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

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

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

inverse laplace

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

2 H (t) - 2 e^{- t} - e^{- t} t

Solution steps

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

Take the partial fraction of

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

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

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

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

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

= 2 H (t) - 2 e^{- t} - e^{- t} t

g (t) = 2 t^{- \frac{3}{4}}

\frac{\partial}{\partial x} (e^{x} + yz)

3 x y^{^{'}} - y (x + 1) + 7 y^{4} = 0

\frac{d y}{d x}, x^{2} + y^{2} = 24

cos^{2} (t)

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