What is inverse oflaplace ((12s+3))/(s^2(2s+8)) ?

The answer to inverse oflaplace ((12s+3))/(s^2(2s+8)) is 45/32 \H(t)+(3t)/8-45/32 e^{-4t}

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inverse oflaplace ((12s+3))/(s^2(2s+8))

inverse laplace

\frac{( 12 s + 3 )}{s ^{2} ( 2 s + 8 )}

\frac{45}{32} H (t) + \frac{3 t}{8} - \frac{45}{32} e^{- 4 t}

Solution steps

L^{- 1} {\frac{( 12 s + 3 )}{s ^{2} ( 2 s + 8 )}}

Take the partial fraction of

\frac{12 s + 3}{s ^{2} ( 2 s + 8 )} : \frac{45}{32 s} + \frac{3}{8 s ^{2}} - \frac{45}{32 ( s + 4 )}

= L^{- 1} {\frac{45}{32 s} + \frac{3}{8 s ^{2}} - \frac{45}{32 ( s + 4 )}}

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{45}{32 s}} + L^{- 1} {\frac{3}{8 s ^{2}}} - L^{- 1} {\frac{45}{32 ( s + 4 )}}

L^{- 1} {\frac{45}{32 s}} : \frac{45}{32} H (t)

L^{- 1} {\frac{3}{8 s ^{2}}} : \frac{3 t}{8}

L^{- 1} {\frac{45}{32 ( s + 4 )}} : \frac{45}{32} e^{- 4 t}

= \frac{45}{32} H (t) + \frac{3 t}{8} - \frac{45}{32} e^{- 4 t}

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

y^{^{'}} + \frac{y}{9.33} = 6.429

y^{^{''}} = x^{4}

\frac{s}{s + 3}

\frac{d x}{d t} = \frac{x ^{2} + t 9 t ^{2} + x ^{2}}{t x}

What is inverse oflaplace ((12s+3))/(s^2(2s+8)) ?
The answer to inverse oflaplace ((12s+3))/(s^2(2s+8)) is 45/32 \H(t)+(3t)/8-45/32 e^{-4t}