The solution of differential equation `x^(2)(x dy + y dx) = (xy - 1)^(2) dx` is (where c is an arbitrary constant)

Options

A. `xy - 1 = cx`
B. `xy - 1 = cx^(2)`
C. `(1)/(xy-1)=(1)/(x)+c`
D. None of these

Explanation

To solve the differential equation \( x^2 (x \, dy + y \, dx) = (xy - 1)^2 \, dx \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ x^2 (x \, dy + y \, dx) = (xy - 1)^2 \, dx \]
We can rearrange this into a more manageable form:
\[ x^2 x \, dy + x^2 y \, dx = (xy - 1)^2 \, dx \]
Now, we can isolate the terms involving \( dy \) and \( dx \):
\[ x^2 x \, dy = (xy - 1)^2 \, dx - x^2 y \, dx \]
This simplifies to:
\[ x^2 x \, dy = ((xy - 1)^2 - x^2 y) \, dx \]

### Step 2: Divide both sides
Next, we can divide both sides by \( (xy - 1)^2 - x^2 y \):
\[ \frac{dy}{dx} = \frac{(xy - 1)^2 - x^2 y}{x^3} \]

### Step 3: Use the product rule
We know that:
\[ d(xy) = x \, dy + y \, dx \]
Thus, we can express \( x \, dy + y \, dx \) as:
\[ d(xy) = x \, dy + y \, dx \]
This allows us to rewrite our equation as:
\[ \frac{d(xy)}{(xy - 1)^2} = \frac{1}{x^2} \, dx \]

### Step 4: Substitute \( xy = t \)
Let \( t = xy \). Then we have:
\[ \frac{dt}{(t - 1)^2} = \frac{1}{x^2} \, dx \]

### Step 5: Integrate both sides
Now we integrate both sides:
\[ \int \frac{1}{(t - 1)^2} \, dt = \int \frac{1}{x^2} \, dx \]
The left side integrates to:
\[ -\frac{1}{t - 1} \]
And the right side integrates to:
\[ -\frac{1}{x} + C \]

### Step 6: Substitute back \( t = xy \)
Substituting back \( t = xy \):
\[ -\frac{1}{xy - 1} = -\frac{1}{x} + C \]

### Step 7: Multiply through by -1
Multiplying through by -1 gives:
\[ \frac{1}{xy - 1} = \frac{1}{x} - C \]

### Step 8: Rearranging the equation
We can rearrange this to:
\[ \frac{1}{xy - 1} = \frac{1}{x} + C \]
where \( C \) is an arbitrary constant.

### Final Answer
Thus, the solution of the differential equation is:
\[ \frac{1}{xy - 1} = \frac{1}{x} + C \]

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