What is inverse oflaplace s/(s^2+\frac{1){lc}} ?

The answer to inverse oflaplace s/(s^2+\frac{1){lc}} is sqrt(lc)cos(sqrt(1/(lc))t)sqrt(1/(lc))

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inverse oflaplace s/(s^2+\frac{1){lc}}

inverse laplace

\frac{s}{s ^{2} + \frac{1}{l c}}

l c cos (\frac{1}{l c} t) \frac{1}{l c}

Solution steps

L^{- 1} {\frac{s}{s ^{2} + \frac{1}{l c}}}

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{s}{s ^{2} + \frac{1}{l c}} : F (s) = \frac{1}{s ^{2} + \frac{1}{l c}}

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

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

\frac{d}{d t} (l c sin (\frac{1}{l c} t)) = l c cos (\frac{1}{l c} t) \frac{1}{l c}

Evaluate

l c sin (\frac{1}{l c} t)

at point

0 = 0

= l c cos (\frac{1}{l c} t) \frac{1}{l c}

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

\int_{0}^{1} x \cdot sin (π x) d x

y = sec (3 x^{4} + 5)

θ \to 0 lim (\frac{cos ( θ ) - 1}{θ})

\int_{0}^{2} π (\frac{1}{x + 4})^{2} d x

What is inverse oflaplace s/(s^2+\frac{1){lc}} ?
The answer to inverse oflaplace s/(s^2+\frac{1){lc}} is sqrt(lc)cos(sqrt(1/(lc))t)sqrt(1/(lc))