Find the solution of the initial value problem given below by Laplace transform. x"+16x=cos(4t) x(0)=0 text{ and } x'(0)=1

Globokim8 2020-12-24 Answered
Find the solution of the initial value problem given below by Laplace transform.
x"+16x=cos(4t)
x(0)=0 and x(0)=1
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Expert Answer

Roosevelt Houghton
Answered 2020-12-25 Author has 106 answers

Step 1
We have to first take Laplace transform of the given differential equation.
step 2
x"+16x=cos(4t)
x(0)=0 and x(0)=1
Taking laplace transform
[s2X(s)0x(0)]+16X(s)=s(s2+16)
X(s)[s2+16]=s(s2+16)+1
X(s)=(s2+s+16)(s2+16)2
Taking Laplace inverse of e X(s)=1(s2+16)+s(s2+16)2
L1{X(s)}=sin4t4+tsin4t8
Answer: x(t)=(sin4t)/4+(tsin4t)/8

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