What is inverse oflaplace (3x)/((x-1)^4) ?

The answer to inverse oflaplace (3x)/((x-1)^4) is (3e^tt^2)/2+(e^tt^3)/2

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inverse oflaplace (3x)/((x-1)^4)

inverse laplace

\frac{3 x}{( x - 1 ) ^{4}}

\frac{3 e ^{t} t ^{2}}{2} + \frac{e ^{t} t ^{3}}{2}

Solution steps

L^{- 1} {\frac{3 x}{( x - 1 ) ^{4}}}

Take the partial fraction of

\frac{3 x}{( x - 1 ) ^{4}} : \frac{3}{( x - 1 ) ^{3}} + \frac{3}{( x - 1 ) ^{4}}

= L^{- 1} {\frac{3}{( x - 1 ) ^{3}} + \frac{3}{( x - 1 ) ^{4}}}

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{3}{( x - 1 ) ^{3}}} + L^{- 1} {\frac{3}{( x - 1 ) ^{4}}}

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

L^{- 1} {\frac{3}{( x - 1 ) ^{4}}} : \frac{e ^{t} t ^{3}}{2}

= \frac{3 e ^{t} t ^{2}}{2} + \frac{e ^{t} t ^{3}}{2}

\frac{d}{d x} (arctan (x^{2} + y^{2}))

\frac{d}{d y} (y - 1)

\int (\frac{1}{cos ( x )}) d x

6 - 5 x^{2}

n = 1 \sum \infty \frac{5}{n ^{2} + 9}

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