Inverse Laplace Transforms Calculator

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Example

Created on 2024-06-20Asked by Ava Lopez (Solvelet student)

Find the inverse Laplace transform of

F(s) = \frac{1}{s^2 + s - 2}

Solution

To find the inverse Laplace transform of

F(s) = \frac{1}{s^2 + s - 2}

: First, factor the denominator:

s^2 + s - 2 = (s + 2)(s - 1)

So, we can rewrite

F(s)

as:

F(s) = \frac{1}{(s + 2)(s - 1)}

Use partial fraction decomposition:

\frac{1}{(s + 2)(s - 1)} = \frac{A}{s + 2} + \frac{B}{s - 1}

Solve for

A

and

B

1 = A(s - 1) + B(s + 2)

Let

s = -2

1 = A(-2 - 1) \Rightarrow 1 = -3A \Rightarrow A = -\frac{1}{3}

Let

s = 1

1 = B(1 + 2) \Rightarrow 1 = 3B \Rightarrow B = \frac{1}{3}

Thus:

F(s) = \frac{-1/3}{s + 2} + \frac{1/3}{s - 1}

Find the inverse Laplace transform of each term:

\mathcal{L}^{-1} \left\{ \frac{-1/3}{s + 2} \right\} = -\frac{1}{3} e^{-2t}

\mathcal{L}^{-1} \left\{ \frac{1/3}{s - 1} \right\} = \frac{1}{3} e^{t}

Combine the results:

\mathcal{L}^{-1} \left\{ \frac{1}{s^2 + s - 2} \right\} = -\frac{1}{3} e^{-2t} + \frac{1}{3} e^{t}

Therefore, the inverse Laplace transform of

F(s) = \frac{1}{s^2 + s - 2}

is:

f(t) = -\frac{1}{3} e^{-2t} + \frac{1}{3} e^{t}

Solved on Solvelet with Basic AI Model

Some of the related questions asked by Harper Campbell on Solvelet

1. Compute the inverse Laplace transform of the function

F(s) = \frac{1}{s + 3}

,2. Determine the principal value of

\arctan(-1)

DefinitionInverse Laplace Transform: A mathematical operation we apply to transform a signal from Laplace domain (frequency domain) to time domain. If you have a function F(s) in the Laplace domain, then the inverse Laplace transform provides you the time domain original function f(t). This is used is in solving differential equations among other places like control theory or signal processing. Example: If F(s)=s2+11, the inverse Laplace transfrom is f(t)=sin(t). The inverse Laplace is typically done by transform tables or maybe by partial fraction decomposition.

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