What is inverse oflaplace (6s+3)/(s^4+5s^2+4) ?

The answer to inverse oflaplace (6s+3)/(s^4+5s^2+4) is 2cos(t)+sin(t)-2cos(2t)-1/2 sin(2t)

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inverse oflaplace (6s+3)/(s^4+5s^2+4)

inverse laplace

\frac{6 s + 3}{s ^{4} + 5 s ^{2} + 4}

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

Solution steps

L^{- 1} {\frac{6 s + 3}{s ^{4} + 5 s ^{2} + 4}}

Take the partial fraction of

\frac{6 s + 3}{s ^{4} + 5 s ^{2} + 4} : \frac{2 s + 1}{s ^{2} + 1} + \frac{- 2 s - 1}{s ^{2} + 4}

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

Expand

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

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)}

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

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

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

L^{- 1} {\frac{s}{s ^{2} + 4}} : cos (2 t)

L^{- 1} {\frac{1}{s ^{2} + 4}} : \frac{1}{2} sin (2 t)

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

x y^{^{'}} + 4 y = 8 x^{4}

\int e^{- t} t d t

\int 27 x^{7} + 3 x^{5} - 45 x^{3} + 2 x d x

x \to 0 lim (sin (x) - x)

\int (\frac{1}{x x}) d x

What is inverse oflaplace (6s+3)/(s^4+5s^2+4) ?
The answer to inverse oflaplace (6s+3)/(s^4+5s^2+4) is 2cos(t)+sin(t)-2cos(2t)-1/2 sin(2t)