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91Ó°ÊÓ

Determine the inverse Laplace transform of the given function.

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

Short Answer

Expert verified

Therefore, the solution is

L−1{2s2+3s−1(s+1)2(s+2)}(t)=e−t−2te−t+e−2t

Step by step solution

01

Given Information

The given function value in s domain is

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

02

Use partial fractions

Make partial fraction of the given function, as:

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

On comparing the coefficients with s=−1,−2,0respectively we get

A=1B=−2C=1

Thus, the partial fractions are obtained as:

2s2+3s−1(s+1)2(s+2)=1s+1−2(s+1)2+1s+2

03

Take Inverse Laplace transform

Take inverse Laplace transform using L−11s−a(t)=eatand L−1n!(s−a)n+1(t)=tneatas:

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

Hence, the required inverse Laplace transform is

L−12s2+3s−1(s+1)2(s+2)(t)=e−t−2te−t+e−2t

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