What is inverse oflaplace (100)/(s(s^2+25)) ?

The answer to inverse oflaplace (100)/(s(s^2+25)) is 4\H(t)-4cos(5t)

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inverse oflaplace (100)/(s(s^2+25))

inverse laplace

\frac{100}{s ( s ^{2} + 25 )}

4 H (t) - 4 cos (5 t)

Solution steps

L^{- 1} {\frac{100}{s ( s ^{2} + 25 )}}

Take the partial fraction of

\frac{100}{s ( s ^{2} + 25 )} : \frac{4}{s} - \frac{4 s}{s ^{2} + 25}

= L^{- 1} {\frac{4}{s} - \frac{4 s}{s ^{2} + 25}}

Use the linearity property of Inverse Laplace Transform:

For functions

f (s), g (s)

and constants

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= L^{- 1} {\frac{4}{s}} - L^{- 1} {\frac{4 s}{s ^{2} + 25}}

L^{- 1} {\frac{4}{s}} : 4 H (t)

L^{- 1} {\frac{4 s}{s ^{2} + 25}} : 4 cos (5 t)

= 4 H (t) - 4 cos (5 t)

x \to 2 - lim (2 x - 4)

\frac{d y}{d x} + 3 x^{2} y = x^{2}

x \to 0 lim (\frac{cos ( x )}{2})

\int s d s

\frac{d}{d x} ((x + \frac{1}{x}) (5 x^{2} - \frac{1}{x ^{2}}))

What is inverse oflaplace (100)/(s(s^2+25)) ?
The answer to inverse oflaplace (100)/(s(s^2+25)) is 4\H(t)-4cos(5t)