What is inverse oflaplace 7/(s(s^2+5s)) ?

The answer to inverse oflaplace 7/(s(s^2+5s)) is -7/25 \H(t)+(7t)/5+7/25 e^{-5t}

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inverse oflaplace 7/(s(s^2+5s))

inverse laplace

\frac{7}{s ( s ^{2} + 5 s )}

- \frac{7}{25} H (t) + \frac{7 t}{5} + \frac{7}{25} e^{- 5 t}

Solution steps

L^{- 1} {\frac{7}{s ( s ^{2} + 5 s )}}

Take the partial fraction of

\frac{7}{s ( s ^{2} + 5 s )} : - \frac{7}{25 s} + \frac{7}{5 s ^{2}} + \frac{7}{25 ( s + 5 )}

= L^{- 1} {- \frac{7}{25 s} + \frac{7}{5 s ^{2}} + \frac{7}{25 ( s + 5 )}}

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{7}{25 s}} + L^{- 1} {\frac{7}{5 s ^{2}}} + L^{- 1} {\frac{7}{25 ( s + 5 )}}

L^{- 1} {\frac{7}{25 s}} : \frac{7}{25} H (t)

L^{- 1} {\frac{7}{5 s ^{2}}} : \frac{7 t}{5}

L^{- 1} {\frac{7}{25 ( s + 5 )}} : \frac{7}{25} e^{- 5 t}

= - \frac{7}{25} H (t) + \frac{7 t}{5} + \frac{7}{25} e^{- 5 t}

f (x) = x^{t a n (x)}

\int_{0}^{0.6} 125 x d x

\int 4 x sin (3 x) d x

\int e^{- y} cos (y) d y

y = (x^{2} - 3 x + 3), (2, 1)

What is inverse oflaplace 7/(s(s^2+5s)) ?
The answer to inverse oflaplace 7/(s(s^2+5s)) is -7/25 \H(t)+(7t)/5+7/25 e^{-5t}