## Solving the Differential Equation: (x − y) dx + x dy = 0

This article will delve into the solution of the differential equation **(x − y) dx + x dy = 0**. We'll explore various techniques and arrive at a general solution.

### Recognizing the Type of Differential Equation

First, let's analyze the given equation to determine its type. We can see that the equation is a **first-order homogeneous differential equation**. This means it can be expressed in the form:

```
dy/dx = f(x,y)
```

where **f(x, y)** is a function of x and y that is homogeneous of degree zero. In other words, **f(tx, ty) = f(x, y)** for any non-zero value of t.

### Solving by Substitution

To solve this equation, we can use the substitution **y = vx**. This implies that:

```
dy/dx = v + x dv/dx
```

Substituting these expressions into the original differential equation, we get:

```
(x - vx) dx + x(v + x dv/dx) = 0
```

Simplifying, we obtain:

```
x dx + x² dv/dx = 0
```

Dividing both sides by x² and rearranging, we get:

```
dv/dx = -1/x
```

### Integration and Finding the General Solution

Now, we can integrate both sides with respect to x:

```
∫ dv = -∫ (1/x) dx
```

This leads to:

```
v = -ln|x| + C
```

where C is the constant of integration.

Finally, substituting back **v = y/x**, we get the general solution:

```
y/x = -ln|x| + C
```

Rearranging, we obtain the final solution:

**y = -x ln|x| + Cx**

### Conclusion

We have successfully solved the differential equation **(x − y) dx + x dy = 0** using the method of substitution. The solution is **y = -x ln|x| + Cx**, where C is an arbitrary constant. This solution represents a family of curves that are solutions to the given differential equation.