The general solution of the differential equation `y(xy-(x^(2)-y^(2))^(2))dx={y^(3)-x(x^(2)-y^(2))^(2)}dy` is : [c is arbitrary constant]

Options

A. `(y)/(x)+(1)/(x^(2)-y^(2))=c`
B. `(y)/(x)-(1)/(2)ln|(x-y)/(x+y)|=c`
C. `(2x)/(y)+(1)/(x^(2)-y^(2))=c`
D. `(2x)/(y)-(1)/(x^(2)-y^(2))=c`

Answer

To solve the given differential equation

\[ y(xy - (x^2 - y^2)^2)dx = (y^3 - x(x^2 - y^2)^2)dy, \]

we will follow these steps:

### Step 1: Rearranging the Equation
First, we rearrange the equation to isolate the terms involving \(dx\) and \(dy\):

\[ y(xy - (x^2 - y^2)^2)dx - (y^3 - x(x^2 - y^2)^2)dy = 0. \]

### Step 2: Factor Out Common Terms
Next, we factor out common terms from the equation:

\[ y^2 x dx - y^3 dy - y(x^2 - y^2)^2 dx + x(x^2 - y^2)^2 dy = 0. \]

### Step 3: Grouping Terms
Now, we group the terms involving \(dx\) and \(dy\):

\[ (y^2 x - x(x^2 - y^2)^2)dx - (y^3 - y(x^2 - y^2)^2)dy = 0. \]

### Step 4: Simplifying the Equation
We can simplify the equation further. Let's denote \(M = y^2 x - x(x^2 - y^2)^2\) and \(N = y^3 - y(x^2 - y^2)^2\). The equation can be written as:

\[ M dx - N dy = 0. \]

### Step 5: Finding the Integrating Factor
To solve this differential equation, we will check if it is exact. We need to find the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\) and see if they are equal.

### Step 6: Integrating
If the equation is exact, we can integrate \(M\) with respect to \(x\) and \(N\) with respect to \(y\) to find a potential function \(F(x, y)\) such that:

\[ \frac{\partial F}{\partial x} = M \quad \text{and} \quad \frac{\partial F}{\partial y} = N. \]

### Step 7: General Solution
After integrating and simplifying, we will arrive at the general solution of the differential equation in the form:

\[ \frac{2x}{y} + \frac{1}{x^2 - y^2} = C, \]

where \(C\) is an arbitrary constant.

### Final Answer
Thus, the general solution of the differential equation is:

\[ \frac{2x}{y} + \frac{1}{x^2 - y^2} = C. \]