Solve Inverse Laplace Transform Using Input Integral Theorem

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Problem

Using the input integral principle below

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s), s > c $$

Find $ mathscrL^-1 left[ frac1s(s^2+1) right](t) $ without using partial fractions.

Attempt

Letting $ f(t) = mathscrL^-1 left[ frac1s(s^2+1) right](t) $,

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ mathscrL^-1 left[ frac1s(s^2+1) right](t) right] (s), s > c $$

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s left[ frac1s(s^2+1) right], s > c $$

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s^2(s^2+1) , s > c $$

Notes

Perhaps I've approached this problem incorrectly, but I'm confused as how to proceed with it. All I'm looking for is a hint or correct first step in solving this problem, with a little bit of explanation as to what the correct method toward solving this problem entails.

Also, I searched up what the input integral principle is and I'm not finding anything on it. Did my professor invent this name or is it an alias for something else?

That being said, any help is appreciated. Thanks!

integration differential-equations definite-integrals laplace-transform

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asked 55 mins ago

Anthony Krivonos

1878

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up vote
1
down vote

favorite

Problem

Using the input integral principle below

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s), s > c $$

Find $ mathscrL^-1 left[ frac1s(s^2+1) right](t) $ without using partial fractions.

Attempt

Letting $ f(t) = mathscrL^-1 left[ frac1s(s^2+1) right](t) $,

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ mathscrL^-1 left[ frac1s(s^2+1) right](t) right] (s), s > c $$

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s left[ frac1s(s^2+1) right], s > c $$

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s^2(s^2+1) , s > c $$

Notes

Perhaps I've approached this problem incorrectly, but I'm confused as how to proceed with it. All I'm looking for is a hint or correct first step in solving this problem, with a little bit of explanation as to what the correct method toward solving this problem entails.

Also, I searched up what the input integral principle is and I'm not finding anything on it. Did my professor invent this name or is it an alias for something else?

That being said, any help is appreciated. Thanks!

integration differential-equations definite-integrals laplace-transform

share|cite|improve this question

asked 55 mins ago

Anthony Krivonos

1878

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Problem

Using the input integral principle below

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s), s > c $$

Find $ mathscrL^-1 left[ frac1s(s^2+1) right](t) $ without using partial fractions.

Attempt

Letting $ f(t) = mathscrL^-1 left[ frac1s(s^2+1) right](t) $,

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ mathscrL^-1 left[ frac1s(s^2+1) right](t) right] (s), s > c $$

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s left[ frac1s(s^2+1) right], s > c $$

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s^2(s^2+1) , s > c $$

Notes

Perhaps I've approached this problem incorrectly, but I'm confused as how to proceed with it. All I'm looking for is a hint or correct first step in solving this problem, with a little bit of explanation as to what the correct method toward solving this problem entails.

Also, I searched up what the input integral principle is and I'm not finding anything on it. Did my professor invent this name or is it an alias for something else?

That being said, any help is appreciated. Thanks!

integration differential-equations definite-integrals laplace-transform

share|cite|improve this question

asked 55 mins ago

Anthony Krivonos

1878

Problem

Using the input integral principle below

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s), s > c $$

Find $ mathscrL^-1 left[ frac1s(s^2+1) right](t) $ without using partial fractions.

Attempt

Letting $ f(t) = mathscrL^-1 left[ frac1s(s^2+1) right](t) $,

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ mathscrL^-1 left[ frac1s(s^2+1) right](t) right] (s), s > c $$

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s left[ frac1s(s^2+1) right], s > c $$

$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s^2(s^2+1) , s > c $$

Notes

Perhaps I've approached this problem incorrectly, but I'm confused as how to proceed with it. All I'm looking for is a hint or correct first step in solving this problem, with a little bit of explanation as to what the correct method toward solving this problem entails.

Also, I searched up what the input integral principle is and I'm not finding anything on it. Did my professor invent this name or is it an alias for something else?

That being said, any help is appreciated. Thanks!

integration differential-equations definite-integrals laplace-transform

integration differential-equations definite-integrals laplace-transform

share|cite|improve this question

asked 55 mins ago

Anthony Krivonos

1878

share|cite|improve this question

asked 55 mins ago

Anthony Krivonos

1878

share|cite|improve this question

share|cite|improve this question

asked 55 mins ago

Anthony Krivonos

1878

asked 55 mins ago

Anthony Krivonos

1878

asked 55 mins ago

Anthony Krivonos

1878

1878

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
3
down vote

Compare these
$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) $$
$$ mathscrL left[ int_0^t colorredsin u du right] (s) = frac1s colorredfrac1(s^2+1) $$

share|cite|improve this answer

answered 47 mins ago

Nosrati

22.9k61951

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 color text makes things easy to understand
  – Isham
  10 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Hint: The implied way is
$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) = frac1sfrac1s^2+1 implies mathscrL left[ f(t) right] (s) = frac1s^2+1.$$

share|cite|improve this answer

answered 48 mins ago

Math Lover

13k31333

add a comment | 

up vote
1
down vote

Note that your $$f(t)= sin t $$ so $$ mathscrL^-1 left[ frac1s(s^2+1) right](t)=int _0^t sin u du = 1- cos t$$

share|cite|improve this answer

answered 38 mins ago

Mohammad Riazi-Kermani

32.2k41853

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote

Compare these
$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) $$
$$ mathscrL left[ int_0^t colorredsin u du right] (s) = frac1s colorredfrac1(s^2+1) $$

share|cite|improve this answer

answered 47 mins ago

Nosrati

22.9k61951

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 color text makes things easy to understand
  – Isham
  10 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

Compare these
$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) $$
$$ mathscrL left[ int_0^t colorredsin u du right] (s) = frac1s colorredfrac1(s^2+1) $$

share|cite|improve this answer

answered 47 mins ago

Nosrati

22.9k61951

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 color text makes things easy to understand
  – Isham
  10 mins ago
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Compare these
$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) $$
$$ mathscrL left[ int_0^t colorredsin u du right] (s) = frac1s colorredfrac1(s^2+1) $$

share|cite|improve this answer

answered 47 mins ago

Nosrati

22.9k61951

Compare these
$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) $$
$$ mathscrL left[ int_0^t colorredsin u du right] (s) = frac1s colorredfrac1(s^2+1) $$

share|cite|improve this answer

answered 47 mins ago

Nosrati

22.9k61951

share|cite|improve this answer

share|cite|improve this answer

answered 47 mins ago

Nosrati

22.9k61951

answered 47 mins ago

Nosrati

22.9k61951

answered 47 mins ago

Nosrati

22.9k61951

22.9k61951

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 color text makes things easy to understand
  – Isham
  10 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 color text makes things easy to understand
  – Isham
  10 mins ago
  

  
  
  
  

+1 color text makes things easy to understand
– Isham
10 mins ago

+1 color text makes things easy to understand
– Isham
10 mins ago

add a comment | 

up vote
1
down vote

Hint: The implied way is
$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) = frac1sfrac1s^2+1 implies mathscrL left[ f(t) right] (s) = frac1s^2+1.$$

share|cite|improve this answer

answered 48 mins ago

Math Lover

13k31333

add a comment | 

up vote
1
down vote

Hint: The implied way is
$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) = frac1sfrac1s^2+1 implies mathscrL left[ f(t) right] (s) = frac1s^2+1.$$

share|cite|improve this answer

answered 48 mins ago

Math Lover

13k31333

add a comment | 

up vote
1
down vote

up vote
1
down vote

Hint: The implied way is
$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) = frac1sfrac1s^2+1 implies mathscrL left[ f(t) right] (s) = frac1s^2+1.$$

share|cite|improve this answer

answered 48 mins ago

Math Lover

13k31333

Hint: The implied way is
$$ mathscrL left[ int_0^t f(u)du right] (s) = frac1s mathscrL left[ f(t) right] (s) = frac1sfrac1s^2+1 implies mathscrL left[ f(t) right] (s) = frac1s^2+1.$$

share|cite|improve this answer

answered 48 mins ago

Math Lover

13k31333

share|cite|improve this answer

share|cite|improve this answer

answered 48 mins ago

Math Lover

13k31333

answered 48 mins ago

Math Lover

13k31333

answered 48 mins ago

Math Lover

13k31333

13k31333

add a comment | 

add a comment | 

up vote
1
down vote

Note that your $$f(t)= sin t $$ so $$ mathscrL^-1 left[ frac1s(s^2+1) right](t)=int _0^t sin u du = 1- cos t$$

share|cite|improve this answer

answered 38 mins ago

Mohammad Riazi-Kermani

32.2k41853

add a comment | 

up vote
1
down vote

Note that your $$f(t)= sin t $$ so $$ mathscrL^-1 left[ frac1s(s^2+1) right](t)=int _0^t sin u du = 1- cos t$$

share|cite|improve this answer

answered 38 mins ago

Mohammad Riazi-Kermani

32.2k41853

add a comment | 

up vote
1
down vote

up vote
1
down vote

Note that your $$f(t)= sin t $$ so $$ mathscrL^-1 left[ frac1s(s^2+1) right](t)=int _0^t sin u du = 1- cos t$$

share|cite|improve this answer

answered 38 mins ago

Mohammad Riazi-Kermani

32.2k41853

Note that your $$f(t)= sin t $$ so $$ mathscrL^-1 left[ frac1s(s^2+1) right](t)=int _0^t sin u du = 1- cos t$$

share|cite|improve this answer

answered 38 mins ago

Mohammad Riazi-Kermani

32.2k41853

share|cite|improve this answer

share|cite|improve this answer

answered 38 mins ago

Mohammad Riazi-Kermani

32.2k41853

answered 38 mins ago

Mohammad Riazi-Kermani

32.2k41853

answered 38 mins ago

Mohammad Riazi-Kermani

32.2k41853

32.2k41853

add a comment | 

add a comment | 

 

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