Solve by using Laplace transform y"+y=cos t ,y(0)=0 , y'(0)=-1

Brittney Lord 2020-10-28 Answered
Solve by using Laplace transform
y"+y=cost ,y(0)=0 , y(0)=1
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Expert Answer

Benedict
Answered 2020-10-29 Author has 108 answers
Step 1
The given initial value problem is y"+y=cost ,y(0)=0 , y(0)=1
Apply Laplace transform on both sides of the given equation as follows.
L{y"}+L{y}=L{cost}
s2Y(s)sy(0)y(0)+Y(s)=ss2+1
Y(s)(s2+1)+1=ss2+1
Y(s)=ss21(s2+1)2
Step 2
By partial fraction decomposition, 
ss21(s2+1)2=As+B(s2+1)+Cs+D(s2+1)2
=(As+B)(s2+1)+Cs+D(s2+1)2
Then, 
ss21=(As+B)(s2+1)+Cs+D
=As3+As+Bs2+B+Cs+D
=As3+Bs2+(A+C)s+(B+D)
Step 3
Equating the coefficients of like terms on both sides, 
A=0
B=1
A+C=10+C=1C=1
B+D=1D=1+1D=0
Therefore, we have, 
ss21(s2+1)2=As+B(s2+1)+Cs+D(s2+1)2
=1(s2+1)+s(s2+1)2
Step 4
Then, 
Y(s)=ss21(s2+1)2
=As+B(s2+1)+Cs+D(s2+1)2
=1(s2+1)+s(s2+1)2
Now take inverse Laplace transform on both sides, 
L1{Y(s)}=L1{1(s2+1)+s(s2+1)2}

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