Solve the following differential equations using the Laplace transform and the unit step function y"+4y=g(t) y(0)=-1 y'(0)=0 , text{ where } g(t)=begin{cases}t &, tleq 25 & ,t > 2end{cases} y"-y=g(t) y(0)=1 y'(0)=2 , text{ where } g(t)=begin{cases}1 &, tleq 3t & ,t > 3end{cases}

Carol Gates 2021-02-02 Answered
Solve the following differential equations using the Laplace transform and the unit step function
y"+4y=g(t)
y(0)=1
y(0)=0, where g(t)={t,t25,t>2
y"y=g(t)
y(0)=1
y(0)=2, where g(t)={1,t3t,t>3
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Expert Answer

Sally Cresswell
Answered 2021-02-03 Author has 91 answers

Step 1 Given :The differential equations using the Laplace Transform and the unit step function
y"+4y=g(t)
y(0)=1
y(0)=0, where g(t)={t,t25,t>2 
Step 2
The given initial value problem is
y"+4y=g(t)
y(0)=1
y(0)=0, where g(t)={t,t25,t>2 
Write the step using the Heaviside step functions.
If f(t) is a function of t for t0 whose Laplace Transform F(s) exists , then for any constant a0 , the function
(g(t)=f(ta)u(ta)={0 if ta 
has the Laplace Transform easF(s)
g(t)=(2u(t2))+5u(t2)=2u(t2)+5u(t2)
g(t)=2+4u(t2) So, the initial value problem can be written as
y"+4y=2+4u(t2),y(0)=1,y(0)=0
Taking the Laplace Transform of both sides , we get
s2y¯sy(0)y(0)+4y¯=2s+4e2s
s2y¯s(1)+4y¯=2s+4e2s
s2y¯+s+4y¯=2s+4e2s
(s2+4)y¯=2s+4e2ss
y¯=2s(s2+4)+4e2ss2+4ss2+4
Taking Inverse Laplace Transform , we get
If L1[F(s)]=f(t) then L1(F(s))=0tf(t)dt
L(1)=1s,L1(ss2+4)=cosat,L1(1s2+4)=1asinat
L1[y¯]=L1[2s(s2+4)]+4L1[e2ss2+4]L1[ss2+4]
y=220tsin2tdt+4e2s2sin2tcos2t
 

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