential equation gives:
\[ v + x \frac{dv}{dx} = \frac{vx(vx+x)}{x(vx-x)} = \frac{v(v+1)x}{v-1} \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ v + x \frac{dv}{dx} = \frac{v(v+1)}{v-1} \]

### Step 5: Separate variables
Now, we can separate the variables:
\[ x \frac{dv}{dx} = \frac{v(v+1)}{v-1} - v \]
Simplifying the right side:
\[ x \frac{dv}{dx} = \frac{v(v+1) - v(v-1)}{v-1} = \frac{v^2 + v - v^2 + v}{v-1} = \frac{2v}{v-1} \]

### Step 6: Further simplify and separate
Now we have:
\[ \frac{dv}{2v} = \frac{dx}{x} \]

### Step 7: Integrate both sides
Integrating both sides:
\[ \int \frac{1}{2v} dv = \int \frac{1}{x} dx \]
This gives:
\[ \frac{1}{2} \ln |v| = \ln |x| + C \]

### Step 8: Solve for \( v \)
Exponentiating both sides:
\[ |v|^{1/2} = |x| e^{2C} \]
Let \( k = e^{2C} \), then:
\[ |v| = k^2 |x|^2 \]

### Step 9: Substitute back for \( y \)
Since \( v = \frac{y}{x} \), we have:
\[ \left|\frac{y}{x}\right| = k^2 |x|^2 \implies |y| = k^2 |x|^3 \]

### Step 10: General solution
Thus, the general solution of the differential equation is:
\[ y = Cx^3 \]
where \( C \) is a constant.