Module II: Inverse Laplace Transform

1. Inverse Laplace Transform

Definition

If \( \mathcal{L}\{f(t)\} = F(s) \), then \( f(t) \) is called the Inverse Laplace Transform of \( F(s) \)

\[ f(t) = \mathcal{L}^{-1}\{F(s)\} \]

2. Linearity Property

\[ \mathcal{L}^{-1}\{aF(s) + bG(s)\} = a\mathcal{L}^{-1}\{F(s)\} + b\mathcal{L}^{-1}\{G(s)\} \]

Find \( \mathcal{L}^{-1}\left\{\frac{3}{s^2+4} + \frac{5}{s-2}\right\} \)

Apply linearity: \( = 3\mathcal{L}^{-1}\left\{\frac{1}{s^2+4}\right\} + 5\mathcal{L}^{-1}\left\{\frac{1}{s-2}\right\} \)
\( \mathcal{L}^{-1}\left\{\frac{1}{s^2+4}\right\} = \mathcal{L}^{-1}\left\{\frac{1}{s^2+2^2}\right\} = \frac{\sin(2t)}{2} \)
\( \mathcal{L}^{-1}\left\{\frac{1}{s-2}\right\} = e^{2t} \)

\( = \frac{3}{2}\sin(2t) + 5e^{2t} \)

3. Standard Inverse Laplace Transform Formulae

\( F(s) \)	\( \mathcal{L}^{-1}\{F(s)\} = f(t) \)
\( \frac{1}{s} \)	\( 1 \)
\( \frac{1}{s^n} \)	\( \frac{t^{n-1}}{(n-1)!} \)
\( \frac{1}{s-a} \)	\( e^{at} \)
\( \frac{1}{(s-a)^n} \)	\( \frac{t^{n-1}e^{at}}{(n-1)!} \)
\( \frac{a}{s^2+a^2} \)	\( \sin(at) \)
\( \frac{s}{s^2+a^2} \)	\( \cos(at) \)
\( \frac{a}{s^2-a^2} \)	\( \sinh(at) \)
\( \frac{s}{s^2-a^2} \)	\( \cosh(at) \)
\( \frac{a}{(s-b)^2+a^2} \)	\( e^{bt}\sin(at) \)
\( \frac{s-b}{(s-b)^2+a^2} \)	\( e^{bt}\cos(at) \)

Find \( \mathcal{L}^{-1}\left\{\frac{s+3}{(s+3)^2+16}\right\} \)

This matches the form \( \frac{s-b}{(s-b)^2+a^2} \) with \( b = -3, a = 4 \)
Using formula: \( \mathcal{L}^{-1}\left\{\frac{s-b}{(s-b)^2+a^2}\right\} = e^{bt}\cos(at) \)

\( = e^{-3t}\cos(4t) \)

4. Inverse Laplace Transform using Derivatives

If \( \mathcal{L}^{-1}\{F(s)\} = f(t) \), then \( \mathcal{L}^{-1}\{F'(s)\} = -t \cdot f(t) \)

In general: \( \mathcal{L}^{-1}\{F^{(n)}(s)\} = (-t)^n f(t) \)

When to use: When \( F(s) \) can be written as derivative of a simpler function

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

Notice that \( \frac{d}{ds}\left(\frac{1}{s^2+4}\right) = \frac{-2s}{(s^2+4)^2} \)
So \( \frac{2s}{(s^2+4)^2} = -\frac{d}{ds}\left(\frac{1}{s^2+4}\right) \)
We know \( \mathcal{L}^{-1}\left\{\frac{1}{s^2+4}\right\} = \frac{\sin(2t)}{2} \)
Using property: \( \mathcal{L}^{-1}\{F'(s)\} = -t \cdot f(t) \)
\( \mathcal{L}^{-1}\left\{-\frac{d}{ds}\left(\frac{1}{s^2+4}\right)\right\} = -(-t) \cdot \frac{\sin(2t)}{2} \)

\( = \frac{t\sin(2t)}{2} \)

Find \( \mathcal{L}^{-1}\left\{\ln\frac{s+1}{s-1}\right\} \)

Let \( F(s) = \ln(s+1) - \ln(s-1) \)
\( F'(s) = \frac{1}{s+1} - \frac{1}{s-1} \)
\( \mathcal{L}^{-1}\{F'(s)\} = \mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} - \mathcal{L}^{-1}\left\{\frac{1}{s-1}\right\} = e^{-t} - e^{t} \)
Since \( \mathcal{L}^{-1}\{F'(s)\} = -t \cdot f(t) \), we have \( -tf(t) = e^{-t} - e^{t} \)
\( f(t) = \frac{e^t - e^{-t}}{t} = \frac{2\sinh(t)}{t} \)

\( \mathcal{L}^{-1}\left\{\ln\frac{s+1}{s-1}\right\} = \frac{2\sinh(t)}{t} \)

5. Partial Fractions Method

Method Overview

Decompose rational function \( F(s) \) into simpler fractions, then find inverse of each

Case 1: Distinct Linear Factors

\[ \frac{P(s)}{(s-a)(s-b)(s-c)} = \frac{A}{s-a} + \frac{B}{s-b} + \frac{C}{s-c} \]

Find \( \mathcal{L}^{-1}\left\{\frac{1}{(s-1)(s-2)}\right\} \)

Partial fractions: \( \frac{1}{(s-1)(s-2)} = \frac{A}{s-1} + \frac{B}{s-2} \)
Multiply by \( (s-1)(s-2) \): \( 1 = A(s-2) + B(s-1) \)
Put \( s = 1 \): \( 1 = A(-1) \Rightarrow A = -1 \)
Put \( s = 2 \): \( 1 = B(1) \Rightarrow B = 1 \)
\( \mathcal{L}^{-1}\left\{\frac{-1}{s-1} + \frac{1}{s-2}\right\} = -e^t + e^{2t} \)

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

Case 2: Repeated Linear Factors

\[ \frac{P(s)}{(s-a)^3} = \frac{A}{s-a} + \frac{B}{(s-a)^2} + \frac{C}{(s-a)^3} \]

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

Write \( s = (s+1) - 1 \): \( \frac{s}{(s+1)^3} = \frac{(s+1)-1}{(s+1)^3} = \frac{1}{(s+1)^2} - \frac{1}{(s+1)^3} \)
\( \mathcal{L}^{-1}\left\{\frac{1}{(s+1)^2}\right\} = te^{-t} \) (using \( \frac{1}{(s-a)^n} \to \frac{t^{n-1}e^{at}}{(n-1)!} \))
\( \mathcal{L}^{-1}\left\{\frac{1}{(s+1)^3}\right\} = \frac{t^2e^{-t}}{2!} = \frac{t^2e^{-t}}{2} \)

\( = te^{-t} - \frac{t^2e^{-t}}{2} = e^{-t}\left(t - \frac{t^2}{2}\right) \)

Case 3: Quadratic Factors

\[ \frac{P(s)}{(s-a)(s^2+b^2)} = \frac{A}{s-a} + \frac{Bs+C}{s^2+b^2} \]

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

Partial fractions: \( \frac{1}{s(s^2+4)} = \frac{A}{s} + \frac{Bs+C}{s^2+4} \)
Multiply: \( 1 = A(s^2+4) + (Bs+C)s \)
Put \( s = 0 \): \( 1 = 4A \Rightarrow A = \frac{1}{4} \)
Compare \( s^2 \) coefficients: \( 0 = A + B \Rightarrow B = -\frac{1}{4} \)
Compare \( s \) coefficients: \( 0 = C \)
\( \frac{1}{s(s^2+4)} = \frac{1/4}{s} - \frac{s/4}{s^2+4} \)
\( \mathcal{L}^{-1} = \frac{1}{4} \cdot 1 - \frac{1}{4}\cos(2t) \)

\( = \frac{1}{4}(1 - \cos(2t)) \)

6. Convolution Theorem

Convolution Definition

The convolution of \( f(t) \) and \( g(t) \) is defined as: \[ (f * g)(t) = \int_0^t f(u) \cdot g(t-u) \, du \]

\[ \mathcal{L}^{-1}\{F(s) \cdot G(s)\} = f(t) * g(t) = \int_0^t f(u) \cdot g(t-u) \, du \]

When to use: When \( F(s) \) is a product of two functions whose inverses are known

Find \( \mathcal{L}^{-1}\left\{\frac{1}{s(s-1)}\right\} \) using convolution

Let \( F(s) = \frac{1}{s} \) and \( G(s) = \frac{1}{s-1} \)
\( f(t) = \mathcal{L}^{-1}\{1/s\} = 1 \) and \( g(t) = \mathcal{L}^{-1}\{1/(s-1)\} = e^t \)
Convolution: \( \int_0^t 1 \cdot e^{t-u} \, du = e^t \int_0^t e^{-u} \, du \)
\( = e^t \left[-e^{-u}\right]_0^t = e^t(-e^{-t} + 1) = e^t - 1 \)

\( \mathcal{L}^{-1}\left\{\frac{1}{s(s-1)}\right\} = e^t - 1 \)

Find \( \mathcal{L}^{-1}\left\{\frac{1}{(s^2+1)^2}\right\} \) using convolution

Let \( F(s) = G(s) = \frac{1}{s^2+1} \), so \( f(t) = g(t) = \sin(t) \)
Convolution: \( \int_0^t \sin(u) \cdot \sin(t-u) \, du \)
Using product formula: \( \sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)] \)
\( = \frac{1}{2}\int_0^t [\cos(2u-t) - \cos(t)] \, du \)
\( = \frac{1}{2}\left[\frac{\sin(2u-t)}{2} - u\cos(t)\right]_0^t \)
\( = \frac{1}{2}\left[\frac{\sin(t)}{2} - t\cos(t) - \frac{\sin(-t)}{2}\right] \)
\( = \frac{1}{2}\left[\frac{\sin(t)}{2} - t\cos(t) + \frac{\sin(t)}{2}\right] = \frac{1}{2}[\sin(t) - t\cos(t)] \)

\( \mathcal{L}^{-1}\left\{\frac{1}{(s^2+1)^2}\right\} = \frac{\sin(t) - t\cos(t)}{2} \)

7. Solving Differential Equations using Laplace Transform

Take L.T. of ODE

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Solve for Y(s)

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Find \( \mathcal{L}^{-1}\{Y(s)\} \)

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Solution y(t)

Key Formulas for ODEs

\( \mathcal{L}\{y'\} = sY(s) - y(0) \)

\( \mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0) \)

First Order ODE

Solve \( y' + 2y = e^{-t} \), given \( y(0) = 1 \)

Take Laplace Transform of both sides: \( \mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = \mathcal{L}\{e^{-t}\} \)
\( sY(s) - y(0) + 2Y(s) = \frac{1}{s+1} \)
\( sY(s) - 1 + 2Y(s) = \frac{1}{s+1} \)
\( Y(s)(s+2) = 1 + \frac{1}{s+1} = \frac{s+1+1}{s+1} = \frac{s+2}{s+1} \)
\( Y(s) = \frac{s+2}{(s+1)(s+2)} = \frac{1}{s+1} \)
Take inverse: \( y(t) = \mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} \)

\( y(t) = e^{-t} \)

Second Order ODE

Solve \( y'' + 4y = 0 \), given \( y(0) = 0, y'(0) = 2 \)

Take Laplace Transform: \( \mathcal{L}\{y''\} + 4\mathcal{L}\{y\} = 0 \)
\( s^2Y(s) - sy(0) - y'(0) + 4Y(s) = 0 \)
\( s^2Y(s) - 0 - 2 + 4Y(s) = 0 \)
\( Y(s)(s^2 + 4) = 2 \)
\( Y(s) = \frac{2}{s^2+4} = \frac{2}{s^2+2^2} \)
Take inverse: \( y(t) = \mathcal{L}^{-1}\left\{\frac{2}{s^2+4}\right\} \)

\( y(t) = \sin(2t) \)

Solve \( y'' - 3y' + 2y = e^{3t} \), given \( y(0) = 0, y'(0) = 0 \)

Take Laplace Transform: \( s^2Y - sy(0) - y'(0) - 3[sY - y(0)] + 2Y = \frac{1}{s-3} \)
\( s^2Y - 3sY + 2Y = \frac{1}{s-3} \)
\( Y(s^2 - 3s + 2) = \frac{1}{s-3} \)
\( Y(s-1)(s-2) = \frac{1}{s-3} \)
\( Y(s) = \frac{1}{(s-1)(s-2)(s-3)} \)
Partial fractions: \( = \frac{A}{s-1} + \frac{B}{s-2} + \frac{C}{s-3} \)
\( A = \frac{1}{(1-2)(1-3)} = \frac{1}{(-1)(-2)} = \frac{1}{2} \)
\( B = \frac{1}{(2-1)(2-3)} = \frac{1}{(1)(-1)} = -1 \)
\( C = \frac{1}{(3-1)(3-2)} = \frac{1}{(2)(1)} = \frac{1}{2} \)
\( Y(s) = \frac{1/2}{s-1} - \frac{1}{s-2} + \frac{1/2}{s-3} \)

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

Solve \( y'' + y = \sin(t) \), given \( y(0) = 1, y'(0) = 0 \)

Take Laplace Transform: \( s^2Y - s(1) - 0 + Y = \frac{1}{s^2+1} \)
\( Y(s^2 + 1) = s + \frac{1}{s^2+1} \)
\( Y = \frac{s}{s^2+1} + \frac{1}{(s^2+1)^2} \)
First term: \( \mathcal{L}^{-1}\left\{\frac{s}{s^2+1}\right\} = \cos(t) \)
Second term (using convolution result): \( \mathcal{L}^{-1}\left\{\frac{1}{(s^2+1)^2}\right\} = \frac{\sin(t) - t\cos(t)}{2} \)

\( y(t) = \cos(t) + \frac{\sin(t) - t\cos(t)}{2} = \frac{2\cos(t) + \sin(t) - t\cos(t)}{2} \)

Quick Reference: All Formulas

Method	Formula/Technique
Linearity	\( \mathcal{L}^{-1}\{aF + bG\} = a\mathcal{L}^{-1}\{F\} + b\mathcal{L}^{-1}\{G\} \)
First Shifting	\( \mathcal{L}^{-1}\{F(s-a)\} = e^{at}f(t) \)
Derivative Method	\( \mathcal{L}^{-1}\{F'(s)\} = -tf(t) \)
Convolution	\( \mathcal{L}^{-1}\{F \cdot G\} = \int_0^t f(u)g(t-u)du \)
For ODE: \( y' \)	\( \mathcal{L}\{y'\} = sY(s) - y(0) \)
For ODE: \( y'' \)	\( \mathcal{L}\{y''\} = s^2Y - sy(0) - y'(0) \)