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

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

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

inverse laplace

\frac{( s + 2 )}{( s + 1 ) ^{2}}

e^{- t} + e^{- t} t

Solution steps

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

Take the partial fraction of

\frac{s + 2}{( s + 1 ) ^{2}} : \frac{1}{s + 1} + \frac{1}{( s + 1 ) ^{2}}

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

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

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

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

\frac{\partial}{\partial z} (x yz + x y^{5} + y z^{5} + z x^{5})

f (x) = 2 e^{3 x}

- 4 e^{4 t}

\frac{\partial}{\partial y} (cos (x))

7 y^{^{''}} + 26 y = 0

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