What is inverse oflaplace s/((s-3)^3) ?

The answer to inverse oflaplace s/((s-3)^3) is e^{3t}t+(3e^{3t}t^2)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace s/((s-3)^3)

inverse laplace

\frac{s}{( s - 3 ) ^{3}}

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

Solution steps

L^{- 1} {\frac{s}{( s - 3 ) ^{3}}}

Take the partial fraction of

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

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

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

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

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

\frac{( x ^{3} )}{3}

f (x) = 9 x^{3} - 7 x

y^{^{''}} = (y^{^{'}})^{2} + 1

\int + 2 t d t

y = x^{2} - 1, y = - x + 2

What is inverse oflaplace s/((s-3)^3) ?
The answer to inverse oflaplace s/((s-3)^3) is e^{3t}t+(3e^{3t}t^2)/2