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

Problem 02 | Division by t

Problem 02
Find the Laplace transform of   f(t)=e4te3tt.
 

Solution 03
f(t)=e4te3tt

f(t)=e4tte3tt

L{f(t)}=L{e4tt}L{e3tt}
 

Since
L(e4t)=1s4   and

L(e3t)=1s+3
 

Then,

Problem 01 | Division by t

Problem 01
Find the Laplace transform of   f(t)=sintt.
 

Solution 01
L(sint)=1s2+1
 

L(sintt)=sduu2+1

L(sintt)=[arctanu]s

L(sintt)=lim

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 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]

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 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}

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 03 | Change of Scale Property of Laplace Transform

Problem 03
Supposed that the Laplace transform of a certain function   f(t)   is   \dfrac{s^2 - s + 1}{(2s + 1)^2 (s - 1)},   find the Laplace transform of   f(2t).
 

Solution 03
If   \mathcal{L} \left\{ f(t) \right\} = F(s),   then by change of scale property,   \mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)
 

\mathcal{L} \left\{ f(t) \right\} = \dfrac{s^2 - s + 1}{(2s + 1)^2 (s - 1)}

Problem 02 | Change of Scale Property of Laplace Transform

Problem 02
Given that   \mathcal{L} \left( \dfrac{\sin t}{t} \right) = \arctan \left( \dfrac{1}{s} \right),   find   \mathcal{L} \left( \dfrac{\sin 3t}{t} \right).
 

Solution 02
\mathcal{L} \left\{ f(at) \right\} = \dfrac{1}{a} F \left( \dfrac{s}{a} \right)
 

\mathcal{L} \left( \dfrac{\sin 3t}{t} \right) = 3\mathcal{L} \left( \dfrac{\sin 3t}{3t} \right),   thus,   a = 3