What is inverse oflaplace 4/((s+1)^2) ?

The answer to inverse oflaplace 4/((s+1)^2) is e^{-t}*4t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse oflaplace 4/((s+1)^2)

inverse laplace

\frac{4}{( s + 1 ) ^{2}}

e^{- t} \cdot 4 t

Solution steps

L^{- 1} {\frac{4}{( s + 1 ) ^{2}}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {F (s - a)} = e^{a t} f (t)

For

\frac{4}{( s + 1 ) ^{2}} : a = - 1, F (s) = \frac{4}{s ^{2}}

= e^{- t} L^{- 1} {\frac{4}{s ^{2}}}

L^{- 1} {\frac{4}{s ^{2}}} : 4 t

= e^{- t} \cdot 4 t

t \cdot y^{^{'}} + t y = 1 - y

t^{2} \cdot e^{a \cdot t}

t^{2} y^{^{''}} + 19 t y^{^{'}} + 81 y = 0

\int 1 + cot^{2} (x) d x

\int_{0}^{5} ∣ 15 - 5 t ∣ d t

What is inverse oflaplace 4/((s+1)^2) ?
The answer to inverse oflaplace 4/((s+1)^2) is e^{-t}*4t