What is inverse oflaplace (1/(s+1)+3/2)/(s+2) ?

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

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inverse oflaplace (1/(s+1)+3/2)/(s+2)

inverse laplace

\frac{\frac{1}{s + 1} + \frac{3}{2}}{s + 2}

e^{- t} + \frac{1}{2} e^{- 2 t}

Solution steps

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

Expand

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

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

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

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

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

Refine

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

= e^{- t} + \frac{1}{2} e^{- 2 t}

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e - 4 x, y (0) = - 6, y^{^{'}} (0) = 1

(x^{2} + 3) \cdot \frac{d y}{d x} = x y

n = 1 \sum \infty \frac{5}{n ^{5}}

\frac{d}{d x} (\frac{1}{x ^{4}} + \frac{1}{x ^{8}})

\frac{d}{d x} (\frac{x ^{3}}{3} - \frac{3}{x ^{3}})

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