What is inverse oflaplace (3s+12)/(s(s^2+4)) ?

The answer to inverse oflaplace (3s+12)/(s(s^2+4)) is 3\H(t)-3cos(2t)+3/2 sin(2t)

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inverse oflaplace (3s+12)/(s(s^2+4))

inverse laplace

\frac{3 s + 12}{s ( s ^{2} + 4 )}

3 H (t) - 3 cos (2 t) + \frac{3}{2} sin (2 t)

Solution steps

L^{- 1} {\frac{3 s + 12}{s ( s ^{2} + 4 )}}

Take the partial fraction of

\frac{3 s + 12}{s ( s ^{2} + 4 )} : \frac{3}{s} + \frac{- 3 s + 3}{s ^{2} + 4}

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

Expand

= L^{- 1} {\frac{3}{s} + \frac{3}{s ^{2} + 4} - \frac{3 s}{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)}

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

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

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

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

= 3 H (t) - 3 cos (2 t) + 3 \cdot \frac{1}{2} sin (2 t)

Refine

3 H (t) - 3 cos (2 t) + 3 \frac{1}{2} sin (2 t) : 3 H (t) - 3 cos (2 t) + \frac{3}{2} sin (2 t)

= 3 H (t) - 3 cos (2 t) + \frac{3}{2} sin (2 t)

What is inverse oflaplace (3s+12)/(s(s^2+4)) ?
The answer to inverse oflaplace (3s+12)/(s(s^2+4)) is 3\H(t)-3cos(2t)+3/2 sin(2t)