What is inverse oflaplace ((5/s-1))/((s+1)^2+9) ?

The answer to inverse oflaplace ((5/s-1))/((s+1)^2+9) is (\H(t)-e^{-t}cos(3t)-e^{-t}sin(3t))/2

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inverse oflaplace ((5/s-1))/((s+1)^2+9)

inverse laplace

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

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

Solution steps

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

Expand

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

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

L^{- 1} {\frac{\frac{1}{s}}{( s + 1 ) ^{2} + 9}} : \frac{1}{10} H (t) - \frac{1}{10} e^{- t} cos (3 t) - \frac{1}{30} e^{- t} sin (3 t)

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

= 5 (\frac{1}{10} H (t) - \frac{1}{10} e^{- t} cos (3 t) - \frac{1}{30} e^{- t} sin (3 t)) - e^{- t} \frac{1}{3} sin (3 t)

Simplify

5 (\frac{1}{10} H (t) - \frac{1}{10} e^{- t} cos (3 t) - \frac{1}{30} e^{- t} sin (3 t)) - e^{- t} \frac{1}{3} sin (3 t) : \frac{H ( t ) - e ^{- t} cos ( 3 t ) - e ^{- t} sin ( 3 t )}{2}

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

\int \frac{1}{1 + 6 x - 3 x ^{2}} d x

y^{^{'}} - 2 y = 3 e^{- 2 t}

\frac{d}{d x} (\frac{e ^{x^{2}}}{2})

(x + 1) \frac{d y}{d x} + (x + 2) y = 6 x e^{- x}

lo g_{10} (n)

What is inverse oflaplace ((5/s-1))/((s+1)^2+9) ?
The answer to inverse oflaplace ((5/s-1))/((s+1)^2+9) is (\H(t)-e^{-t}cos(3t)-e^{-t}sin(3t))/2