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

Explanation

To find the degree of the differential equation satisfying the given 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: Rewrite the given equation
We start with the original equation:

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

### Step 2: Differentiate both sides with respect to \(x\)
We will differentiate the equation implicitly with respect to \(x\).

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

Therefore, the left-hand side becomes:
\[ \frac{x}{\sqrt{1+x^2}} + \frac{y y'}{\sqrt{1+y^2}}. \]

2. The right-hand side becomes:
- The derivative of \(\lambda \left( x \sqrt{1+y^2} \right)\) is \(\lambda \left( \sqrt{1+y^2} + x \cdot \frac{y y'}{\sqrt{1+y^2}} \right)\).
- The derivative of \(-\lambda \left( y \sqrt{1+x^2} \right)\) is \(-\lambda \left( y' \sqrt{1+x^2} + y \cdot \frac{x}{\sqrt{1+x^2}} \right)\).

Therefore, the right-hand side becomes:
\[ \lambda \left( \sqrt{1+y^2} + x \cdot \frac{y y'}{\sqrt{1+y^2}} \right) - \lambda \left( y' \sqrt{1+x^2} + y \cdot \frac{x}{\sqrt{1+x^2}} \right). \]

### Step 3: Set the derivatives equal to each other
Now we equate the left-hand side and right-hand side:

\[ \frac{x}{\sqrt{1+x^2}} + \frac{y y'}{\sqrt{1+y^2}} = \lambda \left( \sqrt{1+y^2} + x \cdot \frac{y y'}{\sqrt{1+y^2}} - y' \sqrt{1+x^2} - y \cdot \frac{x}{\sqrt{1+x^2}} \right). \]

### Step 4: Identify the highest power of \(y'\)
In the resulting equation, we need to identify the highest power of \(y'\) (which is \(dy/dx\)).

From the equation, we see that \(y'\) appears linearly (to the first power) in the equation. There are no higher powers of \(y'\) (like \(y''\) or \(y'^2\)).

### Conclusion
The degree of the differential equation is determined by the highest power of \(y'\) present in the equation. Since the highest power of \(y'\) is 1, we conclude that:

\[ \text{Degree of the differential equation} = 1. \]