use the Laplace transform to solve the given initial-value problem.y"+y=f(t)y(0)=0 , y'(0)=1 wheredisplaystyle f{{left({t}right)}}={leftlbracebegin{matrix}{1}&{0}le{t}<frac{pi}{{2}}{0}&{t}gefrac{pi}{{2}}end{matrix}right.}

bobbie71G 2021-03-04 Answered

use the Laplace transform to solve the given initial-value problem.
y"+y=f(t)
y(0)=0,y(0)=1 where
f(t)={10t<π20tπ2

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Expert Answer

SchulzD
Answered 2021-03-05 Author has 83 answers

Step 1
Given, the initial value problem
y"+y=f(t)
y(0)=0,y(0)=1 where
f(t)={10t<π20tπ2
step 2
The Laplace transform of the two derivatives
L{y}=sY(s)y(0)
L{y}=s2Y(s)sy(0)y(0)
L{1}=1s
L{eat}=1sa
Where y(0) and y’(0) are initial conditions.
Step 3
Now,
when 0tπ2
y"+y=1,y(0)=0.y(0)=1
s2Y(s)sy(0)y(0)+sY(s)y(0)=1
{s2+s}Y(s)s010=1
{s2+s}Y(s)=2
Y(s)=2s2+s=2s(s+1)=2{1s1s+1}
Y(s)=2{1et}
when tπ2
y"+y=0,y(0)=0,y(0)=1
s2Y(s)sy(0)y(0)+sY(s)y(0)=1
{s2+s}Y(s)s010=0
{s2+s}Y(s)=1
Y(s)=1s2+s=1s(s+1)={1s1s+1}
Y(s)={1et}
Step 4
Hence the solution is

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