What is inverse oflaplace (4s-36)/(s^2-8s+15) ?

The answer to inverse oflaplace (4s-36)/(s^2-8s+15) is 12e^{3t}-8e^{5t}

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inverse oflaplace (4s-36)/(s^2-8s+15)

inverse laplace

\frac{4 s - 36}{s ^{2} - 8 s + 15}

12 e^{3 t} - 8 e^{5 t}

Solution steps

L^{- 1} {\frac{4 s - 36}{s ^{2} - 8 s + 15}}

Take the partial fraction of

\frac{4 s - 36}{s ^{2} - 8 s + 15} : \frac{12}{s - 3} - \frac{8}{s - 5}

= L^{- 1} {\frac{12}{s - 3} - \frac{8}{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{12}{s - 3}} - L^{- 1} {\frac{8}{s - 5}}

L^{- 1} {\frac{12}{s - 3}} : 12 e^{3 t}

L^{- 1} {\frac{8}{s - 5}} : 8 e^{5 t}

= 12 e^{3 t} - 8 e^{5 t}

(0.5) (1)

\frac{d}{d u} (ln (u))

(y^{2} - 2 x y) d x + x^{2} d y = 0

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

\frac{1}{s ^{2} ( s + 3 )}

What is inverse oflaplace (4s-36)/(s^2-8s+15) ?
The answer to inverse oflaplace (4s-36)/(s^2-8s+15) is 12e^{3t}-8e^{5t}