Solve: `x(x-y) dy + y^(2) dx= 0`

Answer

To solve the differential equation \( x(x-y) dy + y^2 dx = 0 \), we will follow a systematic approach.

### Step-by-step Solution:

1. **Rearranging the Equation:**
We start with the given equation:
\[ x(x-y) dy + y^2 dx = 0 \]
We can rearrange this to isolate \( dy \):
\[ x(x-y) dy = -y^2 dx \]
Dividing both sides by \( y^2 \) and \( x(x-y) \):
\[ \frac{dy}{dx} = -\frac{y^2}{x(x-y)} \]

**Hint:** Rearranging the equation helps to express it in a standard form suitable for further analysis.

2. **Substituting Variables:**
We can use the substitution \( y = vx \), where \( v \) is a function of \( x \). Then, we have:
\[ dy = v dx + x dv \]
Substituting \( y \) and \( dy \) into the equation gives:
\[ \frac{dy}{dx} = v + x \frac{dv}{dx} \]

**Hint:** The substitution \( y = vx \) simplifies the equation by reducing the number of variables.

3. **Substituting into the Equation:**
Now substituting \( y = vx \) and \( dy = v dx + x dv \) into the rearranged equation:
\[ v + x \frac{dv}{dx} = -\frac{(vx)^2}{x(x - vx)} \]
Simplifying the right-hand side:
\[ v + x \frac{dv}{dx} = -\frac{v^2 x^2}{x^2(1 - v)} = -\frac{v^2}{1 - v} \]

**Hint:** This step involves substituting the expressions for \( y \) and \( dy \) into the differential equation to express everything in terms of \( v \) and \( x \).

4. **Rearranging the Equation:**
Rearranging gives:
\[ x \frac{dv}{dx} = -\frac{v^2}{1 - v} - v \]
This simplifies to:
\[ x \frac{dv}{dx} = -\frac{v^2 + v(1 - v)}{1 - v} = -\frac{v^2 + v - v^2}{1 - v} = -\frac{v}{1 - v} \]

**Hint:** Rearranging helps to isolate \( \frac{dv}{dx} \) on one side, making it easier to integrate.

5. **Separating Variables:**
We can separate the variables:
\[ \frac{1 - v}{v} dv = -\frac{dx}{x} \]

**Hint:** Separating variables is a crucial step that allows us to integrate both sides independently.

6. **Integrating Both Sides:**
Now we integrate both sides:
\[ \int \left( \frac{1}{v} - 1 \right) dv = -\int \frac{1}{x} dx \]
This gives:
\[ \ln |v| - v = -\ln |x| + C \]

**Hint:** Integration is the process of finding the antiderivative, which helps to solve the differential equation.

7. **Solving for \( v \):**
Rearranging gives:
\[ \ln |v| + \ln |x| = v + C \]
This can be rewritten as:
\[ \ln |vx| = v + C \]

**Hint:** Combining logarithmic terms helps to simplify the expression further.

8. **Back Substituting for \( y \):**
Recall that \( v = \frac{y}{x} \), so we have:
\[ \ln |y| + \ln |x| = \frac{y}{x} + C \]
Exponentiating both sides gives:
\[ y = kx e^{\frac{y}{x}} \]
where \( k = e^C \).

**Hint:** Back substitution is essential to express the solution in terms of the original variables.

### Final Solution:
The solution to the differential equation is:
\[ y = kx e^{\frac{y}{x}} \]
where \( k \) is an arbitrary constant.