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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse laplace

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

2 e^{- 3 t} - e^{- 3 t} \cdot 6 t

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{6}{( s + 3 ) ^{2}}} : e^{- 3 t} \cdot 6 t

= 2 e^{- 3 t} - e^{- 3 t} \cdot 6 t

\int e^{- s t} sin (b t) d t

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

x^{3} y y^{^{'}} + x^{2} y^{2} = y^{5}

\frac{\partial}{\partial y} (x^{2} e^{- \frac{x}{y}})

\int 13 e^{u} d u

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