What is inverse oflaplace ((10))/(5*s^2+7s) ?

The answer to inverse oflaplace ((10))/(5*s^2+7s) is 10/7 \H(t)-10/7 e^{-(7t)/5}

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inverse oflaplace ((10))/(5*s^2+7s)

inverse laplace

\frac{( 10 )}{5 \cdot s ^{2} + 7 s}

\frac{10}{7} H (t) - \frac{10}{7} e^{- \frac{7 t}{5}}

Solution steps

L^{- 1} {\frac{( 10 )}{5 s ^{2} + 7 s}}

Take the partial fraction of

\frac{10}{5 s ^{2} + 7 s} : \frac{10}{7 s} - \frac{50}{7 ( 5 s + 7 )}

= L^{- 1} {\frac{10}{7 s} - \frac{50}{7 ( 5 s + 7 )}}

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{10}{7 s}} - L^{- 1} {\frac{50}{7 ( 5 s + 7 )}}

L^{- 1} {\frac{10}{7 s}} : \frac{10}{7} H (t)

L^{- 1} {\frac{50}{7 ( 5 s + 7 )}} : \frac{10}{7} e^{- \frac{7 t}{5}}

= \frac{10}{7} H (t) - \frac{10}{7} e^{- \frac{7 t}{5}}

\int \frac{4 y ^{2} - 7 y - 12}{y ( y + 2 ) ( y - 3 )} d y

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\int sec^{2} (x) + 1 d x

What is inverse oflaplace ((10))/(5*s^2+7s) ?
The answer to inverse oflaplace ((10))/(5*s^2+7s) is 10/7 \H(t)-10/7 e^{-(7t)/5}