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17Calculus - Inverse Laplace Transforms

17Calculus
Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
Precalculus
Functions

Since we use an integral to calculate the Laplace Transform, you might expect to use derivatives to go the other direction. Counterintuitively, this is not the case. Calculating the inverse Laplace Transform involves fewer actual computations. Most of the time, you will use formulas and tables that were derived from calculating the Laplace Transform.

Notation

The notation we used for the Laplace Transform, looked like this. \[ F(s) = \mathcal{L}\{ f(t) \} \] Intuitively, the notation for the Inverse Laplace Transform is written in a logical manner. \[ f(t) = \mathcal{L}^{-1} \{ F(s) \} \]

Rules

Since the Laplace Transform uses integration, which is a linear operation, so too is the inverse Laplace Transform. This means that \[ \mathcal{L}^{-1} \{ F(s) + G(s) \} = \mathcal{L}^{-1} \{ F(s) \} + \mathcal{L}^{-1} \{ G(s) \} \] For the same reason, using the constant \(a\), this equation also holds. \[ \mathcal{L}^{-1} \{ aF(s) \} = a \mathcal{L}^{-1} \{ F(s) \} \]

Calculating the Inverse Laplace Transform

To actually calculate the inverse Laplace Transform, we use tables. However, you will need to make sure your partial fraction expansion skills are sharp.

Okay, here is the table you need to use for calculating inverse Laplace Transforms. The practice problems column contain links to practice problems using this table.

Laplace Transforms Table

Laplace Transforms

\( f(t) \)

\(\displaystyle{ F(s) }\)

Basic Functions

\( t^n, ~ n = 1, 2, 3, \ldots \)

\(\displaystyle{ \frac{n!}{s^{n+1}} }\)

\( e^{at} \)

\(\displaystyle{ \frac{1}{s-a} }\)

\( \sin(\alpha t) \)

\(\displaystyle{ \frac{\alpha}{s^2 + \alpha^2} }\)

\( \cos(at) \)

\(\displaystyle{ \frac{s}{s^2 + a^2} }\)

\( \sinh(at) \)

\(\displaystyle{ \frac{a}{s^2 - a^2} }\)

\( \cosh(at) \)

\(\displaystyle{ \frac{s}{s^2 - a^2} }\)

Special Functions

\( \delta(t) \) unit impulse

\( 1 \)

\( \delta(t-\tau) \) shifted unit impulse

\( e^{-\tau s} \)

\( u(t) \) unit step

\(\displaystyle{ \frac{1}{s} }\)

\( u(t-\tau) \) shifted unit step

\(\displaystyle{ \frac{1}{s} e^{-\tau s} }\)

Combined Functions

\( e^{at}\sin(\alpha t) \)

\(\displaystyle{ \frac{\alpha}{(s-a)^2 + \alpha^2} }\)

\( f(t)u(t-a) \)

\(\displaystyle{ e^{-sa} \mathcal{L}\{ f(t+a) \} }\)

\( t^n e^{at}, ~ n = 1, 2, 3, \ldots \)

\(\displaystyle{ \frac{n!}{(s-a)^{n+1}} \} }\)

Derivatives and Integrals

\( f'(t) \)

\( sF(s) - f(0) \)

\( f''(t) \)

\( s^2F(s) - sf(0) - f'(0) \)

\(\displaystyle{ f^{(n)}(t) }\)

\(\displaystyle{ s^nF(s) - s^{n-1}f(0) - }\) \(\displaystyle{ s^{n-2}f'(0) - . . . - f^{(n-1)}(0) }\)

\( \int_0^t{ f(v)~dv }\)

\(\displaystyle{ \frac{F(s)}{s} }\)

Okay, so why do we need Laplace Transforms? Why are they useful? We use them to solve differential equations that cannot be solved otherwise, sometimes involving some special functions. These special functions also have a purpose. Some that you will run across are the unit step function, unit impulse function and the square wave.

Calculus, Better Explained: A Guide To Developing Lasting Intuition

Practice

Unless otherwise instructed, calculate the inverse Laplace transform \( f(t) = \mathcal{L}^{-1} \{ F(s) \} \) using a table. Give your answer in exact, completely factored form.

Basic

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4} \right\} }\)

Final Answer

\( t^3/6 \)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4} \right\} }\)

Solution

blackpenredpen - 3610 video solution

video by blackpenredpen

Final Answer

\( t^3/6 \)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{6s+3} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{6s+3} \right\} }\)

Final Answer

\(\displaystyle{ \frac{e^{-t/2}}{6} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{6s+3} \right\} }\)

Solution

blackpenredpen - 3624 video solution

video by blackpenredpen

Final Answer

\(\displaystyle{ \frac{e^{-t/2}}{6} }\)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+1}{s^2+2} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+1}{s^2+2} \right\} }\)

Final Answer

\(\displaystyle{ \cos(t\sqrt{2}) + \frac{\sin(t\sqrt{2})}{\sqrt{2}} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+1}{s^2+2} \right\} }\)

Solution

blackpenredpen - 3625 video solution

video by blackpenredpen

Final Answer

\(\displaystyle{ \cos(t\sqrt{2}) + \frac{\sin(t\sqrt{2})}{\sqrt{2}} }\)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^2+2s} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^2+2s} \right\} }\)

Final Answer

\(\displaystyle{ \frac{1}{2} ( 1-e^{-2t} ) }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^2+2s} \right\} }\)

Solution

blackpenredpen - 3626 video solution

video by blackpenredpen

Final Answer

\(\displaystyle{ \frac{1}{2} ( 1-e^{-2t} ) }\)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{(s+2)^2} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{(s+2)^2} \right\} }\)

Final Answer

\( e^{-2t} - 2te^{-2t} \)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{(s+2)^2} \right\} }\)

Solution

blackpenredpen - 3627 video solution

video by blackpenredpen

Final Answer

\( e^{-2t} - 2te^{-2t} \)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{se^{-\pi s/2}}{s^2+1} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{se^{-\pi s/2}}{s^2+1} \right\} }\)

Final Answer

\( \cos(t-\pi/2)u(t-\pi/2) \)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{se^{-\pi s/2}}{s^2+1} \right\} }\)

Solution

blackpenredpen - 3628 video solution

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Final Answer

\( \cos(t-\pi/2)u(t-\pi/2) \)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{s^2+2s+2} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{s^2+2s+2} \right\} }\)

Final Answer

\( e^{-t}( \cos(t) - \sin(t) ) \)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s}{s^2+2s+2} \right\} }\)

Solution

blackpenredpen - 3629 video solution

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Final Answer

\( e^{-t}( \cos(t) - \sin(t) ) \)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{(s+2)^5} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{(s+2)^5} \right\} }\)

Final Answer

\(\displaystyle{ \frac{t^4 e^{-2t}}{24} } \)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{(s+2)^5} \right\} }\)

Solution

blackpenredpen - 3631 video solution

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Final Answer

\(\displaystyle{ \frac{t^4 e^{-2t}}{24} } \)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+8}{s^2+4s+13} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+8}{s^2+4s+13} \right\} }\)

Final Answer

\( e^{-2t}( \cos(3t) + 2\sin(3t) ) \)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s+8}{s^2+4s+13} \right\} }\)

Solution

blackpenredpen - 3633 video solution

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Final Answer

\( e^{-2t}( \cos(3t) + 2\sin(3t) ) \)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4+5s^2+4} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4+5s^2+4} \right\} }\)

Final Answer

\(\displaystyle{ \frac{1}{6} (2\sin(t) - \sin(2t)) }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4+5s^2+4} \right\} }\)

Solution

blackpenredpen - 3634 video solution

video by blackpenredpen

Final Answer

\(\displaystyle{ \frac{1}{6} (2\sin(t) - \sin(2t)) }\)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4 e^{10s}} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4 e^{10s}} \right\} }\)

Final Answer

\(\displaystyle{ \frac{(t-10)^3}{6} u(t-10) }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4 e^{10s}} \right\} }\)

Solution

blackpenredpen - 3635 video solution

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Final Answer

\(\displaystyle{ \frac{(t-10)^3}{6} u(t-10) }\)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \ln \left[ \frac{s^2+9}{s^2+1} \right] \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \ln \left[ \frac{s^2+9}{s^2+1} \right] \right\} }\)

Final Answer

\(\displaystyle{ \frac{2}{t}( \cos(t) -\cos(3t)) }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \ln \left[ \frac{s^2+9}{s^2+1} \right] \right\} }\)

Solution

blackpenredpen - 3637 video solution

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Final Answer

\(\displaystyle{ \frac{2}{t}( \cos(t) -\cos(3t)) }\)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4-16} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4-16} \right\} }\)

Final Answer

\( \frac{1}{32}( e^{2t} - e^{-2t} -2\sin(2t) ) \)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4-16} \right\} }\)

Solution

blackpenredpen - 3638 video solution

video by blackpenredpen

Final Answer

\( \frac{1}{32}( e^{2t} - e^{-2t} -2\sin(2t) ) \)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s^3}{(s^4-16)^2} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s^3}{(s^4-16)^2} \right\} }\)

Hint

Use the result from the previous problem to solve this one.

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{s^3}{(s^4-16)^2} \right\} }\)

Hint

Use the result from the previous problem to solve this one.

Solution

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\(\displaystyle{ F(s) = \frac{1}{s-3} - \frac{16}{s^2+9} }\)

Problem Statement

For \(\displaystyle{ F(s) = \frac{1}{s-3} - \frac{16}{s^2+9} }\), find the inverse Laplace transform \( f(t) = \mathcal{L}^{-1} \{ F(s) \} \).

Solution

PatrickJMT - 651 video solution

video by PatrickJMT

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\(\displaystyle{ F(s) = \frac{s+3}{s^2+4s+13} }\)

Problem Statement

For \(\displaystyle{ F(s) = \frac{s+3}{s^2+4s+13} }\), find the inverse Laplace transform \( f(t) = \mathcal{L}^{-1} \{ F(s) \} \).

Solution

Krista King Math - 654 video solution

video by Krista King Math

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Intermediate

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^3 (s^2+1)} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^3 (s^2+1)} \right\} }\)

Final Answer

\( \cos(t) + t^2/2 - 1 \)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^3 (s^2+1)} \right\} }\)

Solution

blackpenredpen - 3630 video solution

video by blackpenredpen

Final Answer

\( \cos(t) + t^2/2 - 1 \)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \arctan(1/s) \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \arctan(1/s) \right\} }\)

Final Answer

\(\displaystyle{ \frac{\sin(t)}{t} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \arctan(1/s) \right\} }\)

Solution

blackpenredpen - 3636 video solution

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Final Answer

\(\displaystyle{ \frac{\sin(t)}{t} }\)

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Advanced

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{\sqrt{s}} + \frac{1}{\sqrt{e^s}} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{\sqrt{s}} + \frac{1}{\sqrt{e^s}} \right\} }\)

Final Answer

\(\displaystyle{ \frac{1}{\sqrt{\pi t}} + \delta(t-1/2) }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{\sqrt{s}} + \frac{1}{\sqrt{e^s}} \right\} }\)

Solution

blackpenredpen - 3632 video solution

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Final Answer

\(\displaystyle{ \frac{1}{\sqrt{\pi t}} + \delta(t-1/2) }\)

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Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4+4s^2+4} \right\} }\)

Problem Statement

Evaluate \(\displaystyle{ \mathcal{L}^{-1} \left\{ \frac{1}{s^4+4s^2+4} \right\} }\)

Solution

blackpenredpen - 3640 video solution

video by blackpenredpen

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Practice Instructions

Unless otherwise instructed, calculate the inverse Laplace transform \( f(t) = \mathcal{L}^{-1} \{ F(s) \} \) using a table. Give your answer in exact, completely factored form.

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