What is inverse oflaplace (10s)/(s^2-25) ?

The answer to inverse oflaplace (10s)/(s^2-25) is 5e^{-5t}+5e^{5t}

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inverse oflaplace (10s)/(s^2-25)

inverse laplace

\frac{10 s}{s ^{2} - 25}

5 e^{- 5 t} + 5 e^{5 t}

Solution steps

L^{- 1} {\frac{10 s}{s ^{2} - 25}}

Take the partial fraction of

\frac{10 s}{s ^{2} - 25} : \frac{5}{s + 5} + \frac{5}{s - 5}

= L^{- 1} {\frac{5}{s + 5} + \frac{5}{s - 5}}

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{5}{s + 5}} + L^{- 1} {\frac{5}{s - 5}}

L^{- 1} {\frac{5}{s + 5}} : 5 e^{- 5 t}

L^{- 1} {\frac{5}{s - 5}} : 5 e^{5 t}

= 5 e^{- 5 t} + 5 e^{5 t}

x \to - 5 lim (- 1 x)

x \to 0 lim (\frac{1}{x ^{2} + 3})

\int x e^{2} x d x

\int \frac{x ^{3} + 5}{x ^{2}} d x

\frac{x + 1}{x + 5}

What is inverse oflaplace (10s)/(s^2-25) ?
The answer to inverse oflaplace (10s)/(s^2-25) is 5e^{-5t}+5e^{5t}