What is inverse oflaplace (4(-s+16))/(s^2+10s) ?

The answer to inverse oflaplace (4(-s+16))/(s^2+10s) is 32/5 \H(t)-52/5 e^{-10t}

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inverse oflaplace (4(-s+16))/(s^2+10s)

inverse laplace

\frac{4 ( - s + 16 )}{s ^{2} + 10 s}

\frac{32}{5} H (t) - \frac{52}{5} e^{- 10 t}

Solution steps

L^{- 1} {\frac{4 ( - s + 16 )}{s ^{2} + 10 s}}

Take the partial fraction of

\frac{4 ( - s + 16 )}{s ^{2} + 10 s} : \frac{32}{5 s} - \frac{52}{5 ( s + 10 )}

= L^{- 1} {\frac{32}{5 s} - \frac{52}{5 ( s + 10 )}}

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{32}{5 s}} - L^{- 1} {\frac{52}{5 ( s + 10 )}}

L^{- 1} {\frac{32}{5 s}} : \frac{32}{5} H (t)

L^{- 1} {\frac{52}{5 ( s + 10 )}} : \frac{52}{5} e^{- 10 t}

= \frac{32}{5} H (t) - \frac{52}{5} e^{- 10 t}

(- 4.5) (2.3)

\frac{d}{d x} ((x^{2} ln (x))^{4})

y = \frac{1}{3} cos (\frac{6 π x - 1}{2})

\int - e^{x} - \frac{7}{cos ^{2} ( x )} d x

\frac{d}{d x} (sin (x^{2} cos (x)))

What is inverse oflaplace (4(-s+16))/(s^2+10s) ?
The answer to inverse oflaplace (4(-s+16))/(s^2+10s) is 32/5 \H(t)-52/5 e^{-10t}