Differential Equations : Order and Degree of a Differential Equation

Table Content :
1.Differential Equations : Order and Degree of a Differential Equation
2.Formation of Differential Equations
3. Differential Equations with Variables Separable
4.Homogeneous Differential Equations
5.Angle between the PlanesFirst Order Linear Differential Equations
6.General Form of Variable Separation
7.Application of Differential Equations

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}}$

⇒ 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}) $

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