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)

\frac{d}{d x} (\frac{1}{g ( x )})

\int 8^{3 x - 1} d x

24 x - 176

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

\int \frac{1}{1 - sin ( 5 x )} d x

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)