Solve y"+4y=f(t) text{ subject to } y(0)=0 , y'(0)=2 text{ and } f(t)=begin{cases}0 & t< pi1 & tgeqpiend{cases}

abondantQ 2021-02-12 Answered
Solve y"+4y=f(t) subject to y(0)=0,y(0)=2 and 
f(t)={0t<π1tπ
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Expert Answer

firmablogF
Answered 2021-02-13 Author has 92 answers

Step 1 Given differential equation y"+4y=f(t) subject to y(0)=0,y(0)=2 and f(t)={0t<π1tπ
Lf(t)=0f(t)estdt
=0π(0)estdt+π(1)estdt
=0+π(1)estdt
=|ests|π
=|es()ses(π)s|
=eπts
Step 2
Now taking Laplace transform on both sides of the given differential equation
L{y}+4L{y}=L{f(t)}
s2Y(s)s.y(0)y(0)+4Y(s)=eπss
Applying the conditions
s2Y(s)s.02+4Y(s)=eπss
Y(s)(s2+4)=eπss+2
Y(s)=eπss(s2+4)+2s2+4
Now taking Inverse Laplace transform Step 3
L1{Y(s)}=L1eπss(s2+4)+2s2+4
=L1{eπs1s(s2+4)}+2L1{1s2+4}
By partial fraction
1s(s2+4)=14ss4(s2+4)
L1{Y(s)}=L1{eπs(14ss4(s2+4))}+2L1{1s2+22}
Step 4
Apply the rule then L1{easF(s)}=u(ta)f(ta) and from laplace transform table
L1{14ss4(s2+4)}=1414cos(2t)
y(t)=u(1π)(1414cos(2(tπ)))+2(12)sin(2t)

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