Laplace transform

Many scientists and engineers require Laplace transform as a vital part of mathematical background.

because, the methods of this transform is providing an effective and easy way of solution for many problems in engineering and science.

It is having the advantage of giving direct solution for differential equations with boundary values given without finding the general solution first, and then evaluating the arbitrary constant. The transformation of Laplace also incorporates the initial condition and helps to get the particular solution.

Most of the problems in Electric circuits are being solved by using this method.


Given: f(t) is a function of t which is defined for all positive values of "t". If "s" is considered as any parameter, then it could be real or complex number such that s>0

If the integration 0e-stdt exists, then the Laplace transform of f(t) is denoted by L[f(t)] which would be defined as

L[f(t)] = 0e-stdt

L[f(t)]=F(s), since it is clearly the function of "s".

Hence L[f(t)] = F(s) = 0e-stdt

These are the formulas to be used in many problems.

y yx



(-1)nn! an/(ax+b)n+1


(-1)n(n+1)! an/(ax+b)n+2


(-1)n-1(n-1)! an/(ax+b)n


anSin(ax+b+n π/2


anCos(ax+b+n π/2


(a2+b2)n/2eax sin(bx+c+n tan2-1 b/a)


(a2+b2)n/2eax Cos(bx+c+n tan2-1 b/a)

Example 1

Find the nth derivative of 2x+1/(2x-1)(2x+3) Solution:

Let y = 2x+1/(2x-1)(2x+3)
let 2x+1/(2x-1)(2x+3) = A/(2x-1) + B/(2x+3)
2x+1 = A(2x+3)+B(2x-1)
put x = 1/2
2(1/2) + 1 = A(2(1/2) + 3) + B(2(1/2)- 1)
1 + 1 = A(1+3) + B(1-1)
2 = 4A + 0
4A = 2
A = 2/4
A = 1/2
Put x = -3/2
2(-3/2) + 1 = A(2(-3/2) + 3) + B(2(-3/2) - 1)
-2 = 0 + B (-3-1)
-2 = B(-4)
B = -2/(-4)
B = 1/2
y = (1/2)/(2x-1) + (1/2)/(2x+3)
yn = (1/2)(-1)n2nn!/(2x-1)n+1 + (1/2)(-1)n2nn!/(2x-3)n

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