Mathematics, 13.08.2020 14:01 Jgarciar6841
find the response of the function at * (t=4) using Laplace transform (y() + 2y" + y = sint) y(0)=1, y (0)=-2, y"(0)=3 , y"(0)=0
Answers: 1
), I'm assuming the ODE is
, 
, 
, and 
by 
:
![L^{-1}\left[\dfrac s{s^2+1}\right]=\cos t](/tpl/images/0721/3730/d0ab0.png)
![L^{-1}\left[-\dfrac2{s^2+1}\right]=-2\sin t](/tpl/images/0721/3730/019e8.png)
![L\left[(\alpha \ast \beta)(t)\right]=A(s)\cdot B(s)](/tpl/images/0721/3730/7fa75.png)
and 
are the Laplace transforms of 
and 
, respectively, and the convolution is defined by
![L^{-1}\left[\dfrac{4s}{(s^2+1)^2}\right]=(\alpha \ast \beta)(t)=2t\sin t](/tpl/images/0721/3730/71b18.png)
![\implies L^{-1}\left[-\dfrac2{(s^2+1)^2}\right]=t\cos t-\sin t](/tpl/images/0721/3730/38a5c.png)
![L^{-1}\left[\dfrac1{(s^2+1)^2}\right]](/tpl/images/0721/3730/bacf1.png)
![L^{-1}\left[\dfrac1{(s^2+1)^2}\right]=\dfrac{\sin t-t\cos t}2](/tpl/images/0721/3730/c3b0c.png)
![\implies L^{-1}\left[\dfrac1{(s^2+1)^3}\right]=\dfrac{(3-t^2)\sin t-3t\cos t}8](/tpl/images/0721/3730/7c2a0.png)
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find the response of the function at * (t=4) using Laplace transform (y() + 2y" + y = sint) y(0)=1,...
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