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

The answer to inverse oflaplace-1/(s(s^2+1)) is -\H(t)+cos(t)

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

inverse laplace

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

- H (t) + cos (t)

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= - H (t) + cos (t)

y^{^{''}} + 4 y = 8 sin (2 x)

\frac{3 s - 1}{s ^{2} + 2 s + 5}

sin^{2} (x) + cos^{2} (x)

x \to 0.5 lim (\frac{sin ( x )}{x})

(- 5, - 1), (5, 2)

What is inverse oflaplace-1/(s(s^2+1)) ?
The answer to inverse oflaplace-1/(s(s^2+1)) is -\H(t)+cos(t)