If y(a) is a unique function which is continuous on [0, $\infty$] and also satisfy L[y(a)](b) = Y(b), then it is an Inverse Laplace transform of Y(b).

You can select a piecewise continuous function, if all other possible functions, y(a) are discontinuous, to be the inverse transform.

L-1[Y(b)](a)

## Definition of Inverse Laplace Transform

An integral defines the laplace transform Y(b) of a function y(a) defined on [o, $\infty$].

Also, the formula to determine y(a) if Y(b) is given, involves an integral.

## Inverse Laplace transform table

To compute the inverse transform, we will use the table:

 Function y(a) Transform Y(b) b 1 $\frac{1}{b}$ b>0 a $\frac{1}{b^2}$ b>0 Ai , i = integer $\frac{i!}{s^(i+1)}$ b>0 exp (ta), where t = constant $\frac{1}{(b-t)}$ b>t cos (sa), s= constant $\frac{b}{b^2+s^2}$ b>0 Sin (sa), s = constant $\frac{t}{b^2+s^2}$ b>0 exp(ta)cos(sa) $\frac{b-t}{(b-t)^2 +s^2}$ b>t exp(ta)sin(sa) $\frac{s}{(b-t)^2 +s^2}$ b>t

### Example of Inverse Laplace

Example: Find the inverse transform of each of the following.

Y(b)= $\frac{6}{b}$ -$\frac{1}{b-8}$ – $\frac{4}{b-3}$

Solution:

Step 1: The first term is a constant as we can see from the denominator of the first term.

Step 2: Before taking the inverse transform, let’s take the factor 6 out, so the correct numerator is 6.

Step 3: The second term has an exponential t = 8.

Step 4: The numerator is perfect.

Step 5: The third term is also an exponential, t= 3.

Step 6: Now before taking the inverse transforms, we need to factor out 4 first.

Y(b) = 6 $\frac{1}{b}$ -$\frac{1}{b-8}$ – 4$\frac{1}{b-3}$

y(a) = 6(1) – e8a +4 (e3t)

= 6 – e8a +4 e3t

### Inverse Laplace Transform Calculator

The Inverse Laplace Transform Calculator helps in finding the Inverse Laplace Transform Calculator of the given function.

Remember, L-1[Y(b)](a) is a function that y(a) that L(y(a) )= Y(b). So, generally, we use this property of linearity of Laplace transform to find the Inverse Laplace transform. 