What is inverse oflaplace ((s-10))/((s^2-2s-3)) ?

The answer to inverse oflaplace ((s-10))/((s^2-2s-3)) is 11/4 e^{-t}-7/4 e^{3t}

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inverse oflaplace ((s-10))/((s^2-2s-3))

inverse laplace

\frac{( s - 10 )}{( s ^{2} - 2 s - 3 )}

\frac{11}{4} e^{- t} - \frac{7}{4} e^{3 t}

Solution steps

L^{- 1} {\frac{( s - 10 )}{( s ^{2} - 2 s - 3 )}}

Take the partial fraction of

\frac{s - 10}{s ^{2} - 2 s - 3} : \frac{11}{4 ( s + 1 )} - \frac{7}{4 ( s - 3 )}

= L^{- 1} {\frac{11}{4 ( s + 1 )} - \frac{7}{4 ( s - 3 )}}

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{11}{4 ( s + 1 )}} - L^{- 1} {\frac{7}{4 ( s - 3 )}}

L^{- 1} {\frac{11}{4 ( s + 1 )}} : \frac{11}{4} e^{- t}

L^{- 1} {\frac{7}{4 ( s - 3 )}} : \frac{7}{4} e^{3 t}

= \frac{11}{4} e^{- t} - \frac{7}{4} e^{3 t}

\frac{d y}{d x}, x^{6} - 5 x y^{3} = 9 x y

x \to 0 lim (arctan (\frac{1}{x}))

\int \frac{x ^{2} + 2}{x ^{3} ( x + 1 ) ^{3}} d x

\frac{1}{( 1 - 7 x )}, - 6 < x < 0

\frac{\partial}{\partial x} (4 x^{- x^{2} - y^{2} - z^{2}})

What is inverse oflaplace ((s-10))/((s^2-2s-3)) ?
The answer to inverse oflaplace ((s-10))/((s^2-2s-3)) is 11/4 e^{-t}-7/4 e^{3t}