What is inverse oflaplace (360)/(2s^2+10s+8) ?

The answer to inverse oflaplace (360)/(2s^2+10s+8) is 60e^{-t}-60e^{-4t}

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inverse oflaplace (360)/(2s^2+10s+8)

inverse laplace

\frac{360}{2 s ^{2} + 10 s + 8}

60 e^{- t} - 60 e^{- 4 t}

Solution steps

L^{- 1} {\frac{360}{2 s ^{2} + 10 s + 8}}

Take the partial fraction of

\frac{360}{2 s ^{2} + 10 s + 8} : \frac{60}{s + 1} - \frac{60}{s + 4}

= L^{- 1} {\frac{60}{s + 1} - \frac{60}{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{60}{s + 1}} - L^{- 1} {\frac{60}{s + 4}}

L^{- 1} {\frac{60}{s + 1}} : 60 e^{- t}

L^{- 1} {\frac{60}{s + 4}} : 60 e^{- 4 t}

= 60 e^{- t} - 60 e^{- 4 t}

\frac{\partial}{\partial x} (y e^{3 x} \cdot 3)

y = \frac{7}{x ^{2} + 3}, (2, 1)

x \to 2 lim (x^{3} + 4 x^{2} - 3 x - 6)

\int \frac{1}{t ^{2} 5 - t ^{2}} d t

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

What is inverse oflaplace (360)/(2s^2+10s+8) ?
The answer to inverse oflaplace (360)/(2s^2+10s+8) is 60e^{-t}-60e^{-4t}