What is inverse oflaplace ((10s+8))/(s^3) ?

The answer to inverse oflaplace ((10s+8))/(s^3) is 10t+4t^2

Popular >

inverse oflaplace ((10s+8))/(s^3)

inverse laplace

\frac{( 10 s + 8 )}{s ^{3}}

10 t + 4 t^{2}

Solution steps

L^{- 1} {\frac{( 10 s + 8 )}{s ^{3}}}

Expand

= L^{- 1} {\frac{10 s}{s ^{3}} + \frac{8}{s ^{3}}}

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)}

= 10 L^{- 1} {\frac{s}{s ^{3}}} + 8 L^{- 1} {\frac{1}{s ^{3}}}

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

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

= 10 t + 8 \cdot \frac{t ^{2}}{2}

Refine

10 t + 8 \frac{t ^{2}}{2} : 10 t + 4 t^{2}

= 10 t + 4 t^{2}

x \to 0 lim (\frac{c}{x})

x^{8} (x - 1)^{7}

\int t csc (t) cot (t) d t

\frac{d}{d x} (\frac{5}{cos ^{2} ( x )})

x \to 2 - lim (\frac{7}{x - 2})

What is inverse oflaplace ((10s+8))/(s^3) ?
The answer to inverse oflaplace ((10s+8))/(s^3) is 10t+4t^2