What is inverse oflaplace (-s)/((s-4)^2(s-5)) ?

The answer to inverse oflaplace (-s)/((s-4)^2(s-5)) is 5e^{4t}+e^{4t}*4t-5e^{5t}

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inverse oflaplace (-s)/((s-4)^2(s-5))

inverse laplace

\frac{- s}{( s - 4 ) ^{2} ( s - 5 )}

5 e^{4 t} + e^{4 t} \cdot 4 t - 5 e^{5 t}

Solution steps

L^{- 1} {\frac{- s}{( s - 4 ) ^{2} ( s - 5 )}}

Take the partial fraction of

\frac{- s}{( s - 4 ) ^{2} ( s - 5 )} : \frac{5}{s - 4} + \frac{4}{( s - 4 ) ^{2}} - \frac{5}{s - 5}

= L^{- 1} {\frac{5}{s - 4} + \frac{4}{( s - 4 ) ^{2}} - \frac{5}{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{5}{s - 4}} + L^{- 1} {\frac{4}{( s - 4 ) ^{2}}} - L^{- 1} {\frac{5}{s - 5}}

L^{- 1} {\frac{5}{s - 4}} : 5 e^{4 t}

L^{- 1} {\frac{4}{( s - 4 ) ^{2}}} : e^{4 t} \cdot 4 t

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

= 5 e^{4 t} + e^{4 t} \cdot 4 t - 5 e^{5 t}

\frac{d}{d x} (3 x^{2} - 14 x + 15)

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

\int 3 - x d x

\frac{3}{( x - 7 ) ^{4}}

\frac{d}{d x} (\frac{x ^{4}}{8} + \frac{x ^{2}}{6} - \frac{x}{5} + \frac{1}{7})

What is inverse oflaplace (-s)/((s-4)^2(s-5)) ?
The answer to inverse oflaplace (-s)/((s-4)^2(s-5)) is 5e^{4t}+e^{4t}*4t-5e^{5t}