What is inverse oflaplace 1/((s+2/7)^2)*1/s ?

The answer to inverse oflaplace 1/((s+2/7)^2)*1/s is 49/4 \H(t)-49/4 e^{-(2t)/7}-7/2 e^{-(2t)/7}t

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inverse oflaplace 1/((s+2/7)^2)*1/s

inverse laplace

\frac{1}{( s + \frac{2}{7} ) ^{2}} \cdot \frac{1}{s}

\frac{49}{4} H (t) - \frac{49}{4} e^{- \frac{2 t}{7}} - \frac{7}{2} e^{- \frac{2 t}{7}} t

Solution steps

L^{- 1} {\frac{1}{( s + \frac{2}{7} ) ^{2}} \cdot \frac{1}{s}}

Expand

\frac{1}{( s + \frac{2}{7} ) ^{2}} \frac{1}{s} : \frac{49}{49 s ^{3} + 28 s ^{2} + 4 s}

= L^{- 1} {\frac{49}{49 s ^{3} + 28 s ^{2} + 4 s}}

Take the partial fraction of

\frac{49}{49 s ^{3} + 28 s ^{2} + 4 s} : \frac{49}{4 s} - \frac{343}{4 ( 7 s + 2 )} - \frac{343}{2 ( 7 s + 2 ) ^{2}}

= L^{- 1} {\frac{49}{4 s} - \frac{343}{4 ( 7 s + 2 )} - \frac{343}{2 ( 7 s + 2 ) ^{2}}}

Expand

= L^{- 1} {\frac{49}{4 s} - \frac{343}{28 s + 8} - \frac{343}{98 s ^{2} + 56 s + 8}}

Expand

\frac{343}{98 s ^{2} + 56 s + 8} : \frac{7}{2} \cdot \frac{1}{( s + \frac{2}{7} ) ^{2}}

= L^{- 1} {\frac{49}{4 s} - \frac{343}{28 s + 8} - \frac{7}{2} \cdot \frac{1}{( s + \frac{2}{7} ) ^{2}}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{49}{4 s}} - L^{- 1} {\frac{343}{28 s + 8}} - \frac{7}{2} L^{- 1} {\frac{1}{( s + \frac{2}{7} ) ^{2}}}

L^{- 1} {\frac{49}{4 s}} : \frac{49}{4} H (t)

L^{- 1} {\frac{343}{28 s + 8}} : \frac{49}{4} e^{- \frac{2 t}{7}}

L^{- 1} {\frac{1}{( s + \frac{2}{7} ) ^{2}}} : e^{- \frac{2 t}{7}} t

= \frac{49}{4} H (t) - \frac{49}{4} e^{- \frac{2 t}{7}} - \frac{7}{2} e^{- \frac{2 t}{7}} t

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What is inverse oflaplace 1/((s+2/7)^2)*1/s ?
The answer to inverse oflaplace 1/((s+2/7)^2)*1/s is 49/4 \H(t)-49/4 e^{-(2t)/7}-7/2 e^{-(2t)/7}t