What is inverse oflaplace (1+6s^2)/(s(s^2+1)) ?

The answer to inverse oflaplace (1+6s^2)/(s(s^2+1)) is \H(t)+5cos(t)

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inverse oflaplace (1+6s^2)/(s(s^2+1))

inverse laplace

\frac{1 + 6 s ^{2}}{s ( s ^{2} + 1 )}

H (t) + 5 cos (t)

Solution steps

L^{- 1} {\frac{1 + 6 s ^{2}}{s ( s ^{2} + 1 )}}

Take the partial fraction of

\frac{1 + 6 s ^{2}}{s ( s ^{2} + 1 )} : \frac{1}{s} + \frac{5 s}{s ^{2} + 1}

= L^{- 1} {\frac{1}{s} + \frac{5 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{5 s}{s ^{2} + 1}}

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

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

= H (t) + 5 cos (t)

\frac{d ^{2} y}{d x ^{2}} = (4 sin (3 x))

\frac{s}{( s ^{2} + 4 ) ( s + 2 )}

\int - 4 x - 1 d x

y = sin (x^{5})

\int \frac{x ^{2} + 8 x - 7}{x - 3} d x

What is inverse oflaplace (1+6s^2)/(s(s^2+1)) ?
The answer to inverse oflaplace (1+6s^2)/(s(s^2+1)) is \H(t)+5cos(t)