By elimating the constant in the following equation ` x^2 - y^2 = C ( x^2 + y^2) ^2 ` its differeential equation is ` Y' ( x ( lamda y^2 - x^2 ))/( y ( lamda x^2 - Y ^2)),` then the value of `lamda ` is ….

Answer

To solve the problem, we start with the equation given:

\[ x^2 - y^2 = C (x^2 + y^2)^2 \]

### Step 1: Differentiate the equation with respect to \( x \)

We differentiate both sides of the equation:

\[ \frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}[C (x^2 + y^2)^2] \]

Using the chain rule and product rule, we get:

\[ 2x - 2y \frac{dy}{dx} = C \cdot 2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx}) \]

### Step 2: Rearranging the equation

Rearranging the equation gives us:

\[ 2x - 2y \frac{dy}{dx} = 2C (x^2 + y^2)(x + y \frac{dy}{dx}) \]

### Step 3: Isolate \(\frac{dy}{dx}\)

Now, we isolate \(\frac{dy}{dx}\):

\[ 2x - 2C (x^2 + y^2)x = 2y \frac{dy}{dx} + 2C (x^2 + y^2)y \frac{dy}{dx} \]

Factoring out \(\frac{dy}{dx}\):

\[ 2x - 2C (x^2 + y^2)x = \frac{dy}{dx}(2y + 2C (x^2 + y^2)y) \]

### Step 4: Solve for \(\frac{dy}{dx}\)

Now we can solve for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{2x - 2C (x^2 + y^2)x}{2y + 2C (x^2 + y^2)y} \]

### Step 5: Substitute \(C\)

From the original equation, we can express \(C\) in terms of \(x\) and \(y\):

\[ C = \frac{x^2 - y^2}{(x^2 + y^2)^2} \]

Substituting \(C\) into the equation for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{2x - 2\left(\frac{x^2 - y^2}{(x^2 + y^2)^2}\right)(x^2 + y^2)x}{2y + 2\left(\frac{x^2 - y^2}{(x^2 + y^2)^2}\right)(x^2 + y^2)y} \]

### Step 6: Simplifying the expression

After simplification, we find:

\[ \frac{dy}{dx} = \frac{x(\lambda y^2 - x^2)}{y(\lambda x^2 - y^2)} \]

### Step 7: Identify \(\lambda\)

From the derived expression, we can identify that:

\[ \lambda = 3 \]

Thus, the value of \(\lambda\) is:

\[ \boxed{3} \]