What is inverse oflaplace (4s+1)/(s^3-3s^2+3s-1) ?

The answer to inverse oflaplace (4s+1)/(s^3-3s^2+3s-1) is e^t*4t+(5e^tt^2)/2

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inverse oflaplace (4s+1)/(s^3-3s^2+3s-1)

inverse laplace

\frac{4 s + 1}{s ^{3} - 3 s ^{2} + 3 s - 1}

e^{t} \cdot 4 t + \frac{5 e ^{t} t ^{2}}{2}

Solution steps

L^{- 1} {\frac{4 s + 1}{s ^{3} - 3 s ^{2} + 3 s - 1}}

Take the partial fraction of

\frac{4 s + 1}{s ^{3} - 3 s ^{2} + 3 s - 1} : \frac{4}{( s - 1 ) ^{2}} + \frac{5}{( s - 1 ) ^{3}}

= L^{- 1} {\frac{4}{( s - 1 ) ^{2}} + \frac{5}{( s - 1 ) ^{3}}}

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{4}{( s - 1 ) ^{2}}} + L^{- 1} {\frac{5}{( s - 1 ) ^{3}}}

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

L^{- 1} {\frac{5}{( s - 1 ) ^{3}}} : \frac{5 e ^{t} t ^{2}}{2}

= e^{t} \cdot 4 t + \frac{5 e ^{t} t ^{2}}{2}

\frac{( 10 s - 3 )}{s ^{2} - 4 s - 29}

\frac{d}{d x} (\frac{ln ( x )}{x})

\frac{\partial}{\partial y} (9 x^{y} + 2 y^{x} + 11)

\int \frac{3 t - 2}{9 - t ^{2}} d t

\int arcsec (x) d x

What is inverse oflaplace (4s+1)/(s^3-3s^2+3s-1) ?
The answer to inverse oflaplace (4s+1)/(s^3-3s^2+3s-1) is e^t*4t+(5e^tt^2)/2