What is inverse oflaplace 1/(s^2(s^2+25)) ?

The answer to inverse oflaplace 1/(s^2(s^2+25)) is t/(25)-1/125 sin(5t)

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inverse oflaplace 1/(s^2(s^2+25))

inverselaplace

\frac{t}{25} - \frac{1}{125} sin (5 t)

Solution steps

L^{- 1} {\frac{1}{s ^{2} ( s ^{2} + 25 )}}

Take the partial fraction of

\frac{1}{s ^{2} ( s ^{2} + 25 )} : \frac{1}{25 s ^{2}} - \frac{1}{25 ( s ^{2} + 25 )}

= L^{- 1} {\frac{1}{25 s ^{2}} - \frac{1}{25 ( s ^{2} + 25 )}}

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{1}{25 s ^{2}}} - L^{- 1} {\frac{1}{25 ( s ^{2} + 25 )}}

L^{- 1} {\frac{1}{25 s ^{2}}} : \frac{t}{25}

L^{- 1} {\frac{1}{25 ( s ^{2} + 25 )}} : \frac{1}{125} sin (5 t)

= \frac{t}{25} - \frac{1}{125} sin (5 t)

What is inverse oflaplace 1/(s^2(s^2+25)) ?
The answer to inverse oflaplace 1/(s^2(s^2+25)) is t/(25)-1/125 sin(5t)