d x / d t – 2 x = 0 represents the first order differential equation, second order differential equation can be represented as:

d

^{2}x / d t

^{2}– 2 d x / d t + 3 x = 0.

Let’s consider a differential equation to show that the function is a solution of the differential equation as shown below.

Consider the following differential equation:

d

^{2}x / d t

^{2}+ 2 (d x / d t) – 3x = 0,

And also consider two Functions as x1 (t) = e-3t and x

^{2}(t) = e

^{t}.

Since x1 (t) = e

^{-3t}, then differentiating x1 (t) = e

^{-3t}with respect to ‘t’ gives d x

^{1}/ d t = -3 e

^{-3t}, and similarly again differentiating d x1 / d t = -3 e

^{-3t}with respect to ‘t’ gives d

^{2}x

_{1}/d t

^{2}= 9e

^{-3t}. Using the given equation and substituting the value of d

^{2}x

_{1}/d t

^{2}and d x

_{1}/ d t results in d

^{2}x

_{1}/ d t

^{2}+ 2 (d x

_{1}/ d t) – 3x

_{1}= 9e

^{-3t}+ 2 (-3 e

^{-3t}) – 3 e

^{-3t}= 9e

^{-3t}– 6 e

^{-3t}– 3 e

^{-3t}= 9e

^{-3t}– 9 e

^{-3t}= 0.

Hence it has been proved that function x1 (t) = e-3t proves the differential equation d

^{2}x / d t

^{2}+ 2 (d x / d t) – 3x = 0,

Let’s take second function, x

^{2}(t) = et,

d

^{2 }x/ d t

^{2}= e

^{t}.

Substituting the values of d x

_{2}/ d t and d

^{2 }x

_{2 }/ d t

^{2}in differential equation d

^{2}x / d t

^{2}+ 2 (d x / d t) – 3x = 0, we get following result.

d

^{2}x

_{2}/ d t

^{2}+ 2 (d x / d t) – 3x

_{2}= et + 2et - 3et = 3et – 3et = 0, thus second function x2 (t) = et also satisfies the given differential equation.Differential equation can be defined as equation which includes derivatives. Differential equations may be of first or second or third order and are represented as shown below.

d x / d t – 2 x = 0 represents the first order differential equation, second order differential equation can be represented as:

d

^{2}x / d t

^{2}– 2 d x / d t + 3 x = 0.