What is inverse oflaplace (3*s)/(5s^2+125) ?

The answer to inverse oflaplace (3*s)/(5s^2+125) is 3/5 cos(5t)

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inverse oflaplace (3*s)/(5s^2+125)

inverse laplace

\frac{3 \cdot s}{5 s ^{2} + 125}

\frac{3}{5} cos (5 t)

Solution steps

L^{- 1} {\frac{3 s}{5 s ^{2} + 125}}

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{3 s}{5 s ^{2} + 125} : F (s) = \frac{3}{5 s ^{2} + 125}

= \frac{d}{d t} (L^{- 1} {\frac{3}{5 s ^{2} + 125}}) + L^{- 1} {\frac{3}{5 s ^{2} + 125}} (0)

L^{- 1} {\frac{3}{5 s ^{2} + 125}} : \frac{3}{25} sin (5 t)

\frac{d}{d t} (\frac{3}{25} sin (5 t)) = \frac{3}{5} cos (5 t)

Evaluate

\frac{3}{25} sin (5 t)

at point

0 = 0

= \frac{3}{5} cos (5 t)

y^{^{'}} + 2 x y = 10 x

\int cos^{13} (x) sin (x) d x

\frac{d}{d x} ((\frac{x + 8}{x - 8})^{5})

1 + (\frac{a}{x ^{2}})

x \to 0 lim (\frac{f ( x )}{x ^{2}})

What is inverse oflaplace (3*s)/(5s^2+125) ?
The answer to inverse oflaplace (3*s)/(5s^2+125) is 3/5 cos(5t)