What is inverse oflaplace (28s)/((s^2+11)) ?

The answer to inverse oflaplace (28s)/((s^2+11)) is 28cos(sqrt(11)t)

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inverse oflaplace (28s)/((s^2+11))

inverse laplace

\frac{28 s}{( s ^{2} + 11 )}

28 cos (11 t)

Solution steps

L^{- 1} {\frac{28 s}{( s ^{2} + 11 )}}

Apply inverse transform rule:

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

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

\frac{28 s}{s ^{2} + 11} : F (s) = \frac{28}{s ^{2} + 11}

= \frac{d}{d t} (L^{- 1} {\frac{28}{s ^{2} + 11}}) + L^{- 1} {\frac{28}{s ^{2} + 11}} (0)

L^{- 1} {\frac{28}{s ^{2} + 11}} : \frac{28}{11} sin (11 t)

\frac{d}{d t} (\frac{28}{11} sin (11 t)) = 28 cos (11 t)

Evaluate

\frac{28}{11} sin (11 t)

at point

0 = 0

= 28 cos (11 t)

\frac{d y}{d x} + 4 tan (4 x) (y) = 3 cos (4 x)

\int_{- 2}^{3} (x^{2} + 3 x - 13) d x

x \to 0 lim (arctan (ln (x)))

(cos (x) sin (x))^{^{'}}

\frac{\partial}{\partial x} (3 x e^{- x y})

What is inverse oflaplace (28s)/((s^2+11)) ?
The answer to inverse oflaplace (28s)/((s^2+11)) is 28cos(sqrt(11)t)