Solve `(dy)/(dx)=(2x-y+1)/(x+2y-3)`

Explanation

To solve the differential equation

\[ \frac{dy}{dx} = \frac{2x - y + 1}{x + 2y - 3}, \]

we will follow these steps:

### Step 1: Cross-multiply
We start by cross-multiplying to eliminate the fraction:

\[ (x + 2y - 3) dy = (2x - y + 1) dx. \]

### Step 2: Rearrange the equation
Next, we will rearrange the equation to group the terms involving \(y\) on one side and the terms involving \(x\) on the other side:

\[ x \, dy + 2y \, dy - 3 \, dy = 2x \, dx - y \, dx + dx. \]

This simplifies to:

\[ x \, dy + y \, dx + 2y \, dy = (2x + 1) \, dx. \]

### Step 3: Rewrite the equation
Now, we can rewrite the left-hand side in terms of a total differential:

\[ d(xy) + (2y - 3) \, dy = (2x + 1) \, dx. \]

### Step 4: Integrate both sides
Now we will integrate both sides. The left side can be integrated as follows:

\[ \int d(xy) + \int (2y - 3) \, dy = \int (2x + 1) \, dx. \]

The integral of \(d(xy)\) is simply \(xy\), and the integral of \((2y - 3) \, dy\) is:

\[ y^2 - 3y. \]

The integral of \((2x + 1) \, dx\) is:

\[ x^2 + x. \]

Putting it all together, we have:

\[ xy + y^2 - 3y = x^2 + x + C, \]

where \(C\) is the constant of integration.

### Step 5: Final form
Thus, the general solution of the differential equation is:

\[ xy + y^2 - 3y - x^2 - x = C. \]