Use Laplace transforms to solve the following initial value problem. x"+x=3 cos 3t , x(0)=1 , x'(0)=0

tabita57i 2021-09-26 Answered

Use Laplace transforms to solve the following initial value problem.
x+x=3cos3t,x(0)=1,x(0)=0
The solution is x(t)=?

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Expert Answer

cyhuddwyr9
Answered 2021-09-27 Author has 90 answers

Step 1
The given differential equation is
x+x=3cos(3t)
Initial condition: x(0)=1andx(0)=0
Solve the differential equation by using laplace transform.
step 2
Consider
x+x=3cos(3t)
Operate laplace operator on the both sides, then
L{x"(t)}+L{x(t)}=L{3cos(3t)}
s2x(s)sx(0)x(0)+x(s)=3ss2+9
s2x(s)+x(s)1=3ss2+9
(s2+1)x(s)=3ss2+9+1
x(s)=s2+3s+9(s2+9)(s2+1)
Step 3
By using partial fraction method,
x(s)=s2+3s+9(s2+9)(s2+1)
=38(s(s2+9))+38(s(s2+1))+1(s2+1)
Operate laplace inverse on the both sides,then
L1{x(s)}=38L1(s(s2+9))+38L1(s(s2+1))+L1(1(s2+1))
=38cos(3t)+38cos(t)+sin(t)
x(t)=38cos(3t)+38cos(t)+sin(t)
Hence,the required solution is

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