What is inverse oflaplace (3s-15)/(2s^2-4s+10) ?

The answer to inverse oflaplace (3s-15)/(2s^2-4s+10) is 3/2 e^tcos(2t)-3e^tsin(2t)

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inverse oflaplace (3s-15)/(2s^2-4s+10)

inverse laplace

\frac{3 s - 15}{2 s ^{2} - 4 s + 10}

\frac{3}{2} e^{t} cos (2 t) - 3 e^{t} sin (2 t)

Solution steps

L^{- 1} {\frac{3 s - 15}{2 s ^{2} - 4 s + 10}}

Expand

\frac{3 s - 15}{2 s ^{2} - 4 s + 10} : \frac{3}{2} \cdot \frac{s - 1}{( s - 1 ) ^{2} + 4} - 6 \cdot \frac{1}{( s - 1 ) ^{2} + 4}

= L^{- 1} {\frac{3}{2} \cdot \frac{s - 1}{( s - 1 ) ^{2} + 4} - 6 \cdot \frac{1}{( s - 1 ) ^{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)}

= \frac{3}{2} L^{- 1} {\frac{s - 1}{( s - 1 ) ^{2} + 4}} - 6 L^{- 1} {\frac{1}{( s - 1 ) ^{2} + 4}}

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

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

= \frac{3}{2} e^{t} cos (2 t) - 6 e^{t} \frac{1}{2} sin (2 t)

Refine

\frac{3}{2} e^{t} cos (2 t) - 6 e^{t} \frac{1}{2} sin (2 t) : \frac{3}{2} e^{t} cos (2 t) - 3 e^{t} sin (2 t)

= \frac{3}{2} e^{t} cos (2 t) - 3 e^{t} sin (2 t)

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 4 x + 3})

\frac{d}{d x} (cos (x) + sin (x) + x)

\int \frac{x}{25 + x ^{4}} d x

\frac{d}{d x} (e^{x^{3}} \cdot 3 x^{2})

\frac{d}{d x} (2 x^{3} e^{- 0.6 x})

What is inverse oflaplace (3s-15)/(2s^2-4s+10) ?
The answer to inverse oflaplace (3s-15)/(2s^2-4s+10) is 3/2 e^tcos(2t)-3e^tsin(2t)