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## Q. 2.7.3

Two Repeated Roots and One Distinct Root

Obtain the inverse Laplace transform of

$X(s)=\frac{7}{(s+3)^2(s+5)}$

## Verified Solution

The denominator roots are s= −5, s= −3, and s= −3. Thus, the partial-fraction expansion has the form

$X(s)=\frac{7}{(s+3)^2(s+5)}= \frac{C_1}{(s+3)^2}+\frac{C_2}{s+3}+\frac{C_3}{s+5}$

where

$C_1=\underset{s \rightarrow -3}{\text{lim}} [(s+3)^2\frac{7}{(s+3)^2(s+5)}]= \underset{s \rightarrow -3}{\text{lim}}(\frac{7}{s+5}) = \frac{7}{2}$

$C_2=\underset{s \rightarrow -3}{\text{lim}} \frac{d}{ds}[(s+3)^2 \frac{7}{(s+3)^2(s+5)}] = \underset{s \rightarrow -3}{\text{lim}} \frac{d}{ds} (\frac{7}{s+5}) \\ = \underset{s \rightarrow -3}{\text{lim}} [\frac{-7}{(s+5)^2}]=- \frac{7}{4}$

$C_3= \underset{s \rightarrow -5}{\text{lim}} [(s+5) \frac{7}{(s+3)^2(s+5)}] = \underset{s \rightarrow -5}{\text{lim}} [\frac{7}{(s+3)^2}]=\frac{7}{4}$

The inverse transform is

$x(t)=C_1te^{-3t}+C_2e^{-3t}+C_3e^{-5t}= \frac{7}{2}te^{-3t}-\frac{7}{4}e^{-3t}+\frac{7}{4}e^{-5t}$