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

The answer to inverse oflaplace (s^2)/((s-3)^2) is \delta(t)+6e^{3t}+e^{3t}*9t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace (s^2)/((s-3)^2)

inverse laplace

\frac{s ^{2}}{( s - 3 ) ^{2}}

δ (t) + 6 e^{3 t} + e^{3 t} \cdot 9 t

Solution steps

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

Take the partial fraction of

\frac{s ^{2}}{( s - 3 ) ^{2}} : 1 + \frac{6}{s - 3} + \frac{9}{( s - 3 ) ^{2}}

= L^{- 1} {1 + \frac{6}{s - 3} + \frac{9}{( s - 3 ) ^{2}}}

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} {1} + L^{- 1} {\frac{6}{s - 3}} + L^{- 1} {\frac{9}{( s - 3 ) ^{2}}}

L^{- 1} {1} : δ (t)

L^{- 1} {\frac{6}{s - 3}} : 6 e^{3 t}

L^{- 1} {\frac{9}{( s - 3 ) ^{2}}} : e^{3 t} \cdot 9 t

= δ (t) + 6 e^{3 t} + e^{3 t} \cdot 9 t

2 + 5 x^{2} - 2 x^{3}

f (x) = x^{2} - 7 x + 7, at x = 6

f (x) = (6 - 3 x) (6 + 3 x), at x = 3

\int_{- 0.5}^{0.5} 1.5 x^{2} d x

y = (3 x - 2)^{2} (5 x + 4)

What is inverse oflaplace (s^2)/((s-3)^2) ?
The answer to inverse oflaplace (s^2)/((s-3)^2) is \delta(t)+6e^{3t}+e^{3t}*9t