What is inverse oflaplace (3s+3)/((s^2-9)) ?

The answer to inverse oflaplace (3s+3)/((s^2-9)) is e^{-3t}+2e^{3t}

Popular >

inverse oflaplace (3s+3)/((s^2-9))

inverse laplace

\frac{3 s + 3}{( s ^{2} - 9 )}

e^{- 3 t} + 2 e^{3 t}

Solution steps

L^{- 1} {\frac{3 s + 3}{( s ^{2} - 9 )}}

Take the partial fraction of

\frac{3 s + 3}{s ^{2} - 9} : \frac{1}{s + 3} + \frac{2}{s - 3}

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

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

L^{- 1} {\frac{2}{s - 3}} : 2 e^{3 t}

= e^{- 3 t} + 2 e^{3 t}

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

\int \frac{1}{x ^{8}} d x

\frac{\partial}{\partial s} (r - s)

n = 1 \sum \infty \frac{9 ^{n}}{8 ^{n}}

\int_{- 2}^{3} \frac{34}{x ^{4}} d x

What is inverse oflaplace (3s+3)/((s^2-9)) ?
The answer to inverse oflaplace (3s+3)/((s^2-9)) is e^{-3t}+2e^{3t}