Session 16: Convolution Theorem. Laplace and Laplace inverse examples using convolution

In this video we will see that Laplace of product of two functions is not equal to product of Laplace of functions. To prove this claim I will give a counterexample. Then we see the notion of Convolution product. We will definition of convolution product and examples on convolution product.
Then we will see that Laplace of convolution product of two functions is equal to product of Laplace of functions. After seeing this theorem we will see examples on how to use this theorem to find Laplace inverse.
This method is many times very much efficient as compared to method of partial fractions.

[@THEKINGSEKHAR]

loads of love from NIT R

[@DrMathaholic]

Thank you..😊🙏❤️❤️

[@lyricaldust]

1) 1/s^4 2) 1/s^(2n) 3)1/27(3t-sin3t) 4) 1/3(cost-cos2t) great lecture sir thanks

[@DrMathaholic]

Good work Bassam 👏

[@wsar7669]

#3 I got: t/3(cos(3t)) + cos(3t) - 3sin(9t)

[@Ehmedbaz]

thank you very much sir ❤

[@DrMathaholic]

Welcome 😊

[@sachinbalamurugan7030]

Sir u put cos instead of sin in the last problem in taking laplace

[@DrMathaholic]

At what timestamp? I see sin everywhere in the last example which I solve..

[@rahulthorat639]

Ans to homework problem 1) 1/s^(4) 2) 1/s^(2n) 3) (t/9)-(sin(3t)/27) 4) cos(2t)-1 Sir the 4) ans is correct?? It look like wrong.

[@kartikmandhan8850]

4th answer is (cost - cos2t)/3

[@rahulthorat639]

@@kartikmandhan8850 what is your choice for f(t) and g(t)??

[@kartikmandhan8850]

@@rahulthorat639 f(t) = sin(2t)/2 ,g(t) = cos(t)

[@rushikeshsanap5069]

@@kartikmandhan8850 yes this is correct

[@DrMathaholic]

Good discussion guys... Appreciated!!

[@wsar7669]

4:46 why can sin(t - 𝜏) simply become sin(𝜏)

[@DrMathaholic]

How ? Sin ( t - 2 pi) = sin t and sin(2pi -t)= - sin (t)