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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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 (2 t) - 3 \cdot \frac{1}{2} sin (2 t)

Refine

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

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

y^{^{'}} + x y = 1 x y^{3}

\int_{1}^{6} - x^{2} + 7 x - 6 d x

x \to 2 + lim (\frac{6 x}{x - 2})

\frac{\partial}{\partial x} (ln (1 + x^{2}) - 2 e^{5 y})

\int \frac{1}{x ( ln ( x ) ) ^{p}} d x

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