What is inverse oflaplace-1/(s^2)e^{-5s}+1/(s^2) ?

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

inverse oflaplace-1/(s^2)e^{-5s}+1/(s^2)

inverse laplace

- \frac{1}{s ^{2}} e^{- 5 s} + \frac{1}{s ^{2}}

- H (t - 5) (t - 5) + t

Solution steps

L^{- 1} {- e^{- 5 s} \frac{1}{s ^{2}} + \frac{1}{s ^{2}}}

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 ^{2}} e^{- 5 s}} + L^{- 1} {\frac{1}{s ^{2}}}

L^{- 1} {\frac{1}{s ^{2}} e^{- 5 s}} : H (t - 5) (t - 5)

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

= - H (t - 5) (t - 5) + t

\int \frac{3 x ^{2} - 8 x - 9}{x ^{3} - 4 x ^{2} - 9 x + 36} d x

\frac{1 - ln ( x )}{2 x ^{2}}

x \to 1 + lim ((x - 1)^{x - 1})

y = 2 x + 3

\frac{d}{d x} (- x - 5)

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