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

The answer to inverse oflaplace (2s)/((s^2+1)^2(s+1)) is (tsin(t)-tcos(t)+cos(t)-e^{-t})/2

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

inverse laplace

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

\frac{t sin ( t ) - t cos ( t ) + cos ( t ) - e ^{- t}}{2}

Solution steps

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

Take the partial fraction of

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

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

Expand

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

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)}

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

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

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

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

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

L^{- 1} {\frac{1}{s + 1}} : e^{- t}

= \frac{1}{2} cos (t) - \frac{1}{2} sin (t) + \frac{t sin ( t )}{2} + \frac{1}{2} (sin (t) - t cos (t)) - \frac{1}{2} e^{- t}

Simplify

\frac{1}{2} cos (t) - \frac{1}{2} sin (t) + \frac{t sin ( t )}{2} + \frac{1}{2} (sin (t) - t cos (t)) - \frac{1}{2} e^{- t} : \frac{t sin ( t ) - t cos ( t ) + cos ( t ) - e ^{- t}}{2}

= \frac{t sin ( t ) - t cos ( t ) + cos ( t ) - e ^{- t}}{2}

What is inverse oflaplace (2s)/((s^2+1)^2(s+1)) ?
The answer to inverse oflaplace (2s)/((s^2+1)^2(s+1)) is (tsin(t)-tcos(t)+cos(t)-e^{-t})/2