What is inverse oflaplace (3s+7)/(s^2-2s-3) ?

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

Popular >

inverse oflaplace (3s+7)/(s^2-2s-3)

inverse laplace

\frac{3 s + 7}{s ^{2} - 2 s - 3}

- e^{- t} + 4 e^{3 t}

Solution steps

L^{- 1} {\frac{3 s + 7}{s ^{2} - 2 s - 3}}

Take the partial fraction of

\frac{3 s + 7}{s ^{2} - 2 s - 3} : - \frac{1}{s + 1} + \frac{4}{s - 3}

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

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

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

= - e^{- t} + 4 e^{3 t}

\frac{d}{d x} (ln (x e^{x} + 7))

\int_{0}^{2} \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x

\frac{d}{d y} (\frac{y ^{3}}{21} + \frac{7}{4 y})

f (x) = 8 x + ln (x^{2})

\int (5 x^{3} + 6)^{4} x^{2} d x

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