If I use Inverse Laplace transformation using the table do I get cos(t)−sin(t). can I go back to the t domain with the x domain normally ?

memLosycecyjz 2022-09-22 Answered
Let
F ( s ) = L { f ( t ) }
After a Laplace transform of f(t) the expression is :
F ( s ) = 2 s 4 s 2 2 s + 2
which equals to
F ( s ) = 2 ( s 1 ( s 1 ) 2 + 1 1 ( s 1 ) 2 + 1 )
let
x = s 1
Therefore
F ( x ) = x x 2 + 1 1 x 2 + 1
My question is, if I use Inverse Laplace transformation using the table do I get cos ( t ) sin ( t ). Can I go back to the t domain with the x domain normally ?
You can still ask an expert for help

Want to know more about Laplace transform?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Phoenix Morse
Answered 2022-09-23 Author has 10 answers
L 1 { F ( s α ) } = e α t L 1 { F ( s ) }
L 1 { F ( s 1 ) } = e t L 1 { F ( s ) }
L 1 { 2 ( s 1 ( s 1 ) 2 + 1 1 ( s 1 ) 2 + 1 ) } = 2 e t ( cos t sin t ) .
Where does the "frequency shift" property of the Laplace transform come from?:
L { e α t f ( t ) } = 0 e s t e α t f ( t ) d t = 0 e ( s α ) t f ( t ) d t = F ( s α ) .
Did you like this example?
Subscribe for all access
batejavizb
Answered 2022-09-24 Author has 4 answers
F ( s ) = 2 ( s 1 ( s 1 ) 2 + 1 1 ( s 1 ) 2 + 1 )
If you change the variable and substitute x=s−1 you have to substitute s on both sides:
F ( x + 1 ) = 2 ( x x 2 + 1 1 x 2 + 1 )
L 1 { F ( x + 1 ) } = 2 L 1 { ( x x 2 + 1 1 x 2 + 1 ) }
e t f ( t ) = 2 ( cos ( t ) sin ( t ) )
Therefore:
f ( t ) = 2 e t ( cos ( t ) sin ( t ) )
So this line is not correct. Because your substitution on LHS is not correct s=x. When on RHS you substitute s=x−1.
F ( s ) = 2 ( s 1 ( s 1 ) 2 + 1 1 ( s 1 ) 2 + 1 )
F ( x ) x = s = x x 2 + 1 1 x 2 + 1 x = s 1
It should be:
F ( x + 1 ) x = s 1 = x x 2 + 1 1 x 2 + 1 x = s 1
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-11-11
How to solve the following integral:
I = 0 cos ( t l o g ( x ) ) e a x d x ,
where t and a are real.
asked 2022-09-06
Prove that some function is the solution of some equation
Show that
x ( t ) = n = 0 ( 1 ) n ( t / 2 ) 2 n ( n ! ) 2
is the solution of
x x = 0 t x ( u ) x ( t u ) d u = sin t
asked 2022-11-01
Consider the Laplace transform f ~ ( s ) of a function f(t) defined as
f ~ ( s ) = 0 f ( t ) e s t d t .
Can this relation be inverted to obtain f(t) in terms of f ~ ( s )?
The reason I ask this is as follows. The Fourier transform
f ~ ( k ) = f ( x ) e i k x d x
is defined simulataneously with the inversion formula
f ( x ) = f ~ ( k ) e i k x d k .
But I haven't seen the same for the Laplace's transform.
asked 2021-12-07

Take the Laplace transform of the following initial value and solve for Y(s)=L{y(t)}:

              sin(π t), 0 ≤ t <1

y''+9y= {                               y(0)=0, y′(0)=0

                  0, 1 ≤ t

a) Y(s)= ? (Hint: write the right hand side in terms of the Heaviside function)

b) Now find the inverse transform to find y(t)= ? . (Use step(t-c) for uc(t) .)

Note= π/ (s^2+π^2)(s^2+9) = π/ π^2 -9 (1/(s^2+9) - 1/(s^2+ π^2)

asked 2020-10-31
Use the Laplace transform to solve the given initial-value problem
y+2y+y=0,y(0)=1,y(0)=1
asked 2022-11-02
Find:
L ( t 2 e t u ( t 6 ) )
My trying
L ( F ( t a ) u ( t a ) ) = e a s F ( s ) L ( t f ( t ) ) = d d s f ( s ) L ( t 2 f ( t ) ) = d 2 d s 2 f ( s )
Then:
L ( t 2 e t u ( t 6 ) ) d 2 d s 2 ( e t u ( t 6 ) ) e t u ( t 6 )
L ( e t u ( t 6 ) ) = e 6 s 1 s 1
is my attempt correct ??
asked 2021-09-03
Solve y+y=4δ(t2π)
Subject to
y(0)=0
y'(0)=0