What is inverse oflaplace (s+6)/(s^3-3s^2+3s-1) ?

The answer to inverse oflaplace (s+6)/(s^3-3s^2+3s-1) is e^tt+(7e^tt^2)/2

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inverse oflaplace (s+6)/(s^3-3s^2+3s-1)

inverse laplace

\frac{s + 6}{s ^{3} - 3 s ^{2} + 3 s - 1}

e^{t} t + \frac{7 e ^{t} t ^{2}}{2}

Solution steps

L^{- 1} {\frac{s + 6}{s ^{3} - 3 s ^{2} + 3 s - 1}}

Take the partial fraction of

\frac{s + 6}{s ^{3} - 3 s ^{2} + 3 s - 1} : \frac{1}{( s - 1 ) ^{2}} + \frac{7}{( s - 1 ) ^{3}}

= L^{- 1} {\frac{1}{( s - 1 ) ^{2}} + \frac{7}{( 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{1}{( s - 1 ) ^{2}}} + L^{- 1} {\frac{7}{( s - 1 ) ^{3}}}

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

L^{- 1} {\frac{7}{( s - 1 ) ^{3}}} : \frac{7 e ^{t} t ^{2}}{2}

= e^{t} t + \frac{7 e ^{t} t ^{2}}{2}

x \to - 4 lim (- 2 x + 7)

2 y^{2}

\frac{d}{d x} (cos^{3} (sin (2 x)))

15 x^{4} - 4 x^{3}

\int 10 x^{\frac{3}{7}} + 7 x^{- \frac{6}{7}} d x

What is inverse oflaplace (s+6)/(s^3-3s^2+3s-1) ?
The answer to inverse oflaplace (s+6)/(s^3-3s^2+3s-1) is e^tt+(7e^tt^2)/2