July 2011

Multiplication by Power of t | Laplace Transform

Multiplication by Power of $t$
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,
 

$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s) = (-1)^n F^{(n)}(s)$

where   $n = 1, \, 2, \, 3, \, ...$
 

Problem 01 | Multiplication by Power of t

Problem 01
Find the Laplace transform of   $f(t) = t \cos 2t$.
 

Solution 01
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
 

$\mathcal{L} (\cos 2t) = \dfrac{s}{s^2 + 2^2}$

$\mathcal{L} (\cos 2t) = \dfrac{s}{s^2 + 4}$
 

$\mathcal{L} (t \cos 2t) = (-1)^1 \dfrac{d}{ds} \left[ \dfrac{s}{s^2 + 4} \right]$

$\mathcal{L} (t \cos 2t) = -\left[ \dfrac{(s^2 + 4)(1) - s(2s)}{(s^2 + 4)^2} \right]$

$\mathcal{L} (t \cos 2t) = -\dfrac{s^2 + 4 - 2s^2}{(s^2 + 4)^2}$

Problem 02 | Multiplication by Power of t

Problem 02
Find the Laplace transform of   $f(t) = t \sin 3t$.
 

Solution 02
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
 

$\mathcal{L} (\sin 3t) = \dfrac{3}{s^2 + 3^2}$

$\mathcal{L} (\sin 3t) = \dfrac{3}{s^2 + 9}$
 

$\mathcal{L} (t \sin 3t) = (-1)^1 \dfrac{d}{ds} \left[ \dfrac{3}{s^2 + 9} \right]$

$\mathcal{L} (t \sin 3t) = -\left[ \dfrac{-3(2s)}{(s^2 + 9)^2} \right]$

$\mathcal{L} (t \sin 3t) = -\dfrac{-6s}{(s^2 + 9)^2}$

Problem 03 | Multiplication by Power of t

Problem 03
Find the Laplace transform of   $f(t) = t^2 \cos 3t$.
 

Solution 03
$\mathcal{L} \left\{ t^n f(t) \right\} = (-1)^n \dfrac{d^n}{ds^n} F(s)$
 

$\mathcal{L} (\cos 3t) = \dfrac{s}{s^2 + 3^2}$

$\mathcal{L} (\cos 3t) = \dfrac{s}{s^2 + 9}$
 

$\mathcal{L} (t^2 \cos 3t) = (-1)^2 \dfrac{d^2}{ds^2} \left[ \dfrac{s}{s^2 + 9} \right]$

$\mathcal{L} (t^2 \cos 3t) = \dfrac{d}{ds} \left[ \dfrac{(s^2 + 9)(1) - s(2s)}{(s^2 + 9)^2} \right]$

Division by t | Laplace Transform

Division by $t$
If   $\mathcal{L} \left\{ f(t) \right\} = F(s)$,   then,
 

$\displaystyle \mathcal{L} \left\{ \dfrac{f(t)}{t} \right\} = \int_s^\infty F(u) \, du$

 

provided   $\displaystyle \lim_{t \rightarrow 0} \left[ \dfrac{f(t)}{t} \right]$   exists.
 

Problem 01 | Division by t

Problem 01
Find the Laplace transform of   $f(t) = \dfrac{\sin t}{t}$.
 

Solution 01
$\mathcal{L} (\sin t) = \dfrac{1}{s^2 + 1}$
 

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \int_s^\infty \dfrac{du}{u^2 + 1}$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \big[ \arctan u \big]_s^\infty$

$\displaystyle \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \lim_{a \to \infty}\big[ \arctan u \big]_s^a$

Problem 02 | Division by t

Problem 02
Find the Laplace transform of   $f(t) = \dfrac{e^{4t} - e^{-3t}}{t}$.
 

Solution 03
$f(t) = \dfrac{e^{4t} - e^{-3t}}{t}$

$f(t) = \dfrac{e^{4t}}{t} - \dfrac{e^{-3t}}{t}$

$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left\{ \dfrac{e^{4t}}{t} \right\} - \mathcal{L} \left\{ \dfrac{e^{-3t}}{t} \right\}$
 

Since
$\mathcal{L} (e^{4t}) = \dfrac{1}{s - 4}$   and

$\mathcal{L} (e^{-3t}) = \dfrac{1}{s + 3}$
 

Then,

Greek Alphabets

The following are Greek alphabets commonly used in science and mathematics.

Greek Symbol Greek Letter Name English Equivalent
Upper Case Lower Case
Α α Alpha a
Β β Beta b
Γ γ Gamma g
Δ δ Delta d
Ε ε Epsilon e
Ζ ζ Zeta z
Η η Eta h
Θ θ Theta th
Ι ι Iota i
Κ κ Kappa k
Λ λ Lambda l
Μ μ Mu m
Ν ν Nu n
Ξ ξ Xi x
Ο ο Omicron o
Π π Pi p
Ρ ρ Rho r
Σ σ Sigma s
Τ τ Tau t
Υ υ Upsilon u
Φ φ Phi ph
Χ χ Chi ch
Ψ ψ Psi ps
Ω ω Omega o

Pages