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.

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