use the Laplace transform to solve the initial value problem.y"-3y'+2y=begin{cases}0&0leq t<11&1leq t<2 -1&tgeq2end{cases}y(0)=-3y'(0)=1

alesterp 2021-03-07 Answered

use the Laplace transform to solve the initial value problem.
y"3y+2y={00t<111t<21t2
y(0)=3
y(0)=1

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Expert Answer

pierretteA
Answered 2021-03-08 Author has 102 answers

Step 1
It is required to solve the initial value problem using laplace transform:
y"3y+2y={00t<111t<21t2
y(0)=3
y(0)=1
Step 2
The first step in using laplace transform to solve an initial value problem is to take the transform of every term of differential equation:
L(y")3L(y)+2L(y)=0,0t<1
Step 3
Now, use the appropriate formula’s:
s2Ysy(0)y(0)3(sY(s)y(0))+2(Y(s))=L(0),0t<1
s2Y(s)s(3)(1)3sY(s)+3(3)+2Y(s)=0
(s23s+2)Y(s)=103s
Y(s)=103s(s23s+2)Y(s)=103s(s1)(s2)
Step 4
Now, find the partial fraction decomposition for the transform:
103s(s1)(s2)=A(s1)+B(s2)
=A(s2)+B(s1)(s1)(s2)
103s=As2A+BsB
103s=(A+B)s+(2AB)
A+B=3
2AB=10
after solving : A=-7 and B=4
Step 5
Thus,
Y(s)=103s(s1)(s2)=A(s1)+B(s2)=7(s1)+4(s2)
finally taking the inverse transform:
y(t)=7et+4e2t
Step 6
Now, consider the case when f(t)=1,1t<2
s2Y(s)sy(0)y(0)3(sY(s)y(0))+2(Y(s))=L(1),1t<2
s2Y(s)s(3)(1)3sY(s)+3(3)+2Y(s)=1s
(s23s+2)Y(s)=103s+1s
Y(s)=103s+1ss23s+2Y(s)=3s2+10s+1s(s1)(s2)
Step 7
Now, find the partial fraction decomposition for the transform:
3s2+10s+1s(s1)(s2)=As+Bs1+Cs2

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