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

The answer to inverse oflaplace (1+5s^2)/(s^2(s-9)) is -1/81 \H(t)-t/9+406/81 e^{9t}

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

inverse laplace

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

- \frac{1}{81} H (t) - \frac{t}{9} + \frac{406}{81} e^{9 t}

Solution steps

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

Take the partial fraction of

\frac{1 + 5 s ^{2}}{s ^{2} ( s - 9 )} : - \frac{1}{81 s} - \frac{1}{9 s ^{2}} + \frac{406}{81 ( s - 9 )}

= L^{- 1} {- \frac{1}{81 s} - \frac{1}{9 s ^{2}} + \frac{406}{81 ( s - 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)}

= - L^{- 1} {\frac{1}{81 s}} - L^{- 1} {\frac{1}{9 s ^{2}}} + L^{- 1} {\frac{406}{81 ( s - 9 )}}

L^{- 1} {\frac{1}{81 s}} : \frac{1}{81} H (t)

L^{- 1} {\frac{1}{9 s ^{2}}} : \frac{t}{9}

L^{- 1} {\frac{406}{81 ( s - 9 )}} : \frac{406}{81} e^{9 t}

= - \frac{1}{81} H (t) - \frac{t}{9} + \frac{406}{81} e^{9 t}

\int 8 e^{5 x} d x

(2 x + y e^{\frac{y}{x}}) d x - x e^{\frac{y}{x}} d y = 0, y (1) = 4

t an g e n t y = x^{2} - 5 x + 5

\int \frac{x ^{2}}{361 - x ^{2}} d x

\int \frac{4 x + 2}{x ^{2} + x + 5} d x

What is inverse oflaplace (1+5s^2)/(s^2(s-9)) ?
The answer to inverse oflaplace (1+5s^2)/(s^2(s-9)) is -1/81 \H(t)-t/9+406/81 e^{9t}