By a differential equation we mean an equation involving independent variable , dependent variable and the differential coefficients of the dependent variable i.e. it will be an equation in x, y and derivatives of y w .r .t x. e.g.

$ \displaystyle \frac{dy}{dx}+y = x e^x $

$ \displaystyle \frac{d^2y}{dx^2}+y = 0 $

##### Order and Degree of a Differential Equation:

The order of the highest differential coefficient appearing in the differential equation is called the order of the differential equation ,

while the exponent of the highest differential coefficient , when the differential equation is a polynomial in all the differential coefficients, is known as the degree of the differential equation.

Example : Find the order and degree (if defined) of the following differential equations:

(i) $ \displaystyle y = 1 + \frac{dy}{dx}+ \frac{1}{2!} (\frac{dy}{dx})^2 + \frac{1}{3!} (\frac{dy}{dx})^3 +….$

(ii) $ \displaystyle \frac{dy}{dx} = \sqrt{\frac{d^2 y}{dx^2}+ y } $

(iii) $ \displaystyle \frac{d^2 y}{dx^2} = x (ln\frac{dy}{dx}) $

Solution:

(i) The given differential equation can be re-written as $ \displaystyle y = e^{\frac{dy}{dx}}$

⇒ $ \frac{dy}{dx} = ln y $ . Hence its order is 1 and degree 1.

(ii) The given differential equation can be re-written as

$ \displaystyle (\frac{dy}{dx})^2 = \frac{d^2y}{dx^2} + y $

Hence its order is 2 and degree 1.

(iii) Its order is 2. Since the given differential equation cannot be written as a polynomial in all the differential coefficients, the degree of the equation is not defined.

Exercise :

Find the order and the degree of the following differential equations

(i) $ \displaystyle \frac{d^2 y}{dx^2} = \sqrt[3]{1 + (\frac{dy}{dx})^4} $

(ii) $ \displaystyle \sqrt{\frac{dy}{dx}} +\sqrt{y} \frac{d^2 y}{dx^2} = 0 $

(iii)$ \displaystyle \frac{d^2y}{dx^2} = sin(\frac{dy}{dx}) $

(iv) $ \displaystyle [1+(\frac{dy}{dx})^2]^{3/2} = k(\frac{d^2 y}{dx^2}) $