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

The answer to inverse oflaplace (1+s*2)/((1+s*3)^2) is 2/9 e^{-t/3}+(e^{-t/3}t)/(27)

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inverse oflaplace (1+s2)/((1+s3)^2)

inverse laplace

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

\frac{2}{9} e^{- \frac{t}{3}} + \frac{e ^{- \frac{t}{3}} t}{27}

Solution steps

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

Take the partial fraction of

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

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

L^{- 1} {\frac{2}{3 ( 3 s + 1 )}} : \frac{2}{9} e^{- \frac{t}{3}}

L^{- 1} {\frac{1}{3 ( 3 s + 1 ) ^{2}}} : \frac{e ^{- \frac{t}{3}} t}{27}

= \frac{2}{9} e^{- \frac{t}{3}} + \frac{e ^{- \frac{t}{3}} t}{27}

f (x) = x^{2} sin (\frac{1}{x})

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

\int \frac{3 x ^{2}}{( x ^{3} + 1 ) ^{5}} d x

x \cdot e^{- A x}

\frac{d}{d x} (\frac{x}{( 2 x - 1 ) ^{6}})

What is inverse oflaplace (1+s*2)/((1+s*3)^2) ?
The answer to inverse oflaplace (1+s*2)/((1+s*3)^2) is 2/9 e^{-t/3}+(e^{-t/3}t)/(27)