What is inverse oflaplace 1/x*(25)/(x^2+25) ?

The answer to inverse oflaplace 1/x*(25)/(x^2+25) is \H(t)-cos(5t)

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inverse oflaplace 1/x*(25)/(x^2+25)

inverse laplace

\frac{1}{x} \cdot \frac{25}{x ^{2} + 25}

H (t) - cos (5 t)

Solution steps

L^{- 1} {\frac{1}{x} \cdot \frac{25}{x ^{2} + 25}}

Expand

\frac{1}{x} \frac{25}{x ^{2} + 25} : \frac{25}{x ^{3} + 25 x}

= L^{- 1} {\frac{25}{x ^{3} + 25 x}}

Take the partial fraction of

\frac{25}{x ^{3} + 25 x} : \frac{1}{x} - \frac{x}{x ^{2} + 25}

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

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

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

= H (t) - cos (5 t)

What is inverse oflaplace 1/x*(25)/(x^2+25) ?
The answer to inverse oflaplace 1/x*(25)/(x^2+25) is \H(t)-cos(5t)