What is inverse oflaplace (1200)/(s^2(s+10)^2) ?

The answer to inverse oflaplace (1200)/(s^2(s+10)^2) is -12/5 \H(t)+12t+12/5 e^{-10t}+e^{-10t}*12t

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inverse oflaplace (1200)/(s^2(s+10)^2)

inverse laplace

\frac{1200}{s ^{2} ( s + 10 ) ^{2}}

- \frac{12}{5} H (t) + 12 t + \frac{12}{5} e^{- 10 t} + e^{- 10 t} \cdot 12 t

Solution steps

L^{- 1} {\frac{1200}{s ^{2} ( s + 10 ) ^{2}}}

Take the partial fraction of

\frac{1200}{s ^{2} ( s + 10 ) ^{2}} : - \frac{12}{5 s} + \frac{12}{s ^{2}} + \frac{12}{5 ( s + 10 )} + \frac{12}{( s + 10 ) ^{2}}

= L^{- 1} {- \frac{12}{5 s} + \frac{12}{s ^{2}} + \frac{12}{5 ( s + 10 )} + \frac{12}{( s + 10 ) ^{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{12}{5 s}} + L^{- 1} {\frac{12}{s ^{2}}} + L^{- 1} {\frac{12}{5 ( s + 10 )}} + L^{- 1} {\frac{12}{( s + 10 ) ^{2}}}

L^{- 1} {\frac{12}{5 s}} : \frac{12}{5} H (t)

L^{- 1} {\frac{12}{s ^{2}}} : 12 t

L^{- 1} {\frac{12}{5 ( s + 10 )}} : \frac{12}{5} e^{- 10 t}

L^{- 1} {\frac{12}{( s + 10 ) ^{2}}} : e^{- 10 t} \cdot 12 t

= - \frac{12}{5} H (t) + 12 t + \frac{12}{5} e^{- 10 t} + e^{- 10 t} \cdot 12 t

\frac{2 x ^{3} - 6}{x ^{4}}

\frac{d}{d x} (\frac{x cos ( x ) - sin ( x )}{x ^{2}})

\frac{\partial}{\partial x} (30 x^{5} y^{4} - 20 x^{4} y^{3})

x \to - 2 lim (\frac{1}{x ^{2}})

x^{2} + y^{2} = 169, (12, - 5)

What is inverse oflaplace (1200)/(s^2(s+10)^2) ?
The answer to inverse oflaplace (1200)/(s^2(s+10)^2) is -12/5 \H(t)+12t+12/5 e^{-10t}+e^{-10t}*12t