#### Provide solution for RD sharma maths class 12 chapter 21 Differential Equation exercise 21.1 question 14

Order=1, Degree=2, Non- linear

Hint:

The order is the highest numbered derivative in the equation with no negative or fractional power of the dependent variable and its derivatives, while the degree is the highest power to which a derivative is raised.

Given:

$\sqrt{1-y^{2}}dx+\sqrt{1-x^{2}}dy=0$

Solution:

The above equation can be written as

$\sqrt{1-x^{2}}\frac{dy}{dx}=-\sqrt{1-y^{2}}$

Since the power of y cannot be rational so squaring on both sides

$\left (\sqrt{1-x^{2}}\frac{dy}{dx} \right )^{2}=\left (-\sqrt{1-y^{2}} \right )^{2}$

$\left (1-x^{2} \right )\left (\frac{dy}{dx} \right )^{2}=1-y^{2}$

Here in this question, the order of the differential equation is 1 and the degree of the differential equation is 2.

In a differential equation, when the dependent variable and their derivatives are only multiplied by constant or independent variable, then the equation is linear.

Here the dependent variable is y and the term $\frac{dy}{dx}$ is multiplied by itself. So this equation is non-linear differential equation.

Therefore, Order=1, Degree=2, Non-linear