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Determine the inverse Laplace transform of the given function.

s2+16s+9(s+1)(s+3)(s-2)

Short Answer

Expert verified

Therefore, the solution is

L−1{s2+16s+9(s+1)(s+3)(s−2)}(t)=e−t−3e−3t+3e2t

Step by step solution

01

Given Information

The given function value in s domain iss2+16s+9(s+1)(s+3)(s-2)

02

Use partial fractions

Make partial fraction of the given function, as:

s2+16s+9(s+1)(s+3)(s−2)=As+1+Bs+3+Cs−2=A(s+3)(s−2)+B(s+1)(s−2)+C(s+1)(s+3)(s+1)(s+3)(s−2)

On comparing the coefficients with s=−1,−3,2respectively we get

A=1B=−3C=3

Thus, the partial fractions are obtained as:

s2+16s+9(s+1)(s+3)(s−2)=1s+1−3s+3+3s−2

03

Take Inverse Laplace transform

Take inverse Laplace transform usingL−1{1s−a}(t)=eat as:

L−1{1s+1−3s+3+3s−2}=L−1{1s+1}(t)−3L−1{1s+3}(t)+3L−1{1s−2}(t)=e−t−3e−3t+3e2t

Hence, the required inverse Laplace transform is

L−1{s2+16s+9(s+1)(s+3)(s−2)}(t)=e−t−3e−3t+3e2t

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