What is inverse oflaplace (s+4)/(s^2+2s-3) ?

The answer to inverse oflaplace (s+4)/(s^2+2s-3) is 5/4 e^t-1/4 e^{-3t}

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inverse oflaplace (s+4)/(s^2+2s-3)

inverse laplace

\frac{s + 4}{s ^{2} + 2 s - 3}

\frac{5}{4} e^{t} - \frac{1}{4} e^{- 3 t}

Solution steps

L^{- 1} {\frac{s + 4}{s ^{2} + 2 s - 3}}

Take the partial fraction of

\frac{s + 4}{s ^{2} + 2 s - 3} : \frac{5}{4 ( s - 1 )} - \frac{1}{4 ( s + 3 )}

= L^{- 1} {\frac{5}{4 ( s - 1 )} - \frac{1}{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{5}{4 ( s - 1 )}} - L^{- 1} {\frac{1}{4 ( s + 3 )}}

L^{- 1} {\frac{5}{4 ( s - 1 )}} : \frac{5}{4} e^{t}

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

= \frac{5}{4} e^{t} - \frac{1}{4} e^{- 3 t}

5 x^{2} + 2 x + 6

\frac{\partial}{\partial x} (\frac{x ^{2} + y ^{2}}{2})

y^{^{''}} - \frac{2}{x} y^{^{'}} - \frac{10}{x ^{2}} y = 0

cos (4 x)

3 x + 4, at x = 4

What is inverse oflaplace (s+4)/(s^2+2s-3) ?
The answer to inverse oflaplace (s+4)/(s^2+2s-3) is 5/4 e^t-1/4 e^{-3t}