What is inverse oflaplace (2s-7)/((s-4)(s+6)) ?

The answer to inverse oflaplace (2s-7)/((s-4)(s+6)) is 1/10 e^{4t}+19/10 e^{-6t}

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inverse oflaplace (2s-7)/((s-4)(s+6))

inverse laplace

\frac{2 s - 7}{( s - 4 ) ( s + 6 )}

\frac{1}{10} e^{4 t} + \frac{19}{10} e^{- 6 t}

Solution steps

L^{- 1} {\frac{2 s - 7}{( s - 4 ) ( s + 6 )}}

Take the partial fraction of

\frac{2 s - 7}{( s - 4 ) ( s + 6 )} : \frac{1}{10 ( s - 4 )} + \frac{19}{10 ( s + 6 )}

= L^{- 1} {\frac{1}{10 ( s - 4 )} + \frac{19}{10 ( s + 6 )}}

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{1}{10 ( s - 4 )}} + L^{- 1} {\frac{19}{10 ( s + 6 )}}

L^{- 1} {\frac{1}{10 ( s - 4 )}} : \frac{1}{10} e^{4 t}

L^{- 1} {\frac{19}{10 ( s + 6 )}} : \frac{19}{10} e^{- 6 t}

= \frac{1}{10} e^{4 t} + \frac{19}{10} e^{- 6 t}

\int \frac{x + 4}{x ^{2} + 5 x - 6} d x

\int sin (t) sec^{2} (cos (t)) d t

\int_{3}^{0} (5 x - x^{3}) d x

y^{^{'}} = \frac{3 y ^{2} + 3}{2 x y}

\frac{d y}{d x} = 6 x y

What is inverse oflaplace (2s-7)/((s-4)(s+6)) ?
The answer to inverse oflaplace (2s-7)/((s-4)(s+6)) is 1/10 e^{4t}+19/10 e^{-6t}