What is inverse oflaplace (s+5)/((s+10)^2) ?

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

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inverse oflaplace (s+5)/((s+10)^2)

inverse laplace

\frac{s + 5}{( s + 10 ) ^{2}}

e^{- 10 t} - e^{- 10 t} \cdot 5 t

Solution steps

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

Take the partial fraction of

\frac{s + 5}{( s + 10 ) ^{2}} : \frac{1}{s + 10} - \frac{5}{( s + 10 ) ^{2}}

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

L^{- 1} {\frac{1}{s + 10}} : e^{- 10 t}

L^{- 1} {\frac{5}{( s + 10 ) ^{2}}} : e^{- 10 t} \cdot 5 t

= e^{- 10 t} - e^{- 10 t} \cdot 5 t

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

\frac{d}{d x} (\frac{( x ^{2} + 1 )}{2 x ^{2} - 3 x})

f (x) = (9 x^{6} + 8 x^{3})^{4}

\int 5 x^{2} x + 9 d x

f (x) = 9 x + 4 x^{- 1}

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