Mahesh S R

Mahesh S R

Answered question

2022-08-23

 

 

 

Answer & Explanation

xleb123

xleb123

Skilled2023-06-03Added 181 answers

To obtain the inverse Laplace transform for 1(s1)(s2+1) using the convolution theorem, we can express the given function as a product of two Laplace transforms.
Let's rewrite the function as 1s1·1s2+1.
Using the Laplace transform table, we know that the inverse Laplace transform of 1s1 is et, and the inverse Laplace transform of 1s2+1 is sin(t).
According to the convolution theorem, the inverse Laplace transform of the product of two functions can be found by convolving their individual inverse Laplace transforms. Therefore, the inverse Laplace transform of 1(s1)(s2+1) is given by the convolution of et and sin(t).
The convolution integral is defined as:
(f*g)(t)=0tf(tτ)g(τ)dτ
In our case, we have f(t)=et and g(t)=sin(t).
Therefore, the inverse Laplace transform of 1(s1)(s2+1) is given by:
(et*sin(t))(t)=0tetτsin(τ)dτ
Simplifying the integral and evaluating it, we can find the inverse Laplace transform.
0tetτsin(τ)dτ=etτcos(τ)|0t+0tetτcos(τ)dτ
Simplifying further, we get:
(etcos(t)cos(0))+0tetτcos(τ)dτ
=etcos(t)+1+0tetτcos(τ)dτ
The remaining integral can be solved using integration techniques.
Therefore, the inverse Laplace transform of 1(s1)(s2+1) is given by:
etcos(t)+1+0tetτcos(τ)dτ

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