
An equation involving derivatives of the dependent variable with respect to independent variable (variables) is known as a differential equation.

Order of a differential equation is the order of the highest order derivative occurring in the differential equation.

Degree of a differential equation is defined if it is a polynomial equation in its derivatives.

Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it.

A function which satisfies the given differential equation is called its solution.

The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution.

To form a differential equation from a given function we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants.

Variable separable method is used to solve such an equation in which variables can be separated completely i.e. terms containing y should remain with dy and terms containing x should remain with dx.

A differential equation which can be expressed in the form dy/dx = f(x,y) where f(x,y) is a homogenous function of degree zero is called a homogeneous differential equation.

A differential equation of the form (dy/dx)+Py = Q, where P and Q are constants or functions of x only is called a first order linear differential equation.
Examples
Question
Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constants.
Solution
We have y = a sin (x + b)
dy/dx = a cos (x + b)…………………………..(1)
d^{2}y/dx^{2} = – a sin (x + b)……………………….(2)
Eliminating a and b using equations 1 and 2 we get (d^{2}y/dx^{2}) + y = 0.
Question
Find the general solution of the differential equation (dy/dx) = (1 + y^{2})/ (1 + x^{2})
Solution
Integrating both sides,
Question
Find the general solution of the differential equation y dx – (x + 2y^{2}) dy = 0.
Solution
(dx/dy) – (x/y) = 2y
This is of the type (dx/dy) + Px = Q where P = (1/y) Q = 2y
Hence IF =
The solution of differential equation is