What is inverse oflaplace (2250)/(5x^2+x) ?

The answer to inverse oflaplace (2250)/(5x^2+x) is 2250\H(t)-2250e^{-t/5}

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inverse oflaplace (2250)/(5x^2+x)

inverse laplace

\frac{2250}{5 x ^{2} + x}

2250 H (t) - 2250 e^{- \frac{t}{5}}

Solution steps

L^{- 1} {\frac{2250}{5 x ^{2} + x}}

Take the partial fraction of

\frac{2250}{5 x ^{2} + x} : \frac{2250}{x} - \frac{11250}{5 x + 1}

= L^{- 1} {\frac{2250}{x} - \frac{11250}{5 x + 1}}

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{2250}{x}} - L^{- 1} {\frac{11250}{5 x + 1}}

L^{- 1} {\frac{2250}{x}} : 2250 H (t)

L^{- 1} {\frac{11250}{5 x + 1}} : 2250 e^{- \frac{t}{5}}

= 2250 H (t) - 2250 e^{- \frac{t}{5}}

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What is inverse oflaplace (2250)/(5x^2+x) ?
The answer to inverse oflaplace (2250)/(5x^2+x) is 2250\H(t)-2250e^{-t/5}