What is inverse oflaplace (3s+7)/(s^2+4s+3) ?

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

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inverse oflaplace (3s+7)/(s^2+4s+3)

inverse laplace

\frac{3 s + 7}{s ^{2} + 4 s + 3}

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

\frac{d}{d x} (f (x) x)

\frac{d}{d x} (\frac{2 x + 8 x ^{2}}{( 1 + 8 x ) ^{2}})

n = 1 \sum \infty \frac{5 n}{4 n + 3}

h \to 3 lim (\frac{( - 3 h + 5 - 5 )}{h})

x \to 3 + lim (ln (x - 3))

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