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

Options

A. tan xy = log |cx|
B. tan (y/x) = tan log |cx|
C. xy = tan log |cx|
D. None of these

Explanation

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

### Step 1: Rewrite the equation
We start with the given equation:
\[ (1 - xy + x^2 y^2)dx = x^2 dy \]
Rearranging gives us:
\[ (1 - xy + x^2 y^2)dx - x^2 dy = 0 \]

### Step 2: Identify the homogeneous form
We can see that the equation can be expressed in a homogeneous form. To do this, we substitute \(v = xy\), which implies \(y = \frac{v}{x}\).

### Step 3: Differentiate \(y\) with respect to \(x\)
Using the substitution \(y = \frac{v}{x}\), we differentiate \(y\):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{v}{x}\right) = \frac{x \frac{dv}{dx} - v}{x^2} \]

### Step 4: Substitute into the differential equation
Substituting \(y\) and \(\frac{dy}{dx}\) into the original equation:
\[ (1 - v + v^2)dx = x^2 \left(\frac{x \frac{dv}{dx} - v}{x^2}\right)dx \]
This simplifies to:
\[ (1 - v + v^2)dx = (x \frac{dv}{dx} - v)dx \]

### Step 5: Cancel \(dx\) and rearrange
Cancelling \(dx\) from both sides gives:
\[ 1 - v + v^2 = x \frac{dv}{dx} - v \]
Rearranging this leads to:
\[ 1 + v^2 = x \frac{dv}{dx} \]

### Step 6: Separate variables
We can separate variables:
\[ \frac{dv}{1 + v^2} = \frac{dx}{x} \]

### Step 7: Integrate both sides
Integrating both sides:
\[ \int \frac{dv}{1 + v^2} = \int \frac{dx}{x} \]
The left side integrates to \(\tan^{-1}(v)\) and the right side integrates to \(\ln|x| + C\):
\[ \tan^{-1}(v) = \ln|x| + C \]

### Step 8: Substitute back for \(v\)
Substituting back \(v = xy\):
\[ \tan^{-1}(xy) = \ln|x| + C \]

### Step 9: Solve for \(xy\)
Taking the tangent of both sides gives:
\[ xy = \tan(\ln|x| + C) \]

### Final Result
Thus, the solution to the differential equation is:
\[ xy = \tan(\ln|x| + C) \]