What is inverse oflaplace (2-5s)/(s^2+9) ?

The answer to inverse oflaplace (2-5s)/(s^2+9) is 2/3 sin(3t)-5cos(3t)

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inverse oflaplace (2-5s)/(s^2+9)

inverse laplace

\frac{2 - 5 s}{s ^{2} + 9}

\frac{2}{3} sin (3 t) - 5 cos (3 t)

Solution steps

L^{- 1} {\frac{2 - 5 s}{s ^{2} + 9}}

Expand

= L^{- 1} {\frac{2}{s ^{2} + 9} - \frac{5 s}{s ^{2} + 9}}

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{1}{s ^{2} + 9}} - 5 L^{- 1} {\frac{s}{s ^{2} + 9}}

L^{- 1} {\frac{1}{s ^{2} + 9}} : \frac{1}{3} sin (3 t)

L^{- 1} {\frac{s}{s ^{2} + 9}} : cos (3 t)

= 2 \cdot \frac{1}{3} sin (3 t) - 5 cos (3 t)

Refine

2 \frac{1}{3} sin (3 t) - 5 cos (3 t) : \frac{2}{3} sin (3 t) - 5 cos (3 t)

= \frac{2}{3} sin (3 t) - 5 cos (3 t)

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

\frac{d}{d x} (sin (x^{2} + 2 x y))

\int ln (x + 1 + x^{2}) d x

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

x \to - 2 lim (- 2)

What is inverse oflaplace (2-5s)/(s^2+9) ?
The answer to inverse oflaplace (2-5s)/(s^2+9) is 2/3 sin(3t)-5cos(3t)