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

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

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

inverse laplace

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

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

Solution steps

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

Expand

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

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

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

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

\int \frac{7 x}{x ^{2} + 1} d x

f (x) = sec (θ) tan (θ)

\int x^{2} (5 x^{2} + x)^{2} d x

\frac{\partial}{\partial h} (\frac{1}{2} (a + b) h)

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

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