What is inverse oflaplace 1/(s^2(s+1/15)) ?

The answer to inverse oflaplace 1/(s^2(s+1/15)) is -225\H(t)+15t+225e^{-t/(15)}

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inverse oflaplace 1/(s^2(s+1/15))

inverse laplace

\frac{1}{s ^{2} ( s + \frac{1}{15} )}

- 225 H (t) + 15 t + 225 e^{- \frac{t}{15}}

Solution steps

L^{- 1} {\frac{1}{s ^{2} ( s + \frac{1}{15} )}}

Take the partial fraction of

\frac{1}{s ^{2} ( s + \frac{1}{15} )} : - \frac{225}{s} + \frac{15}{s ^{2}} + \frac{3375}{15 s + 1}

= L^{- 1} {- \frac{225}{s} + \frac{15}{s ^{2}} + \frac{3375}{15 s + 1}}

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{225}{s}} + L^{- 1} {\frac{15}{s ^{2}}} + L^{- 1} {\frac{3375}{15 s + 1}}

L^{- 1} {\frac{225}{s}} : 225 H (t)

L^{- 1} {\frac{15}{s ^{2}}} : 15 t

L^{- 1} {\frac{3375}{15 s + 1}} : 225 e^{- \frac{t}{15}}

= - 225 H (t) + 15 t + 225 e^{- \frac{t}{15}}

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\int (1 - 3 y)^{2} - (1 - y^{2})^{2} d y

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\int 2 θ sec^{2} (θ) d θ

What is inverse oflaplace 1/(s^2(s+1/15)) ?
The answer to inverse oflaplace 1/(s^2(s+1/15)) is -225\H(t)+15t+225e^{-t/(15)}