What is inverse oflaplace (s+4)/((s+1)^2) ?

The answer to inverse oflaplace (s+4)/((s+1)^2) is e^{-t}+e^{-t}*3t

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inverse oflaplace (s+4)/((s+1)^2)

inverse laplace

\frac{s + 4}{( s + 1 ) ^{2}}

e^{- t} + e^{- t} \cdot 3 t

Solution steps

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

Take the partial fraction of

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

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

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

L^{- 1} {\frac{3}{( s + 1 ) ^{2}}} : e^{- t} \cdot 3 t

= e^{- t} + e^{- t} \cdot 3 t

\int \frac{1}{x} - \frac{2}{x ^{2} + 1} d x

\frac{( - 2 )}{( 6 s + s ^{2} )}

y d x = 2 (x + y) d y

(x + y) d x + (x + 2 y) d y = 0, y (2) = 3

x \to 0 - lim (\frac{1}{1 + e ^{\frac{1}{x}}})

What is inverse oflaplace (s+4)/((s+1)^2) ?
The answer to inverse oflaplace (s+4)/((s+1)^2) is e^{-t}+e^{-t}*3t