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

Options

A. `sin(xy) = x + c`
B. `sin^(-1)(xy)+x = c`
C. `sin(x+c)=xy`
D. `sin(xy)+x = c`

Explanation

To solve the differential equation given by

\[ \sqrt{1 - x^2 y^2} \, dx = y \, dx + x \, dy, \]

we will follow these steps:

### Step 1: Rearranging the equation
We can rearrange the equation to isolate \(dx\):

\[ \sqrt{1 - x^2 y^2} \, dx = y \, dx + x \, dy. \]

This can be rewritten as:

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

### Step 2: Factor out \(dx\)
We can factor out \(dx\) from the right side:

\[ dx = \frac{dx \cdot y + dy \cdot x}{\sqrt{1 - x^2 y^2}}. \]

### Step 3: Dividing by \(dx\)
Assuming \(dx \neq 0\), we can divide both sides by \(dx\):

\[ 1 = \frac{y + \frac{dy}{dx} \cdot x}{\sqrt{1 - x^2 y^2}}. \]

### Step 4: Cross-multiplying
Cross-multiplying gives us:

\[ \sqrt{1 - x^2 y^2} = y + x \frac{dy}{dx}. \]

### Step 5: Isolating \(\frac{dy}{dx}\)
Rearranging this equation to isolate \(\frac{dy}{dx}\):

\[ x \frac{dy}{dx} = \sqrt{1 - x^2 y^2} - y. \]

Thus,

\[ \frac{dy}{dx} = \frac{\sqrt{1 - x^2 y^2} - y}{x}. \]

### Step 6: Introducing a substitution
Let \(v = xy\). Then, we have:

\[ y = \frac{v}{x} \quad \text{and} \quad \frac{dy}{dx} = \frac{1}{x} \frac{dv}{dx} - \frac{v}{x^2}. \]

### Step 7: Substituting into the equation
Substituting \(y\) and \(\frac{dy}{dx}\) into the equation gives:

\[ \frac{1}{x} \frac{dv}{dx} - \frac{v}{x^2} = \frac{\sqrt{1 - v^2} - \frac{v}{x}}{x}. \]

### Step 8: Simplifying the equation
Multiplying through by \(x\) to eliminate the fractions:

\[ \frac{dv}{dx} - \frac{v}{x} = \sqrt{1 - v^2} - v. \]

### Step 9: Rearranging the equation
Rearranging gives:

\[ \frac{dv}{dx} = \sqrt{1 - v^2}. \]

### Step 10: Integrating both sides
Integrating both sides, we have:

\[ \int \frac{dv}{\sqrt{1 - v^2}} = \int dx. \]

This results in:

\[ \sin^{-1}(v) = x + C. \]

### Step 11: Substituting back for \(v\)
Substituting back for \(v = xy\):

\[ \sin^{-1}(xy) = x + C. \]

### Step 12: Final form of the solution
Thus, we can express the general solution as:

\[ xy = \sin(x + C). \]

### Conclusion
The general solution of the differential equation is:

\[ xy = \sin(x + C). \]