Solve the differential equation : `(dy)/(dx)=(x^(2)-y^(2))/(xy)`.

Options

A. `C=x^(2)(x^(2)+2y^(2))`.
B. `C=x^(2)(x^(2)-2y^(2))`.
C. `C=x^(2)(2x^(2)-2y^(2))`.
D. `C=x^(2)(2x^(2)+2y^(2))`.

Explanation

To solve the differential equation \(\frac{dy}{dx} = \frac{x^2 - y^2}{xy}\), we can follow these steps:

### Step 1: Identify the Type of Differential Equation
The given differential equation is homogeneous because both the numerator and the denominator are of the same degree (2).

**Hint:** Check the degrees of the numerator and the denominator to determine if the equation is homogeneous.

### Step 2: Substitute \(y = ux\)
We will use the substitution \(y = ux\), where \(u\) is a function of \(x\). Consequently, we have:
\[ \frac{dy}{dx} = u + x\frac{du}{dx} \]

**Hint:** This substitution helps to express the equation in terms of a single variable \(u\).

### Step 3: Substitute into the Differential Equation
Substituting \(y = ux\) into the original equation gives:
\[ u + x\frac{du}{dx} = \frac{x^2 - (ux)^2}{x(ux)} = \frac{x^2(1 - u^2)}{ux^2} = \frac{1 - u^2}{u} \]

**Hint:** Simplifying the right-hand side will help in isolating \(\frac{du}{dx}\).

### Step 4: Rearranging the Equation
Rearranging the equation, we get:
\[ x\frac{du}{dx} = \frac{1 - u^2}{u} - u \]
This simplifies to:
\[ x\frac{du}{dx} = \frac{1 - u^2 - u^2}{u} = \frac{1 - 2u^2}{u} \]

**Hint:** Combine like terms to simplify the expression.

### Step 5: Separate Variables
Now, we can separate the variables:
\[ \frac{u}{1 - 2u^2} du = \frac{dx}{x} \]

**Hint:** This step allows us to integrate both sides independently.

### Step 6: Integrate Both Sides
Integrating both sides:
\[ \int \frac{u}{1 - 2u^2} du = \int \frac{dx}{x} \]

For the left side, use the substitution \(v = 1 - 2u^2\), leading to:
\[ \int \frac{-1/2}{v} dv = -\frac{1}{2} \ln |v| + C_1 = -\frac{1}{2} \ln |1 - 2u^2| + C_1 \]
The right side integrates to:
\[ \ln |x| + C_2 \]

**Hint:** Remember to apply appropriate integration techniques for both sides.

### Step 7: Combine the Results
Equating the results from both sides:
\[ -\frac{1}{2} \ln |1 - 2u^2| = \ln |x| + C \]

**Hint:** Combine constants into a single constant \(C\).

### Step 8: Exponentiate Both Sides
Exponentiating both sides gives:
\[ |1 - 2u^2|^{-\frac{1}{2}} = K|x| \quad \text{(where \(K = e^{C}\))} \]

**Hint:** This step transforms the logarithmic equation back into an algebraic form.

### Step 9: Solve for \(u\)
Rearranging gives:
\[ 1 - 2u^2 = \frac{1}{K^2 x^2} \]
Thus,
\[ 2u^2 = 1 - \frac{1}{K^2 x^2} \implies u^2 = \frac{1}{2}\left(1 - \frac{1}{K^2 x^2}\right) \]

**Hint:** Isolate \(u^2\) to express it in terms of \(x\).

### Step 10: Substitute Back for \(y\)
Recalling that \(u = \frac{y}{x}\), we have:
\[ \frac{y^2}{x^2} = \frac{1}{2}\left(1 - \frac{1}{K^2 x^2}\right) \]
Thus,
\[ y^2 = \frac{x^2}{2}\left(1 - \frac{1}{K^2 x^2}\right) \]

**Hint:** Substitute back to express \(y\) in terms of \(x\).

### Final Result
The final solution can be expressed as:
\[ y^2 = \frac{x^2}{2} - \frac{1}{2K^2} \]

This is the general solution to the differential equation.