What is inverse oflaplace (4se^{-3s})/((s^2+4)) ?

The answer to inverse oflaplace (4se^{-3s})/((s^2+4)) is \H(t-3)*4cos(2(t-3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse oflaplace (4se^{-3s})/((s^2+4))

inverse laplace

\frac{4 s e ^{- 3 s}}{( s ^{2} + 4 )}

H (t - 3) \cdot 4 cos (2 (t - 3))

Solution steps

L^{- 1} {\frac{4 e ^{- 3 s} s}{( s ^{2} + 4 )}}

Apply inverse transform rule:

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

then

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

Where

H (t)

is Heaviside step function

= H (t - 3) L^{- 1} {\frac{4 s}{s ^{2} + 4}} (t - 3)

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

= H (t - 3) \cdot 4 cos (2 (t - 3))

im pl i c i t \frac{d y}{d x}, y = 2 x - 5

n = 1 \sum \infty arctan (4 n)

in v erse l a pl a ce \frac{2 s ^{2} + 1}{s + 3}

\frac{x d y}{d x} = y + x e^{- \frac{y}{x}}

\frac{d}{d x} ((ln (x) + 3 x)^{2})

What is inverse oflaplace (4se^{-3s})/((s^2+4)) ?
The answer to inverse oflaplace (4se^{-3s})/((s^2+4)) is \H(t-3)*4cos(2(t-3))