Find the order the degree, if defined, of the following differential equation and state whether they are linear or non-linear :
`sqrt(1-x^(2))dx+sqrt(1-y^(2))dy=0`.

Answer

To solve the given differential equation and determine its order, degree, and whether it is linear or non-linear, we follow these steps:

**Step 1: Write the given differential equation.**
The equation is given as:
\[ \sqrt{1 - x^2} \, dx + \sqrt{1 - y^2} \, dy = 0 \]

**Step 2: Rearrange the equation.**
We can rearrange the equation to isolate \( dy \) on one side:
\[ \sqrt{1 - y^2} \, dy = -\sqrt{1 - x^2} \, dx \]

**Step 3: Separate the variables.**
Now, we separate the variables \( y \) and \( x \):
\[ \frac{dy}{\sqrt{1 - y^2}} = -\frac{dx}{\sqrt{1 - x^2}} \]

**Step 4: Identify the order of the differential equation.**
The order of a differential equation is determined by the highest derivative present. In our rearranged equation, we have \( dy \) and \( dx \), but we can express it as:
\[ \frac{dy}{dx} = -\frac{\sqrt{1 - x^2}}{\sqrt{1 - y^2}} \]
Here, \( \frac{dy}{dx} \) is the first derivative, indicating that the order of the differential equation is 1.

**Step 5: Identify the degree of the differential equation.**
The degree of a differential equation is defined as the power of the highest derivative when the equation is a polynomial in derivatives. In our case, the highest derivative \( \frac{dy}{dx} \) appears to the power of 1. Therefore, the degree is also 1.

**Step 6: Determine if the equation is linear or non-linear.**
A differential equation is linear if it can be expressed in the form:
\[ a_n(x) \frac{d^n y}{dx^n} + a_{n-1}(x) \frac{d^{n-1} y}{dx^{n-1}} + \ldots + a_1(x) \frac{dy}{dx} + a_0(x)y = g(x) \]
where \( a_i(x) \) are functions of \( x \) only and \( g(x) \) is also a function of \( x \). In our case, the equation involves \( \sqrt{1 - y^2} \), which is not a linear function of \( y \). Therefore, the equation is non-linear.

**Final Answer:**
- Order: 1
- Degree: 1
- Type: Non-linear