Solve the initial value problem {(y+16y=cos(4t)),(y(0)=1),(y'(0)=1):} using the Laplace transform.

Wierzycaz 2020-10-23 Answered

Solve the initial value problem {y+16y=cos(4t)y(0)=1y(0)=1 using the Laplace transform.

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Expert Answer

Obiajulu
Answered 2020-10-24 Author has 98 answers
Step 1
Given: y+16y=cos(4t),y(0)=1,y(0)=1
Step 2
Explanation:
y+16y=cos(4t),y(0)=1,y(0)=1
Taking Laplace transform on both sides
L[y+16y]=L[cos(4t)]
L(y(t))+L(16y(t))=L[cos(4t)](i)
We know that if L(y(t))=y(s)
then L(y(t))=s2y(s)sy(0)y(0)
L[cos(at)]=ss2+a2
Using these in equation (i), we get
s2y(s)sy(0)y(0)+16y(s)=ss2+42
Using the given initial conditions y(0)=1,y(0)=1 then
(s2+16)y(s)s(1)1=ss2+42
(s2+42)y(s)=1+s+ss2+42
y(s)=1s2+42+ss2+42+s(s2+42)(s2+42)
y(s)=1s2+42+ss2+42+s(s2+42)2
Taking inverse Laplace transform on both sides
L1(y(s))=L1(1s2+42+ss2+42+s(s2+42)2)

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