The degree of the differential equation satisfying the relation `sqrt(1+x^2) + sqrt(1+y^2) = lambda (x sqrt(1+y^2)- ysqrt(1+x^2))` is

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To find the degree of the differential equation satisfying the relation

\[ \sqrt{1+x^2} + \sqrt{1+y^2} = \lambda \left( x \sqrt{1+y^2} - y \sqrt{1+x^2} \right), \]

we will follow these steps:

### Step 1: Rearranging the Equation
We start by rewriting the equation in a form that is easier to differentiate. We can express it as:

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

### Step 2: Differentiating Both Sides
Next, we differentiate both sides with respect to \(x\). We will use the quotient rule for differentiation, which states that if \(u\) and \(v\) are functions of \(x\), then:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}. \]

Let \(u = \sqrt{1+x^2} + \sqrt{1+y^2}\) and \(v = x \sqrt{1+y^2} - y \sqrt{1+x^2}\).

### Step 3: Applying the Quotient Rule
Using the quotient rule, we differentiate \(u\) and \(v\):

- For \(u\):
- The derivative of \(\sqrt{1+x^2}\) is \(\frac{x}{\sqrt{1+x^2}}\).
- The derivative of \(\sqrt{1+y^2}\) is \(\frac{y}{\sqrt{1+y^2}} \cdot \frac{dy}{dx} = y' \cdot \frac{1}{\sqrt{1+y^2}}\).

Thus,

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

- For \(v\):
- The derivative of \(x \sqrt{1+y^2}\) is \(\sqrt{1+y^2} + x \cdot \frac{y}{\sqrt{1+y^2}} \cdot y'\).
- The derivative of \(-y \sqrt{1+x^2}\) is \(-y' \cdot \sqrt{1+x^2} - y \cdot \frac{x}{\sqrt{1+x^2}}\).

Thus,

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

### Step 4: Setting the Derivative Equal to Zero
After differentiating, we set the derivative equal to zero since the right-hand side (the derivative of \(\lambda\)) is zero:

\[ \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} = 0. \]

This implies that the numerator must be zero:

\[ v \frac{du}{dx} - u \frac{dv}{dx} = 0. \]

### Step 5: Finding the Degree
The degree of a differential equation is defined as the power of the highest derivative term after the equation has been made polynomial in derivatives. In our case, the highest derivative we have is \(y'\), which appears only to the first power.

### Conclusion
Therefore, the degree of the differential equation is:

\[ \text{Degree} = 1. \]