What is inverse oflaplace 1/(s-3)-(16)/(s^2+9) ?

The answer to inverse oflaplace 1/(s-3)-(16)/(s^2+9) is e^{3t}-16/3 sin(3t)

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inverse oflaplace 1/(s-3)-(16)/(s^2+9)

inverse laplace

\frac{1}{s - 3} - \frac{16}{s ^{2} + 9}

e^{3 t} - \frac{16}{3} sin (3 t)

Solution steps

L^{- 1} {\frac{1}{s - 3} - \frac{16}{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)}

= L^{- 1} {\frac{1}{s - 3}} - L^{- 1} {\frac{16}{s ^{2} + 9}}

L^{- 1} {\frac{1}{s - 3}} : e^{3 t}

L^{- 1} {\frac{16}{s ^{2} + 9}} : \frac{16}{3} sin (3 t)

= e^{3 t} - \frac{16}{3} sin (3 t)

x \to 6 lim (ln (x - 6))

y = (3 x^{2} + 4) (3 + x^{3})^{5}

17 x, x^{2}

e^{x^{3} + 10 x}

What is inverse oflaplace 1/(s-3)-(16)/(s^2+9) ?
The answer to inverse oflaplace 1/(s-3)-(16)/(s^2+9) is e^{3t}-16/3 sin(3t)