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

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

Popular >

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

inverse laplace

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

e^{- t} - e^{- 3 t}

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\frac{d}{d x} (- x^{2} sin (x^{2} y))

x \to 3 lim (x^{3} - 3 x^{2} - 9 x)

\frac{x d y}{d x} - y = x^{2} + y^{2}

\int \frac{z ^{- 3}}{3} d z

y^{^{''}} + 16 y = 2 tan (4 x)

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