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In Problems 3-10, determine the Laplace transform of the given function.1(s2+9)2

Short Answer

Expert verified

Therefore, the solution isL−1{1(s2+9)2}=sin3t−3tcos3t54

Step by step solution

01

Given Information

The given function value in s domain is 1(s2+9)2.

02

Use convolution theorem

Letf(t)=sin3tThen

F(s)=L−1{f(t)}(s)=L−1{sin3t}(s)=3s2+32

We can write the given function as

1(s2+9)2=193s2+323s2+32

Now, using convolution theorem we get

L−11(s2+9)2=L−1193s2+323s2+32=19L−1{F(s)F(s)}=19[f(t)*f(t)]=19[sin3t*sin3t]

Simplify further as:

L−11(s2+9)2=19∫0tsin3(t−v)sin3tdv=118∫0t[cos(6v−3t)−cos3t]dv=118(∫0tcos(6v−3t)dv−∫0tcos3tdv)=118([sin(6v−3t)6]0t−cos3t[v]0t)=118(sin3t6+sin3t6−tcos3t)

Further simplify as:

L−11(s2+9)2=118sin3t−3tcos3t3=sin3t−3tcos3t54

Therefore, the required inverse Laplace transform is

L−11(s2+9)2=sin3t−3tcos3t54

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