Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y"+y=g(t) , y(0)=-4 , y'(0)=0 where g{{left({t}right)}}={leftlbracebegin{matrix}{t}&{t}<{4}{5}&{t}>{4}end{matrix}right.} Y(s)-?

sibuzwaW 2020-11-16 Answered
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below.
y"+y=g(t),y(0)=4,y(0)=0
where g(t)={tt<45t>4
Y(s)-?
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Expert Answer

Laaibah Pitt
Answered 2020-11-17 Author has 98 answers
Step 1 Basics
Here we use Second Shifting Property and Laplace Transform of Derivative as follows
Step 1
Laplace transform of derivative
L{f(t)}=s2L{f}0f(0)
second shifting property
L{f(tt0)}=est0L{f(t)}
Step 2 Explanation
y"+y=g(t),y(0)=4,y(0)=0
where g(t)={tt<45t>4
Taking Laplace transform both side of above equation , we get
L{y}+L{y}=L{g(t)}
[s2L{y(t)}sy(0)y(0)]+L{y}=L{tu(t)+5u(tu)]
where u(t) is the step function
[use  L{tnu(n)}=n!sn+1]
shifting property as L{f(tt0)}=est0L{f(t)}
s2L{y}+4s+L{y}=1s2+e4s(5s)
L{y}[s2+1]=1s2+5e4ss4s
L{y}=1s2(s2+1)+5e4ss(s2+1)4ss2+1
Answer
y(s)=1s2(s2+1)+5e4ss(s2+1)4ss2+1
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