What is inverse oflaplace (36)/(s(s^2+1)(s^2+9)) ?

The answer to inverse oflaplace (36)/(s(s^2+1)(s^2+9)) is 4\H(t)-9/2 cos(t)+1/2 cos(3t)

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inverse oflaplace (36)/(s(s^2+1)(s^2+9))

inverse laplace

\frac{36}{s ( s ^{2} + 1 ) ( s ^{2} + 9 )}

4 H (t) - \frac{9}{2} cos (t) + \frac{1}{2} cos (3 t)

Solution steps

L^{- 1} {\frac{36}{s ( s ^{2} + 1 ) ( s ^{2} + 9 )}}

Take the partial fraction of

\frac{36}{s ( s ^{2} + 1 ) ( s ^{2} + 9 )} : \frac{4}{s} - \frac{9 s}{2 ( s ^{2} + 1 )} + \frac{s}{2 ( s ^{2} + 9 )}

= L^{- 1} {\frac{4}{s} - \frac{9 s}{2 ( s ^{2} + 1 )} + \frac{s}{2 ( s ^{2} + 9 )}}

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{4}{s}} - L^{- 1} {\frac{9 s}{2 ( s ^{2} + 1 )}} + L^{- 1} {\frac{s}{2 ( s ^{2} + 9 )}}

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

L^{- 1} {\frac{9 s}{2 ( s ^{2} + 1 )}} : \frac{9}{2} cos (t)

L^{- 1} {\frac{s}{2 ( s ^{2} + 9 )}} : \frac{1}{2} cos (3 t)

= 4 H (t) - \frac{9}{2} cos (t) + \frac{1}{2} cos (3 t)

\int e^{- 23 x} d x

\int \frac{2 y + x}{1 + x} d x

y = (14 x - x^{2})^{3}

n = 1 \sum \infty n^{15}

x (x^{3} - 1)^{6}

What is inverse oflaplace (36)/(s(s^2+1)(s^2+9)) ?
The answer to inverse oflaplace (36)/(s(s^2+1)(s^2+9)) is 4\H(t)-9/2 cos(t)+1/2 cos(3t)